(a)

For \(k\geq 1\), \[ \begin {aligned} \E (X^k) &= \int _{-\infty }^{\infty }x^k f_X(x)\,dx\\ &= \int _0^1 x^k(2x)\,dx\\ &= 2\int _0^1 x^{k+1}\,dx\\ &= 2\left [\frac {x^{k+2}}{k+2}\right ]_0^1\\ &= \frac {2}{k+2}. \end {aligned} \]

Therefore, \[ \E (X^k)=\frac {2}{k+2}. \]

(b)

Recall that, whenever the moment generating function exists in a neighborhood of \(0\), \[ M_X^{(k)}(0)=\E (X^k). \]

From part (a), \[ \E (X^k)=\frac {2}{k+2}. \]

Therefore, \[ M_X^{(k)}(0) = \frac {2}{k+2}, \qquad k=1,2,\ldots \]

In particular, \[ M_X'(0)=\frac 23, \qquad M_X''(0)=\frac 12, \qquad M_X^{(3)}(0)=\frac 25, \qquad M_X^{(4)}(0)=\frac 13. \]

(c)

We have \[ \E (X)=M_X'(0)=\frac 23 \] and \[ \E (X^2)=M_X''(0)=\frac 12. \]

Therefore, \[ \begin {aligned} \operatorname {Var}(X) &= \E (X^2)-[\E (X)]^2\\ &= \frac 12-\left (\frac 23\right )^2\\ &= \frac 12-\frac 49\\ &= \frac 1{18}. \end {aligned} \]

Thus, \[ \sigma ^2=\frac 1{18}, \qquad \sigma =\frac {1}{3\sqrt 2}. \]

(d)

First calculate the CDF. For \(0\leq x\leq 1\), \[ \begin {aligned} F_X(x) &= P(X\leq x)\\ &= \int _0^x 2t\,dt\\ &= x^2. \end {aligned} \]

The median \(m\) satisfies \[ F_X(m)=\frac 12. \]

Hence, \[ m^2=\frac 12. \]

Since \(m\in (0,1)\), \[ m=\frac 1{\sqrt 2}. \]

(e)

The third central moment is \[ \mathbb {E}[(X-\mu )^3]. \]

Expanding, \[ (X-\mu )^3 = X^3-3\mu X^2+3\mu ^2X-\mu ^3. \]

Therefore, \[ \mathbb {E}[(X-\mu )^3] = \E (X^3) -3\mu \E (X^2) +3\mu ^2\E (X) -\mu ^3. \]

Using \[ \mu =\frac 23, \qquad \E (X^2)=\frac 12, \qquad \E (X^3)=\frac 25, \] we obtain \[ \begin {aligned} \mathbb {E}[(X-\mu )^3] &= \frac 25 - 3\left (\frac 23\right )\left (\frac 12\right ) + 3\left (\frac 23\right )^2\left (\frac 23\right ) - \left (\frac 23\right )^3\\ &= \frac 25-1+\frac 89-\frac 8{27}\\ &= -\frac 1{135}. \end {aligned} \]

Also, \[ \sigma ^3 = \left (\frac 1{18}\right )^{3/2} = \frac {1}{54\sqrt 2}. \]

Hence, \[ \begin {aligned} \gamma _1 &= \frac {\mathbb {E}[(X-\mu )^3]}{\sigma ^3}\\ &= \frac {-1/135}{1/(54\sqrt 2)}\\ &= -\frac {2\sqrt 2}{5}. \end {aligned} \]

Therefore, \[ \gamma _1=-\frac {2\sqrt 2}{5}. \]

Since the skewness is negative, the distribution is negatively skewed.

(f)

The fourth central moment is \[ \mathbb {E}[(X-\mu )^4]. \]

Expanding, \[ (X-\mu )^4 = X^4 -4\mu X^3 +6\mu ^2X^2 -4\mu ^3X +\mu ^4. \]

Therefore, \[ \mathbb {E}[(X-\mu )^4] = \E (X^4) -4\mu \E (X^3) +6\mu ^2\E (X^2) -4\mu ^3\E (X) +\mu ^4. \]

Using \[ \mu =\frac 23, \qquad \E (X^2)=\frac 12, \qquad \E (X^3)=\frac 25, \qquad \E (X^4)=\frac 13, \] we obtain \[ \begin {aligned} \mathbb {E}[(X-\mu )^4] &= \frac 13 - 4\left (\frac 23\right )\left (\frac 25\right ) + 6\left (\frac 23\right )^2\left (\frac 12\right )\\ &\qquad - 4\left (\frac 23\right )^3\left (\frac 23\right ) + \left (\frac 23\right )^4\\ &= \frac 13 -\frac {16}{15} +\frac 43 -\frac {64}{81} +\frac {16}{81}\\ &= \frac 1{135}. \end {aligned} \]

Also, \[ \sigma ^4 = \left (\frac 1{18}\right )^2 = \frac 1{324}. \]

Hence, \[ \begin {aligned} \gamma _2 &= \frac {\mathbb {E}[(X-\mu )^4]}{\sigma ^4}\\ &= \frac {1/135}{1/324}\\ &= \frac {12}{5}. \end {aligned} \]

Therefore, \[ \gamma _2=\frac {12}{5}. \]

(a)

Since \(X\) and \(Y\) are independent, \[ \begin {aligned} M_{X+Y}(t) &= \E \left (e^{t(X+Y)}\right )\\ &= \E \left (e^{tX}e^{tY}\right )\\ &= \E \left (e^{tX}\right ) \E \left (e^{tY}\right )\\ &= M_X(t)M_Y(t). \end {aligned} \]

Therefore, \[ M_{X+Y}(t) = e^{2(e^t-1)} \frac {3}{3-t}, \qquad t<3. \]

(b)

Let \[ Z=aX+bY. \]

By definition, \[ \begin {aligned} M_Z(t) &= \E \left (e^{tZ}\right )\\ &= \E \left (e^{t(aX+bY)}\right )\\ &= \E \left (e^{atX}e^{btY}\right ). \end {aligned} \]

Since \(X\) and \(Y\) are independent, \[ \begin {aligned} M_Z(t) &= \E \left (e^{atX}\right ) \E \left (e^{btY}\right )\\ &= M_X(at)M_Y(bt). \end {aligned} \]

Now, \[ M_X(at) = e^{2(e^{at}-1)} \] and \[ M_Y(bt) = \frac {3}{3-bt}. \]

Hence, \[ M_{aX+bY}(t) = e^{2(e^{at}-1)} \frac {3}{3-bt}. \]

The restriction comes from \(M_Y(bt)\), which requires \[ bt<3. \]

Thus, \[ M_{aX+bY}(t) = \frac {3e^{2(e^{at}-1)}}{3-bt}, \qquad bt<3. \]

Equivalently, the domain is \[ \begin {cases} t<3/b, & b>0,\\ t>3/b, & b<0,\\ t\in \mathbb {R}, & b=0. \end {cases} \]

(a)

Since the total probability must equal \(1\), \[ \sum _{x=1}^{3}\sum _{y=1}^{2}c(x+y)=1. \]

Thus, \[ c[(2+3)+(3+4)+(4+5)]=1. \]

Therefore, \[ 21c=1, \] and hence \[ c=\frac 1{21}. \]

(b)

For \(x=1,2,3\), \[ \begin {aligned} p_X(x) &= \sum _{y=1}^{2}p(x,y)\\ &= \frac 1{21}\left [(x+1)+(x+2)\right ]\\ &= \frac {2x+3}{21}. \end {aligned} \]

Thus, \[ p_X(x) = \begin {cases} \dfrac {2x+3}{21}, & x=1,2,3,\\[1mm] 0, & \text {otherwise}. \end {cases} \]

For \(y=1,2\), \[ \begin {aligned} p_Y(y) &= \sum _{x=1}^{3}p(x,y)\\ &= \frac 1{21}\left [(1+y)+(2+y)+(3+y)\right ]\\ &= \frac {6+3y}{21}. \end {aligned} \]

Therefore, \[ p_Y(y) = \begin {cases} \dfrac 37, & y=1,\\[1mm] \dfrac 47, & y=2,\\[1mm] 0, & \text {otherwise}. \end {cases} \]

(c)

\[ P(X\geq 2\mid Y=1) = \frac {P(X\geq 2,Y=1)}{P(Y=1)}. \]

The numerator is \[ \begin {aligned} P(X\geq 2,Y=1) &= p(2,1)+p(3,1)\\ &= \frac 3{21}+\frac 4{21}\\ &= \frac 7{21}. \end {aligned} \]

Also, \[ P(Y=1)=\frac 37=\frac 9{21}. \]

Therefore, \[ P(X\geq 2\mid Y=1) = \frac {7/21}{9/21} = \frac 79. \]

(d)

\[ \begin {aligned} E(X) &= \sum _{x=1}^{3}x\,p_X(x)\\ &= 1\left (\frac 5{21}\right ) + 2\left (\frac 7{21}\right ) + 3\left (\frac 9{21}\right )\\ &= \frac {46}{21}. \end {aligned} \]

Similarly, \[ E(Y) = 1\left (\frac 37\right ) + 2\left (\frac 47\right ) = \frac {11}{7}. \]

Thus, \[ E(X)=\frac {46}{21}, \qquad E(Y)=\frac {11}{7}. \]

The support of the joint density is \[ 0\leq y\leq x\leq 1. \]

(a)

For a fixed \(x\in [0,1]\), the possible values of \(y\) satisfy \[ 0\leq y\leq x. \]

Therefore, \[ \begin {aligned} f_X(x) &= \int _0^x 8xy\,dy\\ &= 8x\left [\frac {y^2}{2}\right ]_0^x\\ &= 4x^3. \end {aligned} \]

Hence, \[ f_X(x) = \begin {cases} 4x^3, & 0\leq x\leq 1,\\ 0, & \text {otherwise}. \end {cases} \]

For a fixed \(y\in [0,1]\), the support requires \[ y\leq x\leq 1. \]

Thus, \[ \begin {aligned} f_Y(y) &= \int _y^1 8xy\,dx\\ &= 8y\left [\frac {x^2}{2}\right ]_y^1\\ &= 4y(1-y^2). \end {aligned} \]

Therefore, \[ f_Y(y) = \begin {cases} 4y(1-y^2), & 0\leq y\leq 1,\\ 0, & \text {otherwise}. \end {cases} \]

(b)

Using the marginal density of \(X\), \[ \begin {aligned} E(X) &= \int _0^1 x f_X(x)\,dx\\ &= \int _0^1 4x^4\,dx\\ &= \frac 45. \end {aligned} \]

Similarly, \[ \begin {aligned} E(Y) &= \int _0^1 y f_Y(y)\,dy\\ &= 4\int _0^1 y^2(1-y^2)\,dy\\ &= 4\left (\frac 13-\frac 15\right )\\ &= \frac 8{15}. \end {aligned} \]

Thus, \[ E(X)=\frac 45, \qquad E(Y)=\frac 8{15}. \]

For \(x>0\), \[ \begin {aligned} f_X(x) &= \int _0^2 \frac 12 ye^{-x}\,dy\\ &= \frac 12e^{-x} \left [\frac {y^2}{2}\right ]_0^2\\ &= e^{-x}. \end {aligned} \]

Thus, \[ f_X(x) = \begin {cases} e^{-x}, & x>0,\\ 0, & \text {otherwise}. \end {cases} \]

For \(0<y<2\), \[ \begin {aligned} f_Y(y) &= \int _0^\infty \frac 12 ye^{-x}\,dx\\ &= \frac y2\int _0^\infty e^{-x}\,dx\\ &= \frac y2. \end {aligned} \]

Therefore, \[ f_Y(y) = \begin {cases} \dfrac y2, & 0<y<2,\\[1mm] 0, & \text {otherwise}. \end {cases} \]

Since the density is equal to \(1\) on the unit square, each desired probability is equal to the area of the corresponding region inside \([0,1]^2\).

First, \[ X+Y\leq \frac 12 \] describes a right triangle with legs of length \(1/2\). Hence, \[ P\left (X+Y\leq \frac 12\right ) = \frac 12\left (\frac 12\right )\left (\frac 12\right ) = \frac 18. \]

Next, \[ X-Y\leq \frac 12. \]

It is easier to subtract the region \[ X-Y>\frac 12. \]

This is a right triangle in the lower-right portion of the unit square with legs of length \(1/2\). Therefore, \[ P\left (X-Y>\frac 12\right )=\frac 18, \] and hence \[ P\left (X-Y\leq \frac 12\right ) = 1-\frac 18 = \frac 78. \]

For the third probability, \[ XY\leq \frac 14. \]

If \(0\leq x\leq 1/4\), every \(0\leq y\leq 1\) satisfies the inequality.

If \(1/4<x\leq 1\), then \[ y\leq \frac {1}{4x}. \]

Thus, \[ \begin {aligned} P\left (XY\leq \frac 14\right ) &= \int _0^{1/4}\int _0^1dy\,dx + \int _{1/4}^{1}\int _0^{1/(4x)}dy\,dx\\ &= \frac 14 + \frac 14\int _{1/4}^{1}\frac 1x\,dx\\ &= \frac 14+\frac 14\ln 4. \end {aligned} \]

Finally, \[ X^2+Y^2\leq 1 \] describes the quarter of the unit disk lying in the first quadrant. Therefore, \[ P(X^2+Y^2\leq 1) = \frac {\pi }{4}. \]

Hence, \[ P\left (X+Y\leq \frac 12\right )=\frac 18, \] \[ P\left (X-Y\leq \frac 12\right )=\frac 78, \] \[ P\left (XY\leq \frac 14\right ) = \frac 14+\frac 14\ln 4, \] and \[ P(X^2+Y^2\leq 1)=\frac {\pi }{4}. \]

Since \[ f(x,y)=h(x)h(y), \] \(X\) and \(Y\) are independent and identically distributed continuous random variables.

