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

First, \begin {align*} E(X) &= \int _0^\infty x f_X(x)\,dx\\ &= \int _0^\infty x^2e^{-x}\,dx\\ &=2. \end {align*}

Next, \begin {align*} E(Y) &= \int _0^\infty y\frac {2}{(1+y)^3}\,dy\\ &=1. \end {align*}

To calculate \(E(XY)\), use the joint density: \begin {align*} E(XY) &= \int _0^\infty \int _0^\infty xy\,x^2e^{-x(y+1)}\,dy\,dx\\ &= \int _0^\infty x^3e^{-x} \left ( \int _0^\infty y e^{-xy}\,dy \right )dx. \end {align*}

For \(x>0\), \[ \int _0^\infty y e^{-xy}\,dy = \frac {1}{x^2}. \]

Therefore, \begin {align*} E(XY) &= \int _0^\infty xe^{-x}\,dx\\ &=1. \end {align*}

Hence \[ \operatorname {Cov}(X,Y) = 1-(2)(1) = -1. \]

(b) Using \[ \operatorname {Cov}(aX+b,cX+d) = ac\,\operatorname {Var}(X), \] we obtain \[ \operatorname {Cov}(3X+1,2X-4) = 6\operatorname {Var}(X). \]

Now \[ E(X)=2, \] and \begin {align*} E(X^2) &= \int _0^\infty x^2 f_X(x)\,dx\\ &= \int _0^\infty x^3e^{-x}\,dx\\ &=6. \end {align*}

Thus \[ \operatorname {Var}(X) = E(X^2)-[E(X)]^2 = 6-4 = 2. \]

Therefore, \[ \operatorname {Cov}(3X+1,2X-4) = 6(2) = 12. \]

(c) If \(X\) and \(Y\) were independent, then we would have \[ f_{X,Y}(x,y)=f_X(x)f_Y(y) \] on their support.

However, \[ f_X(x)f_Y(y) = xe^{-x}\frac {2}{(1+y)^3}, \] whereas \[ f_{X,Y}(x,y) = x^2e^{-x(y+1)}. \]

These are not equal in general. Therefore, \(X\) and \(Y\) are not independent.

Alternatively, part (a) already gives \[ \operatorname {Cov}(X,Y)=-1\neq 0. \] Since independence would imply zero covariance, \(X\) and \(Y\) cannot be independent.

(d) By definition, \begin {align*} M_X(t) &= E(e^{tX})\\ &= \int _0^\infty e^{tx}xe^{-x}\,dx\\ &= \int _0^\infty xe^{-(1-t)x}\,dx. \end {align*}

For \(t<1\), \[ \int _0^\infty xe^{-(1-t)x}\,dx = \frac {1}{(1-t)^2}. \]

Therefore, \[ M_X(t) = \frac {1}{(1-t)^2}, \qquad t<1. \]

(e) We know that \[ E(X)=M_X'(0). \]

Since \[ M_X(t)=(1-t)^{-2}, \] we have \[ M_X'(t)=2(1-t)^{-3}. \]

Therefore, \[ E(X) = M_X'(0) = 2. \]

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

We first find the inverse transformation.

Since \[ v=x \] and \[ u=\frac {x}{y}, \] we obtain \[ x=v, \qquad y=\frac {v}{u}. \]

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

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

Since \[ x=v, \qquad y=\frac vu, \] we have \[ \frac {\partial (x,y)}{\partial (u,v)} = \begin {vmatrix} 0 & 1\\[2mm] -\dfrac {v}{u^2} & \dfrac 1u \end {vmatrix} = \frac {v}{u^2}. \]

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

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

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

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

For \(u\leq 0\), \[ f_U(u)=0. \]

(a) The function \[ g(x)=\log x \] is concave on \((0,\infty )\).

By Jensen’s inequality for a concave function, \[ E[g(X)]\leq g(E[X]). \]

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

(b) Since \(X\) and \(Y\) are i.i.d., the joint distribution of \((X,Y)\) is symmetric under interchange of \(X\) and \(Y\).

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

Also, \[ P(X\leq Y) = P(X<Y)+P(X=Y) \] and \[ P(X\geq Y) = P(X>Y)+P(X=Y). \]

Since \[ P(X<Y)=P(X>Y), \] it follows that \[ P(X\leq Y)=P(X\geq Y). \]

(c) By the Cauchy–Schwarz inequality, \[ |E(XY)| \leq \sqrt {E(X^2)E(Y^2)}. \]

Since \(X\) and \(Y\) are positive, \[ E(XY)\geq 0. \]

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

In fact, because \(X\) and \(Y\) are independent, \[ E(XY)=E(X)E(Y), \] but Cauchy–Schwarz gives the requested comparison directly.