Because \(X\) and \(Y\) have the same distribution, symmetry gives \[ P(X>Y)=P(Y>X). \]

Moreover, since \(X\) and \(Y\) are continuous, \[ P(X=Y)=0. \]

The events \[ \{X>Y\}, \qquad \{Y>X\}, \qquad \{X=Y\} \] partition the sample space. Therefore, \[ P(X>Y)+P(Y>X)+P(X=Y)=1. \]

Hence, \[ 2P(X>Y)=1, \] so \[ P(X>Y)=\frac 12. \]

Finally, \[ P(X\geq Y) = P(X>Y)+P(X=Y) = \frac 12. \]

Introduce the transformation \[ Z=\frac {Y}{X}, \qquad W=X. \]

Then \[ z=\frac {y}{x}, \qquad w=x. \]

Solving for \(x\) and \(y\) in terms of \(z\) and \(w\), \[ x=w, \qquad y=wz. \]

The Jacobian matrix of the inverse transformation is \[ \frac {\partial (x,y)}{\partial (z,w)} = \begin {pmatrix} 0 & 1\\ w & z \end {pmatrix}. \]

Therefore, \[ \det \frac {\partial (x,y)}{\partial (z,w)} = -w, \] and hence \[ \left | \det \frac {\partial (x,y)}{\partial (z,w)} \right | = |w|. \]

By the change-of-variables formula, \[ f_{Z,W}(z,w) = f(w,wz)|w|. \]

To obtain the marginal density of \(Z\), integrate out \(W\): \[ \begin {aligned} f_Z(z) &= \int _{-\infty }^{\infty } f_{Z,W}(z,w)\,dw\\ &= \int _{-\infty }^{\infty } |w|f(w,wz)\,dw. \end {aligned} \]

Renaming the variable of integration from \(w\) to \(x\), \[ f_Z(z) = \int _{-\infty }^{\infty } |x|f(x,xz)\,dx. \]

We first calculate the marginal densities.

For \(0\leq x\leq 1\), \[ \begin {aligned} f_X(x) &= \int _x^1 8xy\,dy\\ &= 8x\left [\frac {y^2}{2}\right ]_x^1\\ &= 4x(1-x^2). \end {aligned} \]

Thus, \[ f_X(x) = \begin {cases} 4x(1-x^2), & 0\leq x\leq 1,\\ 0, & \text {otherwise}. \end {cases} \]

For \(0\leq y\leq 1\), \[ \begin {aligned} f_Y(y) &= \int _0^y8xy\,dx\\ &= 8y\left [\frac {x^2}{2}\right ]_0^y\\ &= 4y^3. \end {aligned} \]

Hence, \[ f_Y(y) = \begin {cases} 4y^3, & 0\leq y\leq 1,\\ 0, & \text {otherwise}. \end {cases} \]

If \(X\) and \(Y\) were independent, we would have \[ f(x,y)=f_X(x)f_Y(y) \] for all \(x,y\).

However, \[ f_X(x)f_Y(y) = 16xy^3(1-x^2), \] which is not equal to \[ f(x,y)=8xy \] on the support.

Therefore, \(X\) and \(Y\) are not independent.

There is also a geometric reason for the dependence. The support \[ 0\leq x\leq y\leq 1 \] is triangular rather than a Cartesian product of one-dimensional supports. In particular, knowing the value of \(Y\) restricts the possible values of \(X\).

The nonzero joint probabilities are \[ p(1,1)=\frac 17, \qquad p(1,2)=\frac 27, \qquad p(2,1)=\frac 47. \]

Therefore, \[ P(X=1) = \frac 17+\frac 27 = \frac 37, \] and \[ P(X=2)=\frac 47. \]

Similarly, \[ P(Y=1) = \frac 17+\frac 47 = \frac 57, \] and \[ P(Y=2)=\frac 27. \]

If \(X\) and \(Y\) were independent, then \[ P(X=2,Y=2) = P(X=2)P(Y=2). \]

However, \[ P(X=2,Y=2)=0, \] whereas \[ P(X=2)P(Y=2) = \frac 47\frac 27 = \frac 8{49}>0. \]

Therefore, \[ P(X=2,Y=2) \neq P(X=2)P(Y=2), \] so \(X\) and \(Y\) are not independent.

We calculate the marginal densities.

For \(x>0\), \[ \begin {aligned} f_X(x) &= \int _0^\infty x^2e^{-x(y+1)}\,dy\\ &= x^2e^{-x} \int _0^\infty e^{-xy}\,dy\\ &= x^2e^{-x}\left (\frac 1x\right )\\ &= xe^{-x}. \end {aligned} \]

Thus, \[ f_X(x) = \begin {cases} xe^{-x}, & x>0,\\ 0, & \text {otherwise}. \end {cases} \]

For \(y\geq 0\), \[ \begin {aligned} f_Y(y) &= \int _0^\infty x^2e^{-(y+1)x}\,dx. \end {aligned} \]

Using \[ \int _0^\infty x^2e^{-ax}\,dx = \frac {2}{a^3}, \qquad a>0, \] we obtain \[ f_Y(y) = \frac {2}{(y+1)^3}, \qquad y\geq 0. \]

Therefore, \[ f_X(x)f_Y(y) = \frac {2xe^{-x}}{(y+1)^3}. \]

But the joint density is \[ f(x,y) = x^2e^{-x(y+1)}. \]

These are not equal in general. Hence, \[ f(x,y)\neq f_X(x)f_Y(y), \] and therefore \(X\) and \(Y\) are not independent.

Since \[ X,Y\overset {\text {i.i.d.}}{\sim }\operatorname {Exp}(\lambda ), \] their joint density is \[ f_{X,Y}(x,y) = \lambda ^2e^{-\lambda (x+y)}, \qquad x>0,\ y>0. \]

Define \[ U=\frac {X}{X+Y}, \qquad V=X+Y. \]

Then \[ X=UV \] and \[ Y=(1-U)V. \]

Since \(X>0\) and \(Y>0\), the support of \((U,V)\) is \[ 0<u<1, \qquad v>0. \]

The Jacobian matrix is \[ \frac {\partial (x,y)}{\partial (u,v)} = \begin {pmatrix} v & u\\ -v & 1-u \end {pmatrix}. \]

Therefore, \[ \det \frac {\partial (x,y)}{\partial (u,v)} = v(1-u)+uv = v. \]

Hence, \[ \left | \det \frac {\partial (x,y)}{\partial (u,v)} \right | = v. \]

The joint density of \(U\) and \(V\) is therefore \[ \begin {aligned} f_{U,V}(u,v) &= f_{X,Y}(uv,(1-u)v)v\\ &= \lambda ^2e^{-\lambda v}v, \end {aligned} \] for \[ 0<u<1,\qquad v>0. \]

Now integrate out \(V\): \[ \begin {aligned} f_U(u) &= \int _0^\infty \lambda ^2ve^{-\lambda v}\,dv\\ &= 1, \qquad 0<u<1. \end {aligned} \]

Thus, \[ f_U(u) = \begin {cases} 1, & 0<u<1,\\ 0, & \text {otherwise}. \end {cases} \]

Therefore, \[ \frac {X}{X+Y} \sim \operatorname {Unif}(0,1). \]

First find the marginal PMF of \(Y\). For \(y=0,1,2\), \[ \begin {aligned} p_Y(y) &= \sum _{x=1}^{2}p(x,y)\\ &= \frac 1{25} \left [ (1+y^2)+(4+y^2) \right ]\\ &= \frac {5+2y^2}{25}. \end {aligned} \]

(a)

By definition, \[ p_{X\mid Y}(x\mid y) = \frac {p(x,y)}{p_Y(y)}. \]

Therefore, \[ p_{X\mid Y}(x\mid y) = \frac {x^2+y^2}{5+2y^2}, \qquad x=1,2,\quad y=0,1,2. \]

(b)

When \(Y=1\), \[ P(X=2\mid Y=1) = \frac {2^2+1^2}{5+2(1)^2} = \frac 57. \]

(c)

When \(Y=1\), \[ P(X=1\mid Y=1) = \frac {1^2+1^2}{7} = \frac 27, \] and \[ P(X=2\mid Y=1) = \frac 57. \]

Hence, \[ \begin {aligned} E(X\mid Y=1) &= 1\left (\frac 27\right ) + 2\left (\frac 57\right )\\ &= \frac {12}{7}. \end {aligned} \]

Substituting \[ y=\frac 34 \] into the conditional density gives \[ f_{X\mid Y}\left (x\mid \frac 34\right ) = \frac { 3\left (x^2+\frac 9{16}\right ) }{ 3\left (\frac 9{16}\right )+1 }. \]

Since \[ 3\left (\frac 9{16}\right )+1 = \frac {43}{16}, \] we obtain \[ f_{X\mid Y}\left (x\mid \frac 34\right ) = \frac {48}{43} \left (x^2+\frac 9{16}\right ). \]

Therefore, \[ \begin {aligned} P\left ( \frac 14<X<\frac 12 \,\middle |\, Y=\frac 34 \right ) &= \int _{1/4}^{1/2} f_{X\mid Y}\left (x\mid \frac 34\right )\,dx\\ &= \frac {48}{43} \int _{1/4}^{1/2} \left (x^2+\frac 9{16}\right )\,dx\\ &= \frac {48}{43} \left [ \frac {x^3}{3} + \frac 9{16}x \right ]_{1/4}^{1/2}\\ &= \frac {17}{86}. \end {aligned} \]

Thus, \[ P\left ( \frac 14<X<\frac 12 \,\middle |\, Y=\frac 34 \right ) = \frac {17}{86}. \]

First calculate the marginal density of \(Y\): \[ \begin {aligned} f_Y(y) &= \int _0^1(x+y)\,dx\\ &= \left [\frac {x^2}{2}+xy\right ]_0^1\\ &= \frac 12+y, \end {aligned} \] for \[ 0\leq y\leq 1. \]

Therefore, \[ \begin {aligned} f_{X\mid Y}(x\mid y) &= \frac {f(x,y)}{f_Y(y)}\\ &= \frac {x+y}{y+\frac 12}, \end {aligned} \] for \[ 0\leq x\leq 1, \qquad 0\leq y\leq 1. \]

Thus, \[ f_{X\mid Y}(x\mid y) = \begin {cases} \dfrac {x+y}{y+\frac 12}, & 0\leq x\leq 1,\quad 0\leq y\leq 1,\\[2mm] 0, & \text {otherwise}. \end {cases} \]

The support is \[ x\geq 0, \qquad -x<y<x. \]

(a)

The density must integrate to \(1\): \[ \int _0^\infty \int _{-x}^{x} ce^{-x}\,dy\,dx = 1. \]

Therefore, \[ \begin {aligned} 1 &= c\int _0^\infty 2xe^{-x}\,dx\\ &= 2c. \end {aligned} \]

Hence, \[ c=\frac 12. \]

Thus, \[ f(x,y) = \frac 12e^{-x} \] on the support.

(b)

First, \[ \begin {aligned} f_X(x) &= \int _{-x}^{x}\frac 12e^{-x}\,dy\\ &= xe^{-x}, \qquad x>0. \end {aligned} \]

Therefore, \[ \begin {aligned} f_{Y\mid X}(y\mid x) &= \frac {f(x,y)}{f_X(x)}\\ &= \frac {\frac 12e^{-x}}{xe^{-x}}\\ &= \frac 1{2x}, \end {aligned} \] for \[ -x<y<x. \]

Thus, \[ f_{Y\mid X}(y\mid x) = \begin {cases} \dfrac 1{2x}, & -x<y<x,\\[1mm] 0, & \text {otherwise}. \end {cases} \]

Hence, conditional on \(X=x\), \[ Y\mid X=x \sim \operatorname {Unif}(-x,x). \]

Next, for a fixed \(y\), the condition \[ |y|<x \] implies \[ x>|y|. \]

Therefore, \[ \begin {aligned} f_Y(y) &= \int _{|y|}^{\infty } \frac 12e^{-x}\,dx\\ &= \frac 12e^{-|y|}. \end {aligned} \]

Hence, \[ \begin {aligned} f_{X\mid Y}(x\mid y) &= \frac {f(x,y)}{f_Y(y)}\\ &= \frac {\frac 12e^{-x}} {\frac 12e^{-|y|}}\\ &= e^{-(x-|y|)}, \end {aligned} \] for \[ x>|y|. \]

Thus, \[ f_{X\mid Y}(x\mid y) = \begin {cases} e^{-(x-|y|)}, & x>|y|,\\ 0, & \text {otherwise}. \end {cases} \]

(c)

Since \[ Y\mid X=x \sim \operatorname {Unif}(-x,x), \] symmetry gives \[ E(Y\mid X=x)=0. \]

The variance of a uniform random variable on \((a,b)\) is \[ \frac {(b-a)^2}{12}. \]

Therefore, \[ \begin {aligned} \operatorname {Var}(Y\mid X=x) &= \frac {(x-(-x))^2}{12}\\ &= \frac {4x^2}{12}\\ &= \frac {x^2}{3}. \end {aligned} \]

Since \[ X,Y\overset {\text {i.i.d.}}{\sim }\operatorname {Unif}(0,1), \] their joint density is \[ f_{X,Y}(x,y)=1, \qquad 0<x<1,\quad 0<y<1. \]

The transformation is \[ u=-2\ln x, \qquad v=-2\ln y. \]

Solving for \(x\) and \(y\), \[ x=e^{-u/2}, \qquad y=e^{-v/2}. \]

Since \(0<x,y<1\), we have \[ u>0, \qquad v>0. \]

The Jacobian matrix of the inverse transformation is \[ \frac {\partial (x,y)}{\partial (u,v)} = \begin {pmatrix} -\dfrac 12e^{-u/2} & 0\\[2mm] 0 & -\dfrac 12e^{-v/2} \end {pmatrix}. \]

Therefore, \[ \left | \det \frac {\partial (x,y)}{\partial (u,v)} \right | = \frac 14e^{-(u+v)/2}. \]

Hence, \[ f_{U,V}(u,v) = f_{X,Y} \left ( e^{-u/2},e^{-v/2} \right ) \frac 14e^{-(u+v)/2}. \]

Since the original joint density equals \(1\) on its support, \[ f_{U,V}(u,v) = \begin {cases} \dfrac 14e^{-(u+v)/2}, & u>0,\quad v>0,\\[2mm] 0, & \text {otherwise}. \end {cases} \]

Define \[ U=\frac {X}{Y}, \qquad V=X. \]

Then \[ u=\frac {x}{y}, \qquad v=x. \]

Solving for \(x\) and \(y\), \[ x=v, \qquad y=\frac vu. \]

Since \(X\) and \(Y\) are positive, \[ u>0, \qquad v>0. \]

The Jacobian matrix is \[ \frac {\partial (x,y)}{\partial (u,v)} = \begin {pmatrix} 0 & 1\\[2mm] -\dfrac {v}{u^2} & \dfrac 1u \end {pmatrix}. \]

Therefore, \[ \left | \det \frac {\partial (x,y)}{\partial (u,v)} \right | = \frac {v}{u^2}. \]

Since \(X\) and \(Y\) are independent, \[ f_{X,Y}(x,y) = f_1(x)f_2(y). \]

Thus, \[ f_{U,V}(u,v) = f_1(v) f_2\left (\frac vu\right ) \frac {v}{u^2}, \qquad u>0,\quad v>0. \]

Integrating out \(V\), \[ f_U(u) = \int _0^\infty f_1(v) f_2\left (\frac vu\right ) \frac {v}{u^2}\,dv, \qquad u>0. \]

Therefore, \[ f_U(u) = \begin {cases} \displaystyle \int _0^\infty \frac {v}{u^2} f_1(v) f_2\left (\frac vu\right )\,dv, & u>0,\\[4mm] 0, & u\leq 0. \end {cases} \]

The transformation is \[ u=\frac xy, \qquad v=xy. \]

Since \(x>0\) and \(y>0\), \[ uv=x^2, \] and \[ \frac vu=y^2. \]

Therefore, \[ x=\sqrt {uv}, \qquad y=\sqrt {\frac vu}. \]

The Jacobian matrix of the inverse transformation is \[ \frac {\partial (x,y)}{\partial (u,v)}. \]

Its absolute determinant is \[ \left | \det \frac {\partial (x,y)}{\partial (u,v)} \right | = \frac 1{2u}. \]

Since \(X\) and \(Y\) are independent, \[ f_{X,Y}(x,y) = \frac 1{x^2y^2}, \qquad x\geq 1,\quad y\geq 1. \]

But \[ x^2y^2=(xy)^2=v^2. \]

Therefore, \[ f_{X,Y}(x,y)=\frac 1{v^2}. \]

It remains to determine the support.

The condition \(x\geq 1\) gives \[ \sqrt {uv}\geq 1, \] so \[ uv\geq 1 \] and therefore \[ v\geq \frac 1u. \]

The condition \(y\geq 1\) gives \[ \sqrt {\frac vu}\geq 1, \] so \[ v\geq u. \]

Thus, \[ u>0, \qquad v\geq \max \left \{u,\frac 1u\right \}. \]

Hence, \[ f_{U,V}(u,v) = \begin {cases} \dfrac {1}{2uv^2}, & u>0,\quad v\geq \max \left \{u,\dfrac 1u\right \},\\[3mm] 0, & \text {otherwise}. \end {cases} \]

The transformation is \[ u=x+y, \qquad v=e^x. \]

From \(v=e^x,\) we obtain \(x=\ln v.\) Therefore, \(y=u-\ln v.\) Since \(x>0\), \(\ln v>0,\) so \( v>1.\) Since \(y>0\), \(u-\ln v>0,\) so \(u>\ln v.\)

The Jacobian matrix of the inverse transformation is \[ \frac {\partial (x,y)}{\partial (u,v)} = \begin {pmatrix} 0 & \dfrac 1v\\[2mm] 1 & -\dfrac 1v \end {pmatrix}. \]

Therefore, \[ \left | \det \frac {\partial (x,y)}{\partial (u,v)} \right | = \frac 1v. \]

Because \(X\) and \(Y\) are independent, \[ f_{X,Y}(x,y) = e^{-x}e^{-y} = e^{-(x+y)}. \]

Since \[ x+y=u, \] we have \[ f_{X,Y}(x,y)=e^{-u}. \]

Hence, \[ f_{U,V}(u,v) = \begin {cases} \dfrac {e^{-u}}{v}, & v>1,\quad u>\ln v,\\[2mm] 0, & \text {otherwise}. \end {cases} \]

Define \[ U=X+Y, \qquad V=X-Y. \]

Solving for \(X\) and \(Y\), \[ X=\frac {U+V}{2}, \qquad Y=\frac {U-V}{2}. \]

The Jacobian matrix of the inverse transformation is \[ \frac {\partial (x,y)}{\partial (u,v)} = \begin {pmatrix} \dfrac 12 & \dfrac 12\\[2mm] \dfrac 12 & -\dfrac 12 \end {pmatrix}. \]

Therefore, \[ \left | \det \frac {\partial (x,y)}{\partial (u,v)} \right | = \frac 12. \]

Since \(X\) and \(Y\) are independent standard Normal random variables, \[ f_{X,Y}(x,y) = \frac 1{2\pi } \exp \left ( -\frac {x^2+y^2}{2} \right ). \]

Now \[ \begin {aligned} x^2+y^2 &= \left (\frac {u+v}{2}\right )^2 + \left (\frac {u-v}{2}\right )^2\\ &= \frac {u^2+v^2}{2}. \end {aligned} \]

Therefore, \[ \begin {aligned} f_{U,V}(u,v) &= \frac 1{2\pi } \exp \left ( -\frac {u^2+v^2}{4} \right ) \frac 12\\ &= \frac 1{4\pi } e^{-u^2/4} e^{-v^2/4}. \end {aligned} \]

But \[ \frac 1{\sqrt {4\pi }}e^{-u^2/4} \] is the density of a \(N(0,2)\) random variable. Hence, \[ f_U(u) = \frac 1{\sqrt {4\pi }}e^{-u^2/4} \] and \[ f_V(v) = \frac 1{\sqrt {4\pi }}e^{-v^2/4}. \]

Consequently, \(f_{U,V}(u,v) = f_U(u)f_V(v)\) and thus the independence.

Define \[ U=X+Y, \qquad V=\frac XY. \]

Since \[ v=\frac xy, \] we have \[ x=vy. \]

Substituting into \[ u=x+y \] gives \[ u=(v+1)y. \]

Therefore, \[ y=\frac {u}{1+v} \] and \[ x=\frac {uv}{1+v}. \]

Since \(X,Y>0\), \[ u>0, \qquad v>0. \]

The Jacobian matrix is \[ \frac {\partial (x,y)}{\partial (u,v)} = \begin {pmatrix} \dfrac {v}{1+v} & \dfrac {u}{(1+v)^2} \\[3mm] \dfrac 1{1+v} & -\dfrac {u}{(1+v)^2} \end {pmatrix}. \]

Therefore, \[ \left | \det \frac {\partial (x,y)}{\partial (u,v)} \right | = \frac {u}{(1+v)^2}. \]

Because \(X\) and \(Y\) are independent exponential random variables, \[ f_{X,Y}(x,y) = \lambda ^2e^{-\lambda (x+y)}, \qquad x,y>0. \]

Since \[ x+y=u, \] the transformed joint density is \[ \begin {aligned} f_{U,V}(u,v) &= \lambda ^2e^{-\lambda u} \frac {u}{(1+v)^2}\\ &= \left ( \lambda ^2u e^{-\lambda u} \right ) \left ( \frac 1{(1+v)^2} \right ), \end {aligned} \] for \[ u>0, \qquad v>0. \]

The first factor is a density on \(u>0\), since \[ \int _0^\infty \lambda ^2u e^{-\lambda u}\,du = 1. \]

The second factor is also a density on \(v>0\), since \[ \int _0^\infty \frac 1{(1+v)^2}\,dv = 1. \]

Thus, \[ f_U(u) = \lambda ^2u e^{-\lambda u}, \qquad u>0, \] and \[ f_V(v) = \frac 1{(1+v)^2}, \qquad v>0. \]

Therefore, \[ f_{U,V}(u,v) = f_U(u)f_V(v). \]

Hence, \[ X+Y \qquad \text {and}\qquad \frac XY \] are independent.

(a)

To obtain the joint marginal PMF of \(X\) and \(Y\), sum over all possible values of \(Z\): \[ \begin {aligned} p_{X,Y}(x,y) &= \sum _{z=1}^{2}p(x,y,z)\\ &= \sum _{z=1}^{2}\frac {xyz}{162}\\ &= \frac {xy}{162}(1+2)\\ &= \frac {xy}{54}. \end {aligned} \]

Thus, \[ p_{X,Y}(x,y) = \begin {cases} \dfrac {xy}{54}, & x=4,5,\quad y=1,2,3,\\[2mm] 0, & \text {otherwise}. \end {cases} \]

Similarly, \[ \begin {aligned} p_{Y,Z}(y,z) &= \sum _{x=4}^{5}p(x,y,z)\\ &= \frac {yz}{162}(4+5)\\ &= \frac {yz}{18}. \end {aligned} \]

Therefore, \[ p_{Y,Z}(y,z) = \begin {cases} \dfrac {yz}{18}, & y=1,2,3,\quad z=1,2,\\[2mm] 0, & \text {otherwise}. \end {cases} \]

Finally, \[ \begin {aligned} p_{X,Z}(x,z) &= \sum _{y=1}^{3}p(x,y,z)\\ &= \frac {xz}{162}(1+2+3)\\ &= \frac {xz}{27}. \end {aligned} \]

Hence, \[ p_{X,Z}(x,z) = \begin {cases} \dfrac {xz}{27}, & x=4,5,\quad z=1,2,\\[2mm] 0, & \text {otherwise}. \end {cases} \]

(b)

Using the joint marginal PMF of \(Y\) and \(Z\), \[ E(YZ) = \sum _{y=1}^{3}\sum _{z=1}^{2} yz\,p_{Y,Z}(y,z). \]

Thus, \[ \begin {aligned} E(YZ) &= \sum _{y=1}^{3}\sum _{z=1}^{2} yz\frac {yz}{18}\\ &= \frac 1{18} \left (\sum _{y=1}^{3}y^2\right ) \left (\sum _{z=1}^{2}z^2\right )\\ &= \frac 1{18}(1+4+9)(1+4)\\ &= \frac {70}{18}\\ &= \frac {35}{9}. \end {aligned} \]

Therefore, \[ E(YZ)=\frac {35}{9}. \]

The support is \[ 0<x<y<z<\infty . \]

(a)

For fixed \(x\) and \(y\) satisfying \(0<x<y\), \[ z>y. \]

Therefore, \[ \begin {aligned} f_{X,Y}(x,y) &= \int _y^\infty 6e^{-x-y-z}\,dz\\ &= 6e^{-x-y} \int _y^\infty e^{-z}\,dz\\ &= 6e^{-x-2y}. \end {aligned} \]

Hence, \[ f_{X,Y}(x,y) = \begin {cases} 6e^{-x-2y}, & 0<x<y,\\ 0, & \text {otherwise}. \end {cases} \]

For fixed \(x\) and \(z\) satisfying \(0<x<z\), \[ x<y<z. \]

Thus, \[ \begin {aligned} f_{X,Z}(x,z) &= \int _x^z 6e^{-x-y-z}\,dy\\ &= 6e^{-x-z} \left (e^{-x}-e^{-z}\right )\\ &= 6\left ( e^{-2x-z}-e^{-x-2z} \right ). \end {aligned} \]

Therefore, \[ f_{X,Z}(x,z) = \begin {cases} 6\left (e^{-2x-z}-e^{-x-2z}\right ), & 0<x<z,\\ 0, & \text {otherwise}. \end {cases} \]

For fixed \(y\) and \(z\) satisfying \(0<y<z\), \[ 0<x<y. \]

Hence, \[ \begin {aligned} f_{Y,Z}(y,z) &= \int _0^y 6e^{-x-y-z}\,dx\\ &= 6e^{-y-z}(1-e^{-y}). \end {aligned} \]

Thus, \[ f_{Y,Z}(y,z) = \begin {cases} 6e^{-y-z}(1-e^{-y}), & 0<y<z,\\ 0, & \text {otherwise}. \end {cases} \]

(b)

We first find the marginal density of \(X\): \[ \begin {aligned} f_X(x) &= \int _x^\infty \int _y^\infty 6e^{-x-y-z}\,dz\,dy\\ &= 6e^{-x} \int _x^\infty e^{-y} \left ( \int _y^\infty e^{-z}\,dz \right )dy\\ &= 6e^{-x} \int _x^\infty e^{-2y}\,dy\\ &= 6e^{-x} \left (\frac 12e^{-2x}\right )\\ &= 3e^{-3x}, \qquad x>0. \end {aligned} \]

Therefore, \[ \begin {aligned} E(X) &= \int _0^\infty x\,3e^{-3x}\,dx\\ &= \frac 13. \end {aligned} \]

Thus, \[ E(X)=\frac 13. \]

We first calculate the marginal densities.

For \(x>0\), \[ \begin {aligned} f_X(x) &= \int _0^\infty \int _0^\infty x^2e^{-x(1+y+z)} \,dy\,dz\\ &= x^2e^{-x} \left ( \int _0^\infty e^{-xy}\,dy \right ) \left ( \int _0^\infty e^{-xz}\,dz \right )\\ &= x^2e^{-x} \left (\frac 1x\right ) \left (\frac 1x\right )\\ &= e^{-x}. \end {aligned} \]

For \(y>0\), \[ \begin {aligned} f_Y(y) &= \int _0^\infty \int _0^\infty x^2e^{-x(1+y+z)} \,dz\,dx\\ &= \int _0^\infty xe^{-x(1+y)}\,dx\\ &= \frac {1}{(1+y)^2}. \end {aligned} \]

By symmetry, \[ f_Z(z)=\frac {1}{(1+z)^2}, \qquad z>0. \]

If \(X,Y,Z\) were mutually independent, then \[ f(x,y,z) = f_X(x)f_Y(y)f_Z(z). \]

However, \[ f_X(x)f_Y(y)f_Z(z) = \frac {e^{-x}} {(1+y)^2(1+z)^2}, \] which is not equal to \[ x^2e^{-x(1+y+z)}. \]

Therefore, the three random variables are not mutually independent.

We now examine pairwise independence.