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

Therefore, \[ E(\widehat p) = E\left ( \frac 1n\sum _{i=1}^nX_i \right ) = p. \]

Since the \(X_i\)’s are independent, \begin {align*} \operatorname {Var}(\widehat p) &= \operatorname {Var} \left ( \frac 1n\sum _{i=1}^nX_i \right )\\ &= \frac 1{n^2} \sum _{i=1}^n \operatorname {Var}(X_i)\\ &= \frac {p(1-p)}{n}. \end {align*}

Chebyshev’s inequality gives \[ P\left ( |\widehat p-p|\geq \varepsilon \right ) \leq \frac {\operatorname {Var}(\widehat p)} {\varepsilon ^2} = \frac {p(1-p)}{n\varepsilon ^2}. \]

We want a bound that holds for every \(0<p<1\).

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

Therefore, \[ P\left ( |\widehat p-p|<\varepsilon \right ) \geq 1-\frac {1}{4n\varepsilon ^2}. \]

To guarantee that this is at least \(1-\alpha \), it is sufficient that \[ \frac {1}{4n\varepsilon ^2} \leq \alpha . \]

Solving for \(n\), \[ n \geq \frac {1}{4\alpha \varepsilon ^2}. \]

Since \(n\) must be an integer, one may take \[ n \geq \left \lceil \frac {1}{4\alpha \varepsilon ^2} \right \rceil . \]

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

Thus, for \(t\leq 0\), \[ f_T(t)=0. \]

Now suppose \(t>0\).

For the integrand \[ f_X(x)f_Y(t-x) \] to be nonzero, we need \[ x>0 \] and \[ t-x>0. \]

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

Using convolution, \begin {align*} f_T(t) &= \int _0^t f_X(x)f_Y(t-x)\,dx\\ &= \int _0^t \lambda e^{-\lambda x} \lambda e^{-\lambda (t-x)} \,dx\\ &= \lambda ^2e^{-\lambda t} \int _0^t dx\\ &= \lambda ^2t e^{-\lambda t}. \end {align*}

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

Thus \(T\) does not have an exponential distribution. It has a Gamma distribution with shape parameter \(2\) and rate parameter \(\lambda \).

(a) To determine pairwise independence, we examine the joint density of each pair.

First, consider \(X\) and \(Y\). Integrating out \(Z\), \begin {align*} f_{X,Y}(x,y) &= \int _0^\infty x^2e^{-x(1+y+z)}\,dz\\ &= x^2e^{-x(1+y)} \int _0^\infty e^{-xz}\,dz\\ &= x^2e^{-x(1+y)} \frac 1x\\ &= xe^{-x(1+y)}, \qquad x,y>0. \end {align*}

On the other hand, \[ f_X(x)f_Y(y) = e^{-x}\frac {1}{(1+y)^2}. \]

These are not equal in general: \[ xe^{-x(1+y)} \neq \frac {e^{-x}}{(1+y)^2}. \]

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

Consequently, \(X,Y,Z\) cannot be pairwise independent.

For completeness, the same phenomenon occurs for the other pairs.

By symmetry, \[ f_{X,Z}(x,z) = xe^{-x(1+z)}, \] which is not equal to \[ f_X(x)f_Z(z) = \frac {e^{-x}}{(1+z)^2}. \]

For \(Y\) and \(Z\), \begin {align*} f_{Y,Z}(y,z) &= \int _0^\infty x^2e^{-x(1+y+z)}\,dx. \end {align*}

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

However, \[ f_Y(y)f_Z(z) = \frac {1}{(1+y)^2(1+z)^2}, \] which is again different in general.

Hence none of the three pairs is independent.

(b) Mutual independence would require \[ f_{X,Y,Z}(x,y,z) = f_X(x)f_Y(y)f_Z(z) \] for all points in the support.

The product of the marginals is \[ f_X(x)f_Y(y)f_Z(z) = \frac {e^{-x}} {(1+y)^2(1+z)^2}. \]

But the joint density is \[ f_{X,Y,Z}(x,y,z) = x^2e^{-x(1+y+z)}. \]

These expressions are not equal in general.

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

Also, since mutual independence implies pairwise independence and part (a) shows that they are not pairwise independent, mutual independence is impossible.

(a) Weak Law of Large Numbers.

The WLLN states that the sample mean converges to the population mean \(\mu \) in probability. That is, for every \(\varepsilon >0\), \[ \lim _{n\to \infty } P\left ( |\overline X_n-\mu |>\varepsilon \right ) = 0. \]

Therefore, the correct answer is

\[ \textbf {B}. \]

(b) Strong Law of Large Numbers.

The SLLN states that \[ \overline X_n\longrightarrow \mu \] almost surely. Equivalently, \[ P\left ( \lim _{n\to \infty }\overline X_n=\mu \right ) = 1. \]

Therefore, the correct answer is

\[ \textbf {A}. \]

(c) Central Limit Theorem.

The CLT states that the standardized sample mean converges in distribution to a standard normal random variable: \[ \frac {\overline X_n-\mu }{\sigma /\sqrt n} \xrightarrow []{d} N(0,1). \]

Equivalently, for every \(z\in \mathbb R\), \[ \lim _{n\to \infty } P\left ( \frac {\overline X_n-\mu }{\sigma /\sqrt n} \leq z \right ) = \Phi (z). \]

Therefore, the correct answer is

\[ \textbf {C}. \]