For \(x,y>0\), \[ \begin {aligned} f_{X,Y}(x,y) &= \int _0^\infty x^2e^{-x(1+y+z)}\,dz\\ &= xe^{-x(1+y)}. \end {aligned} \]

But \[ f_X(x)f_Y(y) = \frac {e^{-x}}{(1+y)^2}, \] which is not equal to \(f_{X,Y}(x,y)\). Hence \(X\) and \(Y\) are not independent.

Similarly, \(X\) and \(Z\) are not independent.

Finally, \[ \begin {aligned} f_{Y,Z}(y,z) &= \int _0^\infty x^2e^{-x(1+y+z)}\,dx\\ &= \frac {2}{(1+y+z)^3}, \end {aligned} \] whereas \[ f_Y(y)f_Z(z) = \frac {1}{(1+y)^2(1+z)^2}. \]

Thus \(Y\) and \(Z\) are not independent either.

Therefore, \(X,Y,Z\) are neither mutually independent nor pairwise independent.

(a)

The joint CDF factors as \[ F(x,y,z) = F_X(x)F_Y(y)F_Z(z), \] where \[ F_X(x)=1-e^{-\lambda _1x}, \] \[ F_Y(y)=1-e^{-\lambda _2y}, \] and \[ F_Z(z)=1-e^{-\lambda _3z}, \] for positive arguments.

Therefore, \[ X\sim \operatorname {Exp}(\lambda _1), \qquad Y\sim \operatorname {Exp}(\lambda _2), \qquad Z\sim \operatorname {Exp}(\lambda _3), \] and \(X,Y,Z\) are mutually independent.

(b)

Differentiating the joint CDF, \[ f(x,y,z) = \frac {\partial ^3} {\partial x\,\partial y\,\partial z} F(x,y,z). \]

Therefore, \[ f(x,y,z) = \lambda _1\lambda _2\lambda _3 e^{-\lambda _1x-\lambda _2y-\lambda _3z}, \qquad x,y,z>0. \]

Thus, \[ f(x,y,z) = \begin {cases} \lambda _1\lambda _2\lambda _3 e^{-\lambda _1x-\lambda _2y-\lambda _3z}, & x,y,z>0,\\ 0, & \text {otherwise}. \end {cases} \]

(c)

We calculate \[ P(X<Y<Z) = \int _0^\infty \int _x^\infty \int _y^\infty f(x,y,z)\,dz\,dy\,dx. \]

Hence, \[ \begin {aligned} P(X<Y<Z) &= \int _0^\infty \int _x^\infty \int _y^\infty \lambda _1\lambda _2\lambda _3 e^{-\lambda _1x-\lambda _2y-\lambda _3z} \,dz\,dy\,dx\\ &= \int _0^\infty \int _x^\infty \lambda _1\lambda _2 e^{-\lambda _1x-(\lambda _2+\lambda _3)y} \,dy\,dx\\ &= \frac {\lambda _1\lambda _2} {\lambda _2+\lambda _3} \int _0^\infty e^{-(\lambda _1+\lambda _2+\lambda _3)x} \,dx\\ &= \frac {\lambda _1\lambda _2} {(\lambda _2+\lambda _3) (\lambda _1+\lambda _2+\lambda _3)}. \end {aligned} \]

Therefore, \[ P(X<Y<Z) = \frac {\lambda _1\lambda _2} {(\lambda _1+\lambda _2+\lambda _3)(\lambda _2+\lambda _3)}. \]

Since \[ f(x,y,z)=h(x)h(y)h(z), \] the random variables \(X,Y,Z\) are independent and identically distributed.

Because their common distribution is continuous, ties occur with probability zero: \[ P(X=Y)=P(X=Z)=P(Y=Z)=0. \]

Therefore, with probability \(1\), the three values occur in exactly one of the six possible strict orderings: \[ X<Y<Z, \] \[ X<Z<Y, \] \[ Y<X<Z, \] \[ Y<Z<X, \] \[ Z<X<Y, \] or \[ Z<Y<X. \]

Since \(X,Y,Z\) are identically distributed, symmetry implies that all six orderings have the same probability.

Their probabilities sum to \(1\), so each must have probability \[ \frac 16. \]

Hence, \[ P(X<Y<Z)=\frac 16. \]

By definition, \[ \operatorname {Var}(aX+bY) = E\left [ (aX+bY)-E(aX+bY) \right ]^2. \]

By linearity of expectation, \[ E(aX+bY)=aE(X)+bE(Y). \]

Therefore, \[ \begin {aligned} \operatorname {Var}(aX+bY) &= E\left [ a(X-E(X)) + b(Y-E(Y)) \right ]^2\\ &= E\Big [ a^2(X-E(X))^2 + b^2(Y-E(Y))^2\\ &\hspace {2.7cm} + 2ab(X-E(X))(Y-E(Y)) \Big ]. \end {aligned} \]

Using linearity of expectation, \[ \begin {aligned} \operatorname {Var}(aX+bY) &= a^2E[(X-E(X))^2] + b^2E[(Y-E(Y))^2]\\ &\qquad + 2abE[(X-E(X))(Y-E(Y))]. \end {aligned} \]

Thus, \[ \operatorname {Var}(aX+bY) = a^2\operatorname {Var}(X) + b^2\operatorname {Var}(Y) + 2ab\operatorname {Cov}(X,Y). \]

If \(X\) and \(Y\) are independent, then \[ \operatorname {Cov}(X,Y)=0. \]

Hence, \[ \operatorname {Var}(aX+bY) = a^2\operatorname {Var}(X) + b^2\operatorname {Var}(Y). \]

Let \[ S=\sum _{i=1}^{n}a_iX_i. \]

Then \[ E(S) = \sum _{i=1}^{n}a_iE(X_i). \]

Therefore, \[ S-E(S) = \sum _{i=1}^{n} a_i\bigl (X_i-E(X_i)\bigr ). \]

Hence, \[ \operatorname {Var}(S) = E\left [ \left ( \sum _{i=1}^{n} a_i(X_i-E(X_i)) \right )^2 \right ]. \]

Expanding the square, \[ \begin {aligned} \operatorname {Var}(S) &= E\left [ \sum _{i=1}^{n} a_i^2(X_i-E(X_i))^2 \right .\\ &\qquad \left . + 2\sum _{i<j} a_ia_j (X_i-E(X_i))(X_j-E(X_j)) \right ]. \end {aligned} \]

By linearity of expectation, \[ \operatorname {Var}(S) = \sum _{i=1}^{n} a_i^2\operatorname {Var}(X_i) + 2\sum _{i<j} a_ia_j\operatorname {Cov}(X_i,X_j). \]

Taking \[ a_i=1, \qquad i=1,\ldots ,n, \] gives \[ \operatorname {Var} \left ( \sum _{i=1}^{n}X_i \right ) = \sum _{i=1}^{n} \operatorname {Var}(X_i) + 2\sum _{i<j} \operatorname {Cov}(X_i,X_j). \]

Since \(X_1,\ldots ,X_n\) form a random sample, \[ E(X_i)=\mu \] and \[ \operatorname {Var}(X_i)=\sigma ^2 \] for every \(i\), and the random variables are independent.

By linearity of expectation, \[ \begin {aligned} E(\overline X) &= E\left ( \frac 1n\sum _{i=1}^{n}X_i \right )\\ &= \frac 1n \sum _{i=1}^{n}E(X_i)\\ &= \frac 1n(n\mu )\\ &= \mu . \end {aligned} \]

For the variance, independence implies \[ \operatorname {Cov}(X_i,X_j)=0 \qquad (i\neq j). \]

Hence, \[ \begin {aligned} \operatorname {Var}(\overline X) &= \operatorname {Var} \left ( \frac 1n\sum _{i=1}^{n}X_i \right )\\ &= \frac 1{n^2} \sum _{i=1}^{n} \operatorname {Var}(X_i)\\ &= \frac 1{n^2}(n\sigma ^2)\\ &= \frac {\sigma ^2}{n}. \end {aligned} \]

Thus, \[ E(\overline X)=\mu , \qquad \operatorname {Var}(\overline X)=\frac {\sigma ^2}{n}. \]

Since \[ Y=X^2, \] the value of \(Y\) is completely determined by \(X\). Therefore, \(X\) and \(Y\) are not independent.

We now calculate their covariance.

By symmetry of the uniform distribution on \((-1,1)\), \[ E(X)=0. \]

Also, \[ E(X^3)=0, \] because \(x^3\) is an odd function and the distribution of \(X\) is symmetric about zero.

Since \[ Y=X^2, \] we have \[ XY=X^3. \]

Thus, \[ \begin {aligned} \operatorname {Cov}(X,Y) &= E(XY)-E(X)E(Y)\\ &= E(X^3)-E(X)E(X^2)\\ &= 0-0\cdot E(X^2)\\ &= 0. \end {aligned} \]

Therefore, \(X\) and \(Y\) are dependent but uncorrelated.

Recall that \[ \operatorname {Cov}(X,Y) = E(XY)-E(X)E(Y). \]

First calculate \(E(X)\): \[ \begin {aligned} E(X) &= \sum _{x=1}^{3} \sum _{y=3}^{4} x\,p(x,y)\\ &= \frac 1{70} \sum _{x=1}^{3} \sum _{y=3}^{4} x^2(x+y). \end {aligned} \]

Evaluating the six terms, \[ E(X) = \frac {156}{70} = \frac {78}{35}. \]

Next, \[ \begin {aligned} E(Y) &= \sum _{x=1}^{3} \sum _{y=3}^{4} y\,p(x,y)\\ &= \frac 1{70} \sum _{x=1}^{3} \sum _{y=3}^{4} xy(x+y)\\ &= \frac {249}{70}. \end {aligned} \]

Finally, \[ \begin {aligned} E(XY) &= \sum _{x=1}^{3} \sum _{y=3}^{4} xy\,p(x,y)\\ &= \frac 1{70} \sum _{x=1}^{3} \sum _{y=3}^{4} x^2y(x+y)\\ &= \frac {558}{70}\\ &= \frac {279}{35}. \end {aligned} \]

Therefore, \[ \begin {aligned} \operatorname {Cov}(X,Y) &= \frac {279}{35} - \left (\frac {78}{35}\right ) \left (\frac {249}{70}\right )\\ &= \frac {279}{35} - \frac {9711}{1225}\\ &= \frac {54}{1225}. \end {aligned} \]

Recall that \[ \operatorname {Cov}(A,B) = E(AB)-E(A)E(B). \]

(a)

\[ \begin {aligned} \operatorname {Cov}(X+Y,Z) &= E[(X+Y)Z] - E(X+Y)E(Z)\\ &= E(XZ)+E(YZ) - [E(X)+E(Y)]E(Z)\\ &= [E(XZ)-E(X)E(Z)]\\ &\qquad + [E(YZ)-E(Y)E(Z)]\\ &= \operatorname {Cov}(X,Z) + \operatorname {Cov}(Y,Z). \end {aligned} \]

(b)

Similarly, \[ \begin {aligned} \operatorname {Cov}(X,Y+Z) &= E[X(Y+Z)] - E(X)E(Y+Z)\\ &= \operatorname {Cov}(X,Y) + \operatorname {Cov}(X,Z). \end {aligned} \]

(c)

Using bilinearity, \[ \begin {aligned} \operatorname {Cov}(X+Y,X-Y) &= \operatorname {Cov}(X,X) - \operatorname {Cov}(X,Y)\\ &\qquad + \operatorname {Cov}(Y,X) - \operatorname {Cov}(Y,Y). \end {aligned} \]

Since \[ \operatorname {Cov}(X,X)=\operatorname {Var}(X), \] \[ \operatorname {Cov}(Y,Y)=\operatorname {Var}(Y), \] and \[ \operatorname {Cov}(X,Y)=\operatorname {Cov}(Y,X), \] the middle terms cancel. Therefore, \[ \operatorname {Cov}(X+Y,X-Y) = \operatorname {Var}(X)-\operatorname {Var}(Y). \]

(d)

Since \[ \operatorname {Var}(X-Y) = \operatorname {Cov}(X-Y,X-Y), \] bilinearity gives \[ \begin {aligned} \operatorname {Var}(X-Y) &= \operatorname {Cov}(X,X) - \operatorname {Cov}(X,Y)\\ &\qquad - \operatorname {Cov}(Y,X) + \operatorname {Cov}(Y,Y)\\ &= \operatorname {Var}(X) + \operatorname {Var}(Y) - 2\operatorname {Cov}(X,Y). \end {aligned} \]

Using bilinearity of covariance repeatedly, \[ \begin {aligned} \operatorname {Cov} \left ( \sum _{i=1}^{n}a_iX_i, \sum _{j=1}^{m}b_jY_j \right ) &= \sum _{i=1}^{n} a_i \operatorname {Cov} \left ( X_i, \sum _{j=1}^{m}b_jY_j \right )\\ &= \sum _{i=1}^{n} a_i \sum _{j=1}^{m} b_j\operatorname {Cov}(X_i,Y_j)\\ &= \sum _{i=1}^{n} \sum _{j=1}^{m} a_ib_j\operatorname {Cov}(X_i,Y_j). \end {aligned} \]

For the special case, \[ \operatorname {Cov}(aX+b,cY+d), \] constants have zero covariance with every random variable. Hence, \[ \begin {aligned} \operatorname {Cov}(aX+b,cY+d) &= ac\,\operatorname {Cov}(X,Y). \end {aligned} \]

We have \[ E(I_A)=P(A), \qquad E(I_B)=P(B). \]

Also, \[ I_AI_B=I_{A\cap B}, \] so \[ E(I_AI_B)=P(A\cap B). \]

Therefore, \[ \operatorname {Cov}(I_A,I_B) = P(A\cap B)-P(A)P(B). \]

The indicators are positively correlated if and only if \[ \operatorname {Cov}(I_A,I_B)>0. \]

Thus, \[ P(A\cap B)>P(A)P(B). \]

Since \(P(B)>0\), dividing by \(P(B)\) gives \[ \frac {P(A\cap B)}{P(B)}>P(A), \] or \[ P(A\mid B)>P(A). \]

Similarly, since \(P(A)>0\), \[ P(A\cap B)>P(A)P(B) \] is equivalent to \[ \frac {P(A\cap B)}{P(A)}>P(B), \] which is \[ P(B\mid A)>P(B). \]

Hence, \[ \operatorname {Cov}(I_A,I_B)>0 \] if and only if \[ P(A\mid B)>P(A), \] and if and only if \[ P(B\mid A)>P(B). \]

Suppose \(X\) and \(Y\) have joint density \(f_{X,Y}(x,y)\).

By definition, \[ E(X\mid Y=y) = \int _{-\infty }^{\infty } x f_{X\mid Y}(x\mid y)\,dx. \]

Therefore, \[ \begin {aligned} E[E(X\mid Y)] &= \int _{-\infty }^{\infty } E(X\mid Y=y)f_Y(y)\,dy\\ &= \int _{-\infty }^{\infty } \left [ \int _{-\infty }^{\infty } x f_{X\mid Y}(x\mid y)\,dx \right ] f_Y(y)\,dy. \end {aligned} \]

Interchanging the order of integration, \[ \begin {aligned} E[E(X\mid Y)] &= \int _{-\infty }^{\infty } x \left [ \int _{-\infty }^{\infty } f_{X\mid Y}(x\mid y)f_Y(y)\,dy \right ]dx. \end {aligned} \]

Since \[ f_{X\mid Y}(x\mid y) = \frac {f_{X,Y}(x,y)}{f_Y(y)}, \] we have \[ f_{X\mid Y}(x\mid y)f_Y(y) = f_{X,Y}(x,y). \]

Hence, \[ \begin {aligned} E[E(X\mid Y)] &= \int _{-\infty }^{\infty } x \left [ \int _{-\infty }^{\infty } f_{X,Y}(x,y)\,dy \right ]dx\\ &= \int _{-\infty }^{\infty } x f_X(x)\,dx\\ &= E(X). \end {aligned} \]

Therefore, \[ E[E(X\mid Y)]=E(X). \]

Recall that \[ \operatorname {Var}(X) = E(X^2)-[E(X)]^2. \]

For each value of \(Y\), \[ \operatorname {Var}(X\mid Y) = E(X^2\mid Y)-[E(X\mid Y)]^2. \]

Taking expectations, \[ \begin {aligned} E[\operatorname {Var}(X\mid Y)] &= E[E(X^2\mid Y)] - E\left ([E(X\mid Y)]^2\right ). \end {aligned} \]

By the Law of Total Expectation, \[ E[E(X^2\mid Y)] = E(X^2). \]

Therefore, \[ E[\operatorname {Var}(X\mid Y)] = E(X^2) - E\left ([E(X\mid Y)]^2\right ). \]

Also, \[ \begin {aligned} \operatorname {Var}(E(X\mid Y)) &= E\left ([E(X\mid Y)]^2\right ) - \left (E[E(X\mid Y)]\right )^2. \end {aligned} \]

Again using the Law of Total Expectation, \[ E[E(X\mid Y)] = E(X). \]

Thus, \[ \operatorname {Var}(E(X\mid Y)) = E\left ([E(X\mid Y)]^2\right ) - [E(X)]^2. \]

Adding the two expressions, \[ \begin {aligned} &E[\operatorname {Var}(X\mid Y)] + \operatorname {Var}(E(X\mid Y))\\ &= E(X^2) - E\left ([E(X\mid Y)]^2\right ) + E\left ([E(X\mid Y)]^2\right ) - [E(X)]^2\\ &= E(X^2)-[E(X)]^2\\ &= \operatorname {Var}(X). \end {aligned} \]

Therefore, \[ \operatorname {Var}(X) = E[\operatorname {Var}(X\mid Y)] + \operatorname {Var}(E(X\mid Y)). \]

We first find the marginal density of \(Y\).

For \(0<y<1\), \[ \begin {aligned} f_Y(y) &= \int _0^1 \frac 32(x^2+y^2)\,dx\\ &= \frac 32 \left ( \frac 13+y^2 \right )\\ &= \frac 12+\frac 32y^2\\ &= \frac {3y^2+1}{2}. \end {aligned} \]

Thus, \[ \begin {aligned} f_{X\mid Y}(x\mid y) &= \frac {f(x,y)}{f_Y(y)}\\ &= \frac { \frac 32(x^2+y^2) }{ \frac 12(3y^2+1) }\\ &= \frac {3(x^2+y^2)} {3y^2+1}, \end {aligned} \] for \(0<x<1\).

Therefore, \[ \begin {aligned} E(X\mid Y=y) &= \int _0^1 x f_{X\mid Y}(x\mid y)\,dx\\ &= \frac {3}{3y^2+1} \int _0^1 (x^3+xy^2)\,dx\\ &= \frac {3}{3y^2+1} \left ( \frac 14+\frac {y^2}{2} \right )\\ &= \frac {3(2y^2+1)} {4(3y^2+1)}. \end {aligned} \]

Hence the conditional expectation as a random variable is \[ E(X\mid Y) = \frac {3(2Y^2+1)} {4(3Y^2+1)}. \]

Let \(I\) be the indicator of the event that \(Y\) is selected. Then \[ P(I=1)=p, \qquad P(I=0)=1-p, \] and \[ X=IY+(1-I)Z. \]

Since \(I\) is independent of \(Y\) and \(Z\), \[ E(IY)=E(I)E(Y)=pE(Y), \] and \[ E[(1-I)Z] = E(1-I)E(Z) = (1-p)E(Z). \]

Therefore, \[ \begin {aligned} E(X) &= E[IY+(1-I)Z]\\ &= pE(Y)+(1-p)E(Z). \end {aligned} \]

Hence, \[ E(X) = pE(Y)+(1-p)E(Z). \]

(a)

By definition, \[ M_X(t) = E(e^{tX}) = \int _{-\infty }^{\infty } e^{tx}f(x)\,dx. \]

Therefore, \[ M_X(t) = 6\int _0^1 x(1-x)e^{tx}\,dx. \]

Writing \[ x(1-x)=x-x^2, \] we have \[ M_X(t) = 6\left [ \int _0^1 xe^{tx}\,dx - \int _0^1 x^2e^{tx}\,dx \right ]. \]

After integration by parts, \[ M_X(t) = \frac { 6\left [ t+(t-2)e^t+2 \right ] }{t^3}, \qquad t\neq 0. \]

At \(t=0\), \[ M_X(0)=1. \]

Thus, \[ M_X(t) = \begin {cases} \dfrac { 6\left [t+(t-2)e^t+2\right ] }{t^3}, & t\neq 0,\\[3mm] 1, & t=0. \end {cases} \]

(b)

Recall that \[ E(X)=M_X'(0). \]

To evaluate this derivative conveniently, expand \(M_X(t)\) around \(t=0\). Using \[ e^t = 1+t+\frac {t^2}{2} +\frac {t^3}{6} +\frac {t^4}{24} +\cdots , \] we obtain \[ M_X(t) = 1+\frac 12t+\frac {3}{20}t^2+\cdots . \]

Therefore, \[ M_X'(0)=\frac 12. \]

Hence, \[ E(X)=\frac 12. \]

Since \[ X\sim \operatorname {Unif}(a,b), \] its density is \[ f_X(x) = \frac 1{b-a}, \qquad a<x<b. \]

Therefore, \[ \begin {aligned} M_X(t) &= E(e^{tX})\\ &= \frac 1{b-a} \int _a^b e^{tx}\,dx. \end {aligned} \]

For \(t\neq 0\), \[ \begin {aligned} M_X(t) &= \frac 1{b-a} \left [ \frac {e^{tx}}{t} \right ]_a^b\\ &= \frac {e^{bt}-e^{at}} {(b-a)t}. \end {aligned} \]

Also, \[ M_X(0)=1. \]

Thus, \[ M_X(t) = \begin {cases} \dfrac {e^{bt}-e^{at}} {(b-a)t}, & t\neq 0,\\[3mm] 1, & t=0. \end {cases} \]

(a)

By definition, \[ \begin {aligned} M_X(t) &= E(e^{tX})\\ &= \sum _{k=1}^{\infty } e^{tk}p q^{k-1}. \end {aligned} \]

Factor out \(pe^t\): \[ M_X(t) = pe^t \sum _{k=1}^{\infty } (qe^t)^{k-1}. \]

The geometric series converges when \[ |qe^t|<1. \]

Since \(q>0\), this is equivalent to \[ t<-\ln q. \]

Therefore, \[ M_X(t) = \frac {pe^t}{1-qe^t}, \qquad t<-\ln q. \]

(b)

Differentiate: \[ M_X'(t) = \frac {pe^t} {(1-qe^t)^2}. \]

Hence, \[ E(X) = M_X'(0) = \frac {p}{(1-q)^2}. \]

Since \[ 1-q=p, \] we obtain \[ E(X)=\frac 1p. \]

Differentiate again: \[ M_X''(t) = \frac { pe^t(1+qe^t) }{ (1-qe^t)^3 }. \]

Therefore, \[ E(X^2) = M_X''(0) = \frac {1+q}{p^2}. \]

Thus, \[ \begin {aligned} \operatorname {Var}(X) &= E(X^2)-[E(X)]^2\\ &= \frac {1+q}{p^2} - \frac 1{p^2}\\ &= \frac {q}{p^2}. \end {aligned} \]

Since \(q=1-p\), \[ \operatorname {Var}(X) = \frac {1-p}{p^2}. \]

By definition, \[ M_Y(t) = E(e^{tY}). \]

Since \[ Y=2X+1, \] we have \[ \begin {aligned} M_Y(t) &= E\left (e^{t(2X+1)}\right )\\ &= e^tE(e^{2tX})\\ &= e^tM_X(2t). \end {aligned} \]

Therefore, \[ M_Y(t) = e^t \frac 1{1-2t}. \]

The MGF \(M_X(2t)\) is finite provided \[ 2t<1, \] that is, \[ t<\frac 12. \]

Hence, \[ M_Y(t) = \frac {e^t}{1-2t}, \qquad t<\frac 12. \]

Write \[ M_X(t) = 8(2-t)^{-3}. \]

Differentiating, \[ M_X'(t) = 24(2-t)^{-4}. \]

Therefore, \[ E(X) = M_X'(0) = \frac {24}{16} = \frac 32. \]

Differentiating again, \[ M_X''(t) = 96(2-t)^{-5}. \]

Thus, \[ E(X^2) = M_X''(0) = \frac {96}{32} = 3. \]

Hence, \[ \begin {aligned} \operatorname {Var}(X) &= E(X^2)-[E(X)]^2\\ &= 3-\left (\frac 32\right )^2\\ &= 3-\frac 94\\ &= \frac 34. \end {aligned} \]

Therefore, \[ E(X)=\frac 32, \qquad \operatorname {Var}(X)=\frac 34. \]

The moment-generating function has the power-series representation \[ M_Z(t) = \sum _{n=0}^{\infty } \frac {E(Z^n)}{n!}t^n. \]

On the other hand, \[ M_Z(t) = e^{t^2/2}. \]

Expanding the exponential, \[ \begin {aligned} e^{t^2/2} &= \sum _{k=0}^{\infty } \frac 1{k!} \left (\frac {t^2}{2}\right )^k\\ &= \sum _{k=0}^{\infty } \frac {t^{2k}}{2^k k!}. \end {aligned} \]

There are no odd powers of \(t\). Therefore, \[ E(Z^n)=0 \] whenever \(n\) is odd.

If \[ n=2k, \] then comparison of the coefficient of \(t^{2k}\) gives \[ \frac {E(Z^{2k})}{(2k)!} = \frac 1{2^k k!}. \]

Thus, \[ E(Z^{2k}) = \frac {(2k)!}{2^k k!}. \]

Equivalently, \[ E(Z^{2k}) = (2k-1)!!. \]

Therefore, \[ E(Z^n) = \begin {cases} 0, & n\text { odd},\\[2mm] \dfrac {n!} {2^{n/2}(n/2)!}, & n\text { even}. \end {cases} \]

For \[ X_i\sim N(\mu _i,\sigma _i^2), \] the MGF is \[ M_{X_i}(t) = \exp \left ( \mu _i t+\frac 12\sigma _i^2t^2 \right ). \]

Let \[ S=\sum _{i=1}^{n}a_iX_i. \]

Then \[ M_S(t) = E\left [ e^{t\sum _{i=1}^{n}a_iX_i} \right ]. \]

Since the \(X_i\)’s are independent, \[ \begin {aligned} M_S(t) &= \prod _{i=1}^{n} E(e^{ta_iX_i})\\ &= \prod _{i=1}^{n} M_{X_i}(a_it). \end {aligned} \]

Therefore, \[ \begin {aligned} M_S(t) &= \prod _{i=1}^{n} \exp \left ( a_i\mu _i t + \frac 12a_i^2\sigma _i^2t^2 \right )\\ &= \exp \left [ t\sum _{i=1}^{n}a_i\mu _i + \frac {t^2}{2} \sum _{i=1}^{n}a_i^2\sigma _i^2 \right ]. \end {aligned} \]

This is the MGF of a Normal random variable with mean \[ \sum _{i=1}^{n}a_i\mu _i \] and variance \[ \sum _{i=1}^{n}a_i^2\sigma _i^2. \]

Hence, \[ \sum _{i=1}^{n}a_iX_i \sim N\left ( \sum _{i=1}^{n}a_i\mu _i, \, \sum _{i=1}^{n}a_i^2\sigma _i^2 \right ). \]

For the sample mean, take \[ a_i=\frac 1n, \qquad \mu _i=\mu , \qquad \sigma _i^2=\sigma ^2. \]

Then \[ E(\overline X) = \sum _{i=1}^{n} \frac 1n\mu = \mu , \] and \[ \operatorname {Var}(\overline X) = \sum _{i=1}^{n} \frac 1{n^2}\sigma ^2 = \frac {\sigma ^2}{n}. \]

Therefore, \[ \overline X \sim N\left ( \mu ,\frac {\sigma ^2}{n} \right ). \]

Let \[ q=1-p. \]

The MGF of a geometric random variable under this convention is \[ M_{X_i}(t) = \frac {pe^t}{1-qe^t}, \qquad t<-\ln q. \]

Define \[ S_n=X_1+\cdots +X_n. \]

Since \(X_1,\ldots ,X_n\) are independent, \[ \begin {aligned} M_{S_n}(t) &= \prod _{i=1}^{n}M_{X_i}(t)\\ &= \left ( \frac {pe^t}{1-qe^t} \right )^n. \end {aligned} \]

Now recall that if \[ W\sim \operatorname {NegBin}(n,p), \] where \(W\) denotes the trial on which the \(n\)-th success occurs, then \[ P(W=k) = \binom {k-1}{n-1} p^nq^{k-n}, \qquad k=n,n+1,\ldots . \]

Its MGF is \[ M_W(t) = \left ( \frac {pe^t}{1-qe^t} \right )^n. \]

Thus, \[ M_{S_n}(t)=M_W(t) \] in a neighborhood of \(0\).

By uniqueness of moment-generating functions, \[ S_n = X_1+\cdots +X_n \sim \operatorname {NegBin}(n,p). \]

Since \(X\) and \(Y\) are independent, \[ X+Y \sim \operatorname {Bin}(n+m,p). \]

Therefore, \[ P(X=i\mid X+Y=j) = \frac { P(X=i,Y=j-i) }{ P(X+Y=j) }. \]

The numerator is \[ \begin {aligned} P(X=i,Y=j-i) &= P(X=i)P(Y=j-i)\\ &= \binom ni p^i(1-p)^{n-i}\\ &\qquad \times \binom m{j-i} p^{j-i}(1-p)^{m-j+i}. \end {aligned} \]

Thus, \[ P(X=i,Y=j-i) = \binom ni \binom m{j-i} p^j(1-p)^{n+m-j}. \]

Also, \[ P(X+Y=j) = \binom {n+m}{j} p^j(1-p)^{n+m-j}. \]

Hence, \[ P(X=i\mid X+Y=j) = \frac { \binom ni \binom m{j-i} }{ \binom {n+m}{j} }, \] for \[ \max \{0,j-m\} \leq i \leq \min \{n,j\}. \]

This is the probability mass function of a hypergeometric random variable.

Conditional on there being exactly \(j\) total successes among the \(n+m\) Bernoulli trials, all choices of the \(j\) successful trials are equally likely. The random variable \(X\) counts how many of those \(j\) successes occur among the first \(n\) trials.

Therefore, \[ X\mid (X+Y=j) \sim \operatorname {Hypergeom}(n+m,n,j). \]

Let \[ T=X+Y+Z. \]

Since independent Poisson random variables add, \[ T \sim \operatorname {Pois} (\lambda _1+\lambda _2+\lambda _3). \]

Also, \[ X+Z \sim \operatorname {Pois}(\lambda _1+\lambda _3), \] and \(X+Z\) is independent of \(Y\).

Therefore, \[ \begin {aligned} P(Y=y\mid T=t) &= \frac { P(Y=y,\ X+Z=t-y) }{ P(T=t) }\\ &= \frac { P(Y=y) P(X+Z=t-y) }{ P(T=t) }. \end {aligned} \]

Substituting the Poisson probabilities, \[ \begin {aligned} P(Y=y\mid T=t) &= \frac { e^{-\lambda _2} \dfrac {\lambda _2^y}{y!} \, e^{-(\lambda _1+\lambda _3)} \dfrac {(\lambda _1+\lambda _3)^{t-y}}{(t-y)!} }{ e^{-(\lambda _1+\lambda _2+\lambda _3)} \dfrac { (\lambda _1+\lambda _2+\lambda _3)^t }{t!} }. \end {aligned} \]

The exponential terms cancel, giving \[ P(Y=y\mid T=t) = \binom ty \frac { \lambda _2^y (\lambda _1+\lambda _3)^{t-y} }{ (\lambda _1+\lambda _2+\lambda _3)^t }. \]

Equivalently, \[ P(Y=y\mid T=t) = \binom ty \left ( \frac {\lambda _2} {\lambda _1+\lambda _2+\lambda _3} \right )^y \left ( \frac {\lambda _1+\lambda _3} {\lambda _1+\lambda _2+\lambda _3} \right )^{t-y}. \]

Thus, \[ Y\mid (X+Y+Z=t) \sim \operatorname {Bin} \left ( t, \frac {\lambda _2} {\lambda _1+\lambda _2+\lambda _3} \right ). \]

(a)

\[ E(\log X)\leq \log (E X). \]

Since \(\log x\) is concave on \((0,\infty )\), Jensen’s inequality gives \[ E(\log X)\leq \log (E X). \]

(b)

\[ E(X)\leq \sqrt {E(X^2)}. \]

By Cauchy–Schwarz, \[ [E(X)]^2\leq E(X^2). \]

Since \(X>0\), \[ E(X)\leq \sqrt {E(X^2)}. \]

(c)

\[ E(\sin ^2 X)+E(\cos ^2 X)=1. \]

Indeed, \[ \sin ^2 X+\cos ^2 X=1 \] for every value of \(X\). Therefore, \[ \begin {aligned} E(\sin ^2 X)+E(\cos ^2 X) &= E(\sin ^2 X+\cos ^2 X)\\ &= E(1)\\ &= 1. \end {aligned} \]

(d)

\[ E(|X|)\leq \sqrt {E(X^2)}. \]

By Cauchy–Schwarz, \[ [E(|X|)]^2 \leq E(|X|^2) = E(X^2). \]

Hence, \[ E(|X|) \leq \sqrt {E(X^2)}. \]

(e)

\[ P(X>c)\leq \frac {E(X^3)}{c^3}. \]

Since \(X>0\), \[ X>c \quad \Longleftrightarrow \quad X^3>c^3. \]

Applying Markov’s inequality to the nonnegative random variable \(X^3\), \[ P(X>c) = P(X^3>c^3) \leq \frac {E(X^3)}{c^3}. \]

(f)

\[ P(X\leq Y)=P(X\geq Y). \]

Since \(X\) and \(Y\) are identically distributed and independent, the joint distribution of \((X,Y)\) is unchanged when \(X\) and \(Y\) are exchanged.

Therefore, \[ P(X\leq Y)=P(Y\leq X)=P(X\geq Y). \]

(g)

\[ E(XY)\leq \sqrt {E(X^2)E(Y^2)}. \]

This follows directly from the Cauchy–Schwarz inequality: \[ |E(XY)| \leq \sqrt {E(X^2)E(Y^2)}. \]

Since \(X\) and \(Y\) are positive, \[ E(XY)\geq 0, \] and hence \[ E(XY) \leq \sqrt {E(X^2)E(Y^2)}. \]

(h)

\[ P(X+Y>10) \leq P(X>5\text { or }Y>5). \]

Indeed, \[ \{X+Y>10\} \subseteq \{X>5\}\cup \{Y>5\}. \]

Therefore, \[ P(X+Y>10) \leq P(X>5\text { or }Y>5). \]

(i)

\[ E(\min (X,Y)) \leq \min (E X,E Y). \]

Pointwise, \[ \min (X,Y)\leq X \] and \[ \min (X,Y)\leq Y. \]

Taking expectations gives \[ E(\min (X,Y))\leq E(X) \] and \[ E(\min (X,Y))\leq E(Y). \]

Therefore, \[ E(\min (X,Y)) \leq \min (E X,E Y). \]

(j)

\[ E\left (\frac {X}{Y}\right ) \ ?\ \frac {E X}{E Y}. \]

No fixed relation holds in general.

By independence, \[ E\left (\frac {X}{Y}\right ) = E(X)E\left (\frac 1Y\right ), \] provided the expectations exist. This is generally not equal to \[ \frac {E(X)}{E(Y)}. \]

(k)

\[ E\left (X^2(X^2+1)\right ) \geq E\left (X^2(Y^2+1)\right ). \]

For the left-hand side, \[ E\left (X^2(X^2+1)\right ) = E(X^4)+E(X^2). \]

For the right-hand side, independence gives \[ \begin {aligned} E\left (X^2(Y^2+1)\right ) &= E(X^2Y^2)+E(X^2)\\ &= E(X^2)E(Y^2)+E(X^2). \end {aligned} \]

Since \(X\) and \(Y\) are identically distributed, \[ E(Y^2)=E(X^2), \] so \[ E\left (X^2(Y^2+1)\right ) = [E(X^2)]^2+E(X^2). \]

Finally, \[ E(X^4)\geq [E(X^2)]^2 \] because \[ \operatorname {Var}(X^2)\geq 0. \]

Therefore, \[ E\left (X^2(X^2+1)\right ) \geq E\left (X^2(Y^2+1)\right ). \]

(l)

\[ E\left (\frac {X^3}{X^3+Y^3}\right ) = E\left (\frac {Y^3}{X^3+Y^3}\right ). \]

Since \(X\) and \(Y\) are i.i.d., exchanging \(X\) and \(Y\) does not change their joint distribution. Hence, \[ E\left (\frac {X^3}{X^3+Y^3}\right ) = E\left (\frac {Y^3}{X^3+Y^3}\right ). \]

In fact, since \[ \frac {X^3}{X^3+Y^3} + \frac {Y^3}{X^3+Y^3} = 1, \] the two equal expectations must each be \[ \frac 12. \]

(a)

Markov’s inequality states that, for a nonnegative random variable, \[ P(X\geq a) \leq \frac {E(X)}{a}. \]

Since \[ E(X)=1, \] we obtain \[ P(X\geq 10) \leq \frac {1}{10}. \]

Thus, the Markov bound is \[ P(X\geq 10)\leq 0.1. \]

(b)

To apply Chebyshev’s inequality, observe that \[ X\geq 10 \] implies \[ X-E(X)\geq 9. \]

Therefore, \[ \{X\geq 10\} \subseteq \{|X-E(X)|\geq 9\}. \]

Hence, \[ \begin {aligned} P(X\geq 10) &\leq P(|X-E(X)|\geq 9)\\ &\leq \frac {\operatorname {Var}(X)}{9^2}\\ &= \frac {1}{81}. \end {aligned} \]

Thus, the Chebyshev bound is \[ P(X\geq 10) \leq \frac {1}{81} \approx 0.01235. \]

(c)

For \(t>0\), the Chernoff bound gives \[ P(X\geq 10) \leq e^{-10t}M_X(t). \]

Since \[ M_X(t)=\frac {1}{1-t}, \qquad t<1, \] we have \[ P(X\geq 10) \leq \frac {e^{-10t}}{1-t}, \qquad 0<t<1. \]

We now minimize \[ h(t)=\frac {e^{-10t}}{1-t}. \]

It is convenient to minimize its logarithm: \[ \log h(t) = -10t-\log (1-t). \]

Differentiating, \[ \frac {d}{dt}\log h(t) = -10+\frac {1}{1-t}. \]

Setting this equal to zero, \[ -10+\frac {1}{1-t}=0. \]

Thus, \[ \frac {1}{1-t}=10, \] so \[ 1-t=\frac {1}{10}, \] and therefore \[ t=\frac {9}{10}. \]

Substituting this value into the Chernoff bound, \[ \begin {aligned} P(X\geq 10) &\leq \frac {e^{-10(9/10)}}{1-9/10}\\ &= \frac {e^{-9}}{1/10}\\ &= 10e^{-9}. \end {aligned} \]

Thus, the optimized Chernoff bound is \[ P(X\geq 10) \leq 10e^{-9} \approx 0.001234. \]

(d)

The three bounds are \[ \begin {array}{c|c} \text {Method} & \text {Upper bound}\\ \hline \text {Markov} & \dfrac {1}{10}\\[2mm] \text {Chebyshev} & \dfrac {1}{81}\\[2mm] \text {Chernoff} & 10e^{-9} \end {array} \]

Numerically, \[ \frac {1}{10}=0.1, \] \[ \frac {1}{81}\approx 0.01235, \] and \[ 10e^{-9}\approx 0.001234. \]

Therefore, \[ 10e^{-9} < \frac {1}{81} < \frac {1}{10}. \]

Hence, among these three inequalities, the Chernoff bound gives the sharpest upper bound.

The exact probability can be calculated directly from the exponential distribution: \[ \begin {aligned} P(X\geq 10) &= \int _{10}^{\infty }e^{-x}\,dx\\ &= e^{-10}. \end {aligned} \]

Numerically, \[ e^{-10}\approx 0.0000454. \]

Thus, \[ \underbrace {e^{-10}}_{\text {exact}} < \underbrace {10e^{-9}}_{\text {Chernoff}} < \underbrace {\frac {1}{81}}_{\text {Chebyshev}} < \underbrace {\frac {1}{10}}_{\text {Markov}}. \]

(e)

Define \[ g(x)=\frac {1}{1+x}, \qquad x>0. \]

We have \[ g'(x) = -\frac {1}{(1+x)^2} \] and \[ g''(x) = \frac {2}{(1+x)^3}. \]

Since \[ g''(x)>0 \qquad \text {for }x>0, \] the function \(g\) is convex on the support of \(X\).

Therefore, Jensen’s inequality gives \[ g(E(X)) \leq E(g(X)). \]

Hence, \[ \frac {1}{1+E(X)} \leq E\left (\frac {1}{1+X}\right ). \]

Equivalently, \[ E\left (\frac {1}{1+X}\right ) \geq \frac {1}{1+E(X)}. \]

Since \(E(X)=1\), \[ E\left (\frac {1}{1+X}\right ) \geq \frac 12. \]

We can write \[ 8000<X<12{,}000 \] as \[ -2000<X-10{,}000<2000. \]

Thus, \[ P(8000<X<12{,}000) = P(|X-10{,}000|<2000). \]

Using the complement, \[ P(|X-10{,}000|<2000) = 1- P(|X-10{,}000|\geq 2000). \]

By Chebyshev’s inequality, \[ P(|X-10{,}000|\geq 2000) \leq \frac {2000}{2000^2}. \]

Therefore, \[ P(|X-10{,}000|\geq 2000) \leq 0.0005. \]

Hence, \[ P(8000<X<12{,}000) \geq 1-0.0005 = 0.9995. \]

Thus, \[ P(8000<X<12{,}000) \geq 0.9995. \]

(a)

Since each outcome has probability \(1/6\), \[ E(X) = \frac 16(1+2+3+4+5+6) = \frac {21}{6} = \frac 72. \]

Also, \[ E(X^2) = \frac 16 (1^2+2^2+3^2+4^2+5^2+6^2) = \frac {91}{6}. \]

Therefore, \[ \begin {aligned} \operatorname {Var}(X) &= E(X^2)-[E(X)]^2\\ &= \frac {91}{6} - \frac {49}{4}\\ &= \frac {35}{12}. \end {aligned} \]

(b)

By Markov’s inequality, \[ P(X\geq 6) \leq \frac {E(X)}6. \]

Hence, \[ P(X\geq 6) \leq \frac {7/2}{6} = \frac 7{12} \approx 0.5833. \]

(c)

By Chebyshev’s inequality, \[ P\left ( \left |X-\frac 72\right | \geq \frac 32 \right ) \leq \frac { 35/12 }{ (3/2)^2 }. \]

Therefore, \[ P\left ( \left |X-\frac 72\right | \geq \frac 32 \right ) \leq \frac {35}{27} \approx 1.296. \]

Since probabilities cannot exceed \(1\), this bound is trivial. The useful upper bound is simply \[ 1. \]

(d)

Exactly, \[ P(X\geq 6) = P(X=6) = \frac 16. \]

Also, \[ \left |X-\frac 72\right | \geq \frac 32 \] occurs when \[ X\leq 2 \qquad \text {or}\qquad X\geq 5. \]

Thus, \[ P\left ( \left |X-\frac 72\right | \geq \frac 32 \right ) = \frac 46 = \frac 23. \]

Therefore, both inequalities give valid upper bounds, but the bounds may be far from the exact probabilities.

Since the sample consists of independent observations, \[ E(\overline X)=500 \] and \[ \operatorname {Var}(\overline X) = \frac {\sigma ^2}{n} = \frac {100^2}{10} = 1000. \]

Now, \[ \begin {aligned} P(460<\overline X<540) &= P(-40<\overline X-500<40)\\ &= P(|\overline X-500|<40). \end {aligned} \]

Using the complement, \[ P(|\overline X-500|<40) = 1- P(|\overline X-500|\geq 40). \]

By Chebyshev’s inequality, \[ P(|\overline X-500|\geq 40) \leq \frac {1000}{40^2}. \]

Thus, \[ P(|\overline X-500|\geq 40) \leq \frac {1000}{1600} = \frac 58. \]

Therefore, \[ P(460<\overline X<540) \geq 1-\frac 58 = \frac 38. \]

Hence, \[ P(460<\overline X<540) \geq \frac 38. \]

Since \[ X_i\sim \operatorname {Bernoulli}(p), \] we have \[ E(X_i)=p \] and \[ \operatorname {Var}(X_i)=p(1-p). \]

Therefore, \[ E(\widehat p)=p, \] and by independence, \[ \operatorname {Var}(\widehat p) = \frac {p(1-p)}{n}. \]

By Chebyshev’s inequality, \[ P(|\widehat p-p|\geq \varepsilon ) \leq \frac { p(1-p) }{ n\varepsilon ^2 }. \]

Hence, \[ P(|\widehat p-p|<\varepsilon ) \geq 1- \frac { p(1-p) }{ n\varepsilon ^2 }. \]

To guarantee \[ P(|\widehat p-p|<\varepsilon ) \geq 1-\alpha , \] it is sufficient that \[ \frac { p(1-p) }{ n\varepsilon ^2 } \leq \alpha . \]

Equivalently, \[ n \geq \frac { p(1-p) }{ \varepsilon ^2\alpha }. \]

However, \(p\) is unknown. Since \[ p(1-p) \leq \frac 14 \] for every \(0<p<1\), a sufficient condition that does not depend on \(p\) is \[ n \geq \frac { 1 }{ 4\varepsilon ^2\alpha }. \]

Thus, choosing \[ n \geq \frac 1{4\varepsilon ^2\alpha } \] guarantees \[ P(|\widehat p-p|<\varepsilon ) \geq 1-\alpha \] for every Bernoulli success probability \(p\).

From the general Bernoulli estimation bound, \[ n \geq \frac 1{4\varepsilon ^2\alpha } \] is sufficient to guarantee \[ P(|\widehat p-p|<\varepsilon ) \geq 1-\alpha . \]

Here, \[ \varepsilon =0.03 \] and \[ 1-\alpha =0.98, \] so \[ \alpha =0.02. \]

Therefore, \[ \begin {aligned} n &\geq \frac { 1 }{ 4(0.03)^2(0.02) }\\ &= 13888.888\ldots . \end {aligned} \]

Since the sample size must be an integer, \[ n\geq 13889. \]

Thus, at least \[ 13{,}889 \] nails should be inspected.

Let \[ X_i = \begin {cases} 1, & \text {if the \(i\)-th flip is heads},\\ 0, & \text {otherwise}. \end {cases} \]

Then \[ X_i\sim \operatorname {Bernoulli}(p) \] and \[ \widehat p = \frac 1{3000} \sum _{i=1}^{3000}X_i. \]

Therefore, \[ E(\widehat p)=p \] and \[ \operatorname {Var}(\widehat p) = \frac {p(1-p)}{3000}. \]

By Chebyshev’s inequality, \[ P(|\widehat p-p|\geq 0.03) \leq \frac { p(1-p) }{ (0.03)^2(3000) }. \]

Since \[ p(1-p)\leq \frac 14, \] we obtain \[ P(|\widehat p-p|\geq 0.03) \leq \frac { 1 }{ 4(0.03)^2(3000) }. \]

Thus, \[ \begin {aligned} P(|\widehat p-p|<0.03) &\geq 1- \frac { 1 }{ 4(0.03)^2(3000) }\\ &= 1-\frac 1{10.8}\\ &\approx 0.9074. \end {aligned} \]

Therefore, \[ P(|\widehat p-p|<0.03) \geq 0.9074 > 0.90. \]

Hence, \[ P(|\widehat p-p|<0.03) \geq 0.90. \]

Since the exponential function is increasing and \(k>0\), \[ X>t \] implies \[ e^{kX}>e^{kt}. \]

Therefore, \[ P(X>t) = P(e^{kX}>e^{kt}). \]

The random variable \[ e^{kX} \] is nonnegative. Hence, by Markov’s inequality, \[ P(e^{kX}\geq e^{kt}) \leq \frac {E(e^{kX})}{e^{kt}}. \]

Since \[ \{e^{kX}>e^{kt}\} \subseteq \{e^{kX}\geq e^{kt}\}, \] we obtain \[ P(X>t) \leq \frac {E(e^{kX})}{e^{kt}}. \]

The random variable \[ (X-Y)^2 \] is nonnegative.

For every \(\varepsilon >0\), Markov’s inequality gives \[ P\left ((X-Y)^2\geq \varepsilon \right ) \leq \frac {E[(X-Y)^2]}{\varepsilon }. \]

Since \[ E[(X-Y)^2]=0, \] we obtain \[ P\left ((X-Y)^2\geq \varepsilon \right )=0 \] for every \(\varepsilon >0\).

Now \[ \{X\neq Y\} = \{(X-Y)^2>0\}. \]

Also, \[ \{(X-Y)^2>0\} = \bigcup _{n=1}^{\infty } \left \{ (X-Y)^2\geq \frac 1n \right \}. \]

Therefore, \[ \begin {aligned} P(X\neq Y) &\leq \sum _{n=1}^{\infty } P\left ( (X-Y)^2\geq \frac 1n \right )\\ &= 0. \end {aligned} \]

Hence, \[ P(X\neq Y)=0, \] and therefore \[ P(X=Y)=1. \]

As written, the proposed function is not a probability density function, because \[ \begin {aligned} \int _0^1 4x(1-x)\,dx &= 4\int _0^1(x-x^2)\,dx\\ &= 4\left ( \frac 12-\frac 13 \right )\\ &= \frac 23 \neq 1. \end {aligned} \]

Therefore, the problem as stated does not define a valid distribution.

If the intended normalized density is \[ f(x)=6x(1-x), \qquad 0\leq x\leq 1, \] then \[ \begin {aligned} E(X_i) &= \int _0^1 x\,6x(1-x)\,dx\\ &= 6\int _0^1(x^2-x^3)\,dx\\ &= 6\left ( \frac 13-\frac 14 \right )\\ &= \frac 12. \end {aligned} \]

By the Strong Law of Large Numbers, \[ \frac {X_1+\cdots +X_n}{n} \longrightarrow E(X_1) \] with probability \(1\).

Thus, under the normalized density, \[ \frac {X_1+\cdots +X_n}{n} \longrightarrow \frac 12 \qquad \text {almost surely}. \]

Let \[ X_1,\ldots ,X_{15} \] be the completion times and let \[ \overline X = \frac 1{15} \sum _{i=1}^{15}X_i. \]

First calculate the population mean: \[ \begin {aligned} E(X) &= \int _1^2 x\,6(x-1)(2-x)\,dx\\ &= \frac 32. \end {aligned} \]

Also, \[ \begin {aligned} E(X^2) &= \int _1^2 x^2\,6(x-1)(2-x)\,dx\\ &= \frac {23}{10}. \end {aligned} \]

Hence, \[ \begin {aligned} \operatorname {Var}(X) &= E(X^2)-[E(X)]^2\\ &= \frac {23}{10} - \frac 94\\ &= \frac 1{20}. \end {aligned} \]

Thus, \[ \mu =\frac 32, \qquad \sigma ^2=\frac 1{20}. \]

By the Central Limit Theorem, \[ \overline X \approx N\left ( \frac 32, \frac {1}{20(15)} \right ) = N\left ( \frac 32, \frac 1{300} \right ). \]

Now \[ 1\text { hour }25\text { minutes} = 1+\frac {25}{60} = \frac {17}{12} \] hours.

Therefore, \[ \begin {aligned} P\left ( \overline X<\frac {17}{12} \right ) &\approx P\left ( Z< \frac { \frac {17}{12}-\frac 32 }{ 1/\sqrt {300} } \right )\\ &= P(Z<-1.443). \end {aligned} \]

Hence, \[ P\left ( \overline X<\frac {17}{12} \right ) \approx \Phi (-1.443) \approx 0.075. \]

Thus, the approximate probability is \[ 0.075. \]

Let \[ X_1,\ldots ,X_{20} \overset {\text {i.i.d.}}{\sim } \operatorname {Unif}(0,1) \] and define \[ S_{20} = \sum _{i=1}^{20}X_i. \]

For a \(\operatorname {Unif}(0,1)\) random variable, \[ E(X_i)=\frac 12 \] and \[ \operatorname {Var}(X_i)=\frac 1{12}. \]

Therefore, \[ E(S_{20}) = 20\left (\frac 12\right ) = 10, \] and \[ \operatorname {Var}(S_{20}) = 20\left (\frac 1{12}\right ) = \frac 53. \]

By the Central Limit Theorem, \[ S_{20} \approx N\left ( 10,\frac 53 \right ). \]

Hence, \[ \begin {aligned} P(S_{20}\geq 8) &\approx P\left ( Z\geq \frac {8-10}{\sqrt {5/3}} \right )\\ &= P(Z\geq -1.549). \end {aligned} \]

Therefore, \[ \begin {aligned} P(S_{20}\geq 8) &\approx 1-\Phi (-1.549)\\ &= \Phi (1.549)\\ &\approx 0.939. \end {aligned} \]

Thus, the approximate probability is \[ 0.939. \]

For a sample of size \(n\), \(E(\overline X)=\ell \) and \(\operatorname {Var}(\overline X) = \frac {1.5}{n}.\) By the Central Limit Theorem, \[ \frac { \overline X-\ell }{ \sqrt {1.5/n} } \approx N(0,1). \]

Thus, \[ \begin {aligned} P(|\overline X-\ell |<0.2) &\approx P\left ( \left | Z \right | < \frac {0.2\sqrt n}{\sqrt {1.5}} \right ). \end {aligned} \]

We want this probability to be approximately \(0.98\). Therefore, \[ P(-z<Z<z)=0.98. \]

This requires \[ \Phi (z)=0.99. \]

Using \[ z_{0.99}\approx 2.33, \] we set \[ \frac {0.2\sqrt n}{\sqrt {1.5}} \approx 2.33. \]

Hence, \[ \sqrt n \approx \frac { 2.33\sqrt {1.5} }{0.2}. \]

Squaring, \[ n \approx \left ( \frac { 2.33\sqrt {1.5} }{0.2} \right )^2 \approx 203.58. \]

Therefore, the sample size should be rounded up to \[ n=204. \]

First calculate the population mean: \[ \begin {aligned} \mu &= E(X)\\ &= \int _1^3 x \frac 19 \left (x+\frac 52\right ) \,dx\\ &= \frac {56}{27}. \end {aligned} \]

Next, \[ \begin {aligned} E(X^2) &= \int _1^3 x^2 \frac 19 \left (x+\frac 52\right ) \,dx\\ &= \frac {125}{27}. \end {aligned} \]

Therefore, \[ \begin {aligned} \sigma ^2 &= E(X^2)-[E(X)]^2\\ &= \frac {125}{27} - \left (\frac {56}{27}\right )^2\\ &= \frac {239}{729}. \end {aligned} \]

For \(n=24\), \[ E(\overline X) = \frac {56}{27}, \] and \[ \operatorname {Var}(\overline X) = \frac {239}{729(24)}. \]

Thus, \[ \operatorname {SD}(\overline X) = \sqrt { \frac {239}{729(24)} }. \]

By the Central Limit Theorem, \[ \overline X \approx N\left ( \frac {56}{27}, \frac {239}{729(24)} \right ). \]

Therefore, \[ \begin {aligned} P(2<\overline X<2.15) &\approx P\left ( \frac { 2-\frac {56}{27} }{ \sqrt {239/[729(24)]} } < Z < \frac { 2.15-\frac {56}{27} }{ \sqrt {239/[729(24)]} } \right )\\ &= P(-0.634<Z<0.650). \end {aligned} \]

Hence, \[ \begin {aligned} P(2<\overline X<2.15) &\approx \Phi (0.650)-\Phi (-0.634)\\ &\approx 0.479. \end {aligned} \]

Thus, the approximate probability is \[ 0.479. \]

The density satisfies \[ f(x)=f(-x), \] so the distribution of each \(X_i\) is symmetric about \(0\).

Since the \(X_i\)’s are independent, their sum \[ S_n=X_1+\cdots +X_n \] is also symmetric about \(0\).

Because the distribution is continuous, \[ P(S_n=0)=0. \]

By symmetry, \[ P(S_n>0) = P(S_n<0). \]

These two probabilities sum to \(1\), so \[ P(S_n>0)=\frac 12. \]

Since \[ \overline X>0 \] if and only if \[ S_n>0, \] we obtain \[ P(\overline X>0)=\frac 12. \]

For a sample of size \[ n=35, \] the sample mean satisfies \[ E(\overline X)=500 \] and \[ \operatorname {SD}(\overline X) = \frac {100}{\sqrt {35}}. \]

By the Central Limit Theorem, \[ \overline X \approx N\left ( 500, \frac {100^2}{35} \right ). \]

Therefore, \[ \begin {aligned} P(460<\overline X<540) &\approx P\left ( \frac {460-500}{100/\sqrt {35}} < Z < \frac {540-500}{100/\sqrt {35}} \right )\\ &= P(-2.366<Z<2.366). \end {aligned} \]

Hence, \[ \begin {aligned} P(460<\overline X<540) &\approx \Phi (2.366)-\Phi (-2.366)\\ &= 2\Phi (2.366)-1\\ &\approx 0.982. \end {aligned} \]

Thus, the approximate probability is \[ 0.982. \]

Since \[ 0<X<1, \] we have \[ -\ln X>0. \] Therefore, \[ Y>0. \]

(a)

For \(y\leq 0\), \[ F_Y(y)=0. \]

Now suppose \(y>0\). Then \[ \begin {aligned} F_Y(y) &= P(Y\leq y)\\ &= P(-\ln X\leq y). \end {aligned} \]

Since \[ -\ln X\leq y \] is equivalent to \[ \ln X\geq -y, \] and hence \[ X\geq e^{-y}, \] we obtain \[ \begin {aligned} F_Y(y) &= P(X\geq e^{-y})\\ &= 1-P(X<e^{-y}). \end {aligned} \]

Because \[ X\sim \operatorname {Unif}(0,1), \] we have \[ P(X<e^{-y})=e^{-y}. \]

Therefore, \[ F_Y(y) = 1-e^{-y}, \qquad y>0. \]

Thus, \[ F_Y(y) = \begin {cases} 0, & y\leq 0,\\[1mm] 1-e^{-y}, & y>0. \end {cases} \]

(b)

Differentiating the CDF, \[ f_Y(y) = F_Y'(y). \]

Hence, \[ f_Y(y) = \begin {cases} e^{-y}, & y>0,\\ 0, & \text {otherwise}. \end {cases} \]

(c)

This is the density of an exponential random variable with parameter \(1\). Therefore, \[ Y\sim \operatorname {Exp}(1). \]

(a)

The transformation \[ g(x)=x^2 \] is not one-to-one because, for every \(x\neq 0\), \[ g(x)=g(-x). \]

Thus, a given positive value of \(Y\) has two possible preimages: \[ x=\sqrt y \qquad \text {and}\qquad x=-\sqrt y. \]

(b)

Since \[ Y=X^2, \] we have \[ Y\geq 0. \]

Therefore, \[ F_Y(y)=0, \qquad y<0. \]

Now suppose \(y\geq 0\). Then \[ \begin {aligned} F_Y(y) &= P(Y\leq y)\\ &= P(X^2\leq y). \end {aligned} \]

The event \[ X^2\leq y \] is equivalent to \[ -\sqrt y\leq X\leq \sqrt y. \]

Hence, \[ \begin {aligned} F_Y(y) &= P(-\sqrt y\leq X\leq \sqrt y)\\ &= \Phi (\sqrt y)-\Phi (-\sqrt y). \end {aligned} \]

Using \[ \Phi (-a)=1-\Phi (a), \] we obtain \[ F_Y(y) = 2\Phi (\sqrt y)-1. \]

Thus, \[ F_Y(y) = \begin {cases} 0, & y<0,\\[1mm] 2\Phi (\sqrt y)-1, & y\geq 0. \end {cases} \]

(c)

For \(y>0\), \[ f_Y(y) = \frac {d}{dy} \left [ 2\Phi (\sqrt y)-1 \right ]. \]

Using the chain rule, \[ \begin {aligned} f_Y(y) &= 2\phi (\sqrt y) \frac {1}{2\sqrt y}\\ &= \frac {\phi (\sqrt y)}{\sqrt y}. \end {aligned} \]

Since \[ \phi (x) = \frac 1{\sqrt {2\pi }} e^{-x^2/2}, \] we have \[ \phi (\sqrt y) = \frac 1{\sqrt {2\pi }} e^{-y/2}. \]

Therefore, \[ f_Y(y) = \frac {1}{\sqrt {2\pi y}} e^{-y/2}, \qquad y>0. \]

Thus, \[ f_Y(y) = \begin {cases} \dfrac {1}{\sqrt {2\pi y}}e^{-y/2}, & y>0,\\[2mm] 0, & \text {otherwise}. \end {cases} \]

Since \[ X,Y\sim \operatorname {Exp}(\lambda ), \] their common density is \[ f_X(x)=f_Y(x) = \begin {cases} \lambda e^{-\lambda x}, & x>0,\\ 0, & \text {otherwise}. \end {cases} \]

Since \(X>0\) and \(Y>0\), \[ T=X+Y>0. \]

Therefore, \[ f_T(t)=0, \qquad t\leq 0. \]

Now suppose \(t>0\). The convolution formula gives \[ f_T(t) = \int _{-\infty }^{\infty } f_X(x)f_Y(t-x)\,dx. \]

For the integrand to be nonzero, we need both \[ x>0 \] and \[ t-x>0. \]

Thus, \[ 0<x<t. \]

Therefore, \[ \begin {aligned} f_T(t) &= \int _0^t \lambda e^{-\lambda x} \lambda e^{-\lambda (t-x)} \,dx\\ &= \lambda ^2 e^{-\lambda t} \int _0^t dx\\ &= \lambda ^2 t e^{-\lambda t}. \end {aligned} \]

Hence, \[ f_T(t) = \begin {cases} \lambda ^2 t e^{-\lambda t}, & t>0,\\ 0, & \text {otherwise}. \end {cases} \]

Since \[ X,Y\sim \operatorname {Unif}(0,1), \] we have \[ f_X(x)=f_Y(x) = \begin {cases} 1, & 0<x<1,\\ 0, & \text {otherwise}. \end {cases} \]

Also, \[ 0<X<1 \qquad \text {and}\qquad 0<Y<1, \] so \[ 0<T<2. \]

Therefore, \[ f_T(t)=0 \] outside the interval \((0,2)\).

For \(0<t\leq 1\), the convolution formula gives \[ f_T(t) = \int _{-\infty }^{\infty } f_X(x)f_Y(t-x)\,dx. \]

For the integrand to be nonzero, we need \[ 0<x<1 \] and \[ 0<t-x<1. \]

When \(0<t\leq 1\), these conditions reduce to \[ 0<x<t. \]

Therefore, \[ f_T(t) = \int _0^t1\,dx = t. \]

Now suppose \[ 1<t<2. \]

Again, we need \[ 0<x<1 \] and \[ 0<t-x<1. \]

The second inequality gives \[ t-1<x<t. \]

Combining this with \(0<x<1\), we obtain \[ t-1<x<1. \]

Hence, \[ f_T(t) = \int _{t-1}^{1}1\,dx = 2-t. \]

Therefore, \[ f_T(t) = \begin {cases} t, & 0<t\leq 1,\\[1mm] 2-t, & 1<t<2,\\[1mm] 0, & \text {otherwise}. \end {cases} \]

The four possible values of \((X,Y)\) are \[ (1,1),\quad (1,-1),\quad (-1,1),\quad (-1,-1), \] each with probability \(1/4\).

Since \[ Z=XY, \] we obtain the table \[ \begin {array}{c|c|c} X & Y & Z\\ \hline 1 & 1 & 1\\ 1 & -1 & -1\\ -1 & 1 & -1\\ -1 & -1 & 1 \end {array} \]

(a)

We have \[ P(Z=1) = P(X=Y) = \frac 12, \] and \[ P(Z=-1) = P(X\neq Y) = \frac 12. \]

Thus, \[ p_Z(z) = \begin {cases} \dfrac 12, & z=1,-1,\\ 0, & \text {otherwise}. \end {cases} \]

(b)

To show that \(X\) and \(Z\) are independent, consider their joint probabilities.

For example, \[ P(X=1,Z=1) = P(X=1,Y=1) = \frac 14. \]

Also, \[ P(X=1)P(Z=1) = \frac 12\cdot \frac 12 = \frac 14. \]

Similarly, \[ P(X=1,Z=-1) = P(X=1,Y=-1) = \frac 14, \] and \[ P(X=1)P(Z=-1) = \frac 14. \]

The same calculation holds for \(X=-1\). Therefore, \[ P(X=x,Z=z) = P(X=x)P(Z=z) \] for every \[ x,z\in \{-1,1\}. \]

Hence, \(X\) and \(Z\) are independent.

(c)

By symmetry, the same argument shows that \[ Y \] and \[ Z \] are independent.

Explicitly, \[ P(Y=y,Z=z) = P(Y=y)P(Z=z) \] for all \[ y,z\in \{-1,1\}. \]

(d)

We are given that \(X\) and \(Y\) are independent. We have also shown that \(X\) and \(Z\) are independent and that \(Y\) and \(Z\) are independent.

Therefore, \(X,Y,Z\) are pairwise independent.

(e)

However, \[ XYZ = XY(XY) = X^2Y^2 = 1 \] with probability \(1\).

Thus, the three random variables satisfy a deterministic relationship.

For example, \[ P(X=1,Y=1,Z=-1)=0. \]

But \[ P(X=1)P(Y=1)P(Z=-1) = \frac 12\cdot \frac 12\cdot \frac 12 = \frac 18. \]

Therefore, \[ P(X=1,Y=1,Z=-1) \neq P(X=1)P(Y=1)P(Z=-1). \]

Hence, \(X,Y,Z\) are not mutually independent.

The support is the triangular region \[ 0<y<x<1. \]

(a)

We calculate \[ F_{X,Y}(x,y) = P(X\leq x,Y\leq y). \]

If \[ x\leq 0 \qquad \text {or}\qquad y\leq 0, \] then \[ F_{X,Y}(x,y)=0. \]

Now consider \[ 0<x<1, \qquad 0<y<x. \]

For a fixed value \(v\) of \(Y\), the support requires \[ v<u<x, \] where \(u\) denotes the value of \(X\).

Thus, \[ \begin {aligned} F_{X,Y}(x,y) &= \int _0^y \int _v^x 2\,du\,dv\\ &= 2\int _0^y(x-v)\,dv\\ &= 2xy-y^2. \end {aligned} \]

Next consider \[ 0<x<1, \qquad y\geq x. \]

In this case, the condition \(Y\leq y\) does not further restrict the support below \(X=x\). Therefore, \[ \begin {aligned} F_{X,Y}(x,y) &= P(X\leq x)\\ &= \int _0^x \int _0^u 2\,dv\,du\\ &= \int _0^x2u\,du\\ &= x^2. \end {aligned} \]

Now suppose \[ x\geq 1. \]

If \[ 0<y<1, \] then the condition \(X\leq x\) imposes no restriction, so \[ \begin {aligned} F_{X,Y}(x,y) &= P(Y\leq y)\\ &= \int _0^y \int _v^1 2\,du\,dv\\ &= 2\int _0^y(1-v)\,dv\\ &= 2y-y^2. \end {aligned} \]

Finally, if \[ x\geq 1 \qquad \text {and}\qquad y\geq 1, \] then \[ F_{X,Y}(x,y)=1. \]

Thus, \[ F_{X,Y}(x,y) = \begin {cases} 0, & x\leq 0\text { or }y\leq 0,\\[1mm] 2xy-y^2, & 0<y<x<1,\\[1mm] x^2, & 0<x<1,\ y\geq x,\\[1mm] 2y-y^2, & x\geq 1,\ 0<y<1,\\[1mm] 1, & x\geq 1,\ y\geq 1. \end {cases} \]

(b)

The marginal CDF of \(X\) is \[ F_X(x) = \lim _{y\to \infty } F_{X,Y}(x,y). \]

Hence, \[ F_X(x) = \begin {cases} 0, & x\leq 0,\\[1mm] x^2, & 0<x<1,\\[1mm] 1, & x\geq 1. \end {cases} \]

Similarly, \[ F_Y(y) = \lim _{x\to \infty } F_{X,Y}(x,y). \]

Thus, \[ F_Y(y) = \begin {cases} 0, & y\leq 0,\\[1mm] 2y-y^2, & 0<y<1,\\[1mm] 1, & y\geq 1. \end {cases} \]

(c)

For a joint CDF, \[ \begin {aligned} &P(a<X\leq b,\ c<Y\leq d)\\ &= F_{X,Y}(b,d) - F_{X,Y}(a,d) - F_{X,Y}(b,c) + F_{X,Y}(a,c). \end {aligned} \]

Here, \[ a=\frac 14, \qquad b=\frac 34, \qquad c=\frac 14, \qquad d=\frac 12. \]

First, \[ F_{X,Y}\left (\frac 34,\frac 12\right ) = 2\left (\frac 34\right )\left (\frac 12\right ) - \left (\frac 12\right )^2 = \frac 12. \]

Next, since \[ \frac 12\geq \frac 14, \] we use the \(y\geq x\) case: \[ F_{X,Y}\left (\frac 14,\frac 12\right ) = \left (\frac 14\right )^2 = \frac 1{16}. \]

Also, \[ F_{X,Y}\left (\frac 34,\frac 14\right ) = 2\left (\frac 34\right )\left (\frac 14\right ) - \left (\frac 14\right )^2 = \frac 5{16}. \]

Finally, \[ F_{X,Y}\left (\frac 14,\frac 14\right ) = \left (\frac 14\right )^2 = \frac 1{16}. \]

Therefore, \[ \begin {aligned} &P\left ( \frac 14<X\leq \frac 34,\, \frac 14<Y\leq \frac 12 \right )\\ &= \frac 12 - \frac 1{16} - \frac 5{16} + \frac 1{16}\\ &= \frac 3{16}. \end {aligned} \]

Thus, \[ P\left ( \frac 14<X\leq \frac 34,\, \frac 14<Y\leq \frac 12 \right ) = \frac 3{16}. \]