Let \(\phi (z)=(2\pi )^{-1/2}e^{-z^2/2}\) denote the standard normal density.

(a) Put \(Y=e^{-X}\). Since \(X>0\), we have \(0<Y<1\). The inverse transformation is \(x=-\log y\), and \[ \left |\frac {dx}{dy}\right |=\frac 1y. \] Thus \[ f_Y(y)=e^{-(-\log y)}\frac 1y=1, \qquad 0<y<1. \]

(b) Put \(Y=X^7\). The inverse is \(x=y^{1/7}\) for \(y>0\), and \[ \left |\frac {dx}{dy}\right |=\frac {1}{7}y^{-6/7}. \] Therefore \[ f_Y(y)=\frac {\lambda }{7}y^{-6/7}e^{-\lambda y^{1/7}}, \qquad y>0. \]

(c) Put \(Y=Z^3\). This transformation is one-to-one. For \(y\ne 0\), the inverse is \(z=y^{1/3}\) and \[ \left |\frac {dz}{dy}\right |=\frac {1}{3|y|^{2/3}}. \] Hence \[ f_Y(y)=\frac {\phi (y^{1/3})}{3|y|^{2/3}}, \qquad y\ne 0. \] The value assigned to the density at the single point \(y=0\) is immaterial.

(d) Put \(Y=Z^4\). The support is \(y>0\). For each \(y>0\), the two preimages are \(y^{1/4}\) and \(-y^{1/4}\). Adding the two contributions gives \[ f_Y(y) = \phi (y^{1/4})\frac {1}{4y^{3/4}} + \phi (-y^{1/4})\frac {1}{4y^{3/4}} = \frac {\phi (y^{1/4})}{2y^{3/4}}, \qquad y>0. \]

(e) Put \(Y=|Z|\). For \(y>0\), the two preimages are \(y\) and \(-y\), and therefore \[ f_Y(y)=\phi (y)+\phi (-y)=2\phi (y), \qquad y>0. \]

For real \(t\), \[ F_T(t)=P\!\left (\log (X/Y)\le t\right )=P(X\le e^tY). \] Conditioning on \(Y=y\) gives \[ F_T(t)=\int _0^\infty P(X\le e^t y)\lambda e^{-\lambda y}\dd y. \] Since \(X\sim \operatorname {Expo}(\lambda )\), \[ P(X\le e^t y)=1-e^{-\lambda e^t y}. \] Thus \[ F_T(t)=\int _0^\infty \left (1-e^{-\lambda e^t y}\right )\lambda e^{-\lambda y}\dd y =1-\frac {1}{1+e^t} =\frac {e^t}{1+e^t}. \] Differentiating the CDF gives \[ f_T(t)=\frac {e^t}{(1+e^t)^2}, \qquad -\infty <t<\infty . \] The parameter \(\lambda \) disappears because only the ratio \(X/Y\) matters.

(a) Let \[ D=ad-bc. \] Solving the linear system gives \[ x=\frac {dt-bw}{D}, \qquad y=\frac {-ct+aw}{D}. \] The Jacobian determinant of the inverse transformation is \[ \left |\frac {\partial (x,y)}{\partial (t,w)}\right |=\frac {1}{|D|}. \] Therefore \[ f_{T,W}(t,w) = \frac {1}{|ad-bc|} f_{X,Y}\left (\frac {dt-bw}{ad-bc},\frac {-ct+aw}{ad-bc}\right ), \] for all \((t,w)\) whose inverse point lies in the support of \((X,Y)\).

(b) Here \(a=1\), \(b=1\), \(c=1\), and \(d=-1\), so \[ ad-bc=-2. \] Also \[ x=\frac {t+w}{2}, \qquad y=\frac {t-w}{2}. \] Thus \[ f_{T,W}(t,w) = \frac 12 f_{X,Y}\left (\frac {t+w}{2},\frac {t-w}{2}\right ). \]

(a) For \(s>0\), the convolution density of \(S=X+Y\) is \[ f_S(s)=\int _0^s \frac {\lambda ^a}{\Gamma (a)}x^{a-1}e^{-\lambda x} \frac {\lambda ^b}{\Gamma (b)}(s-x)^{b-1}e^{-\lambda (s-x)}\dd x. \] This becomes \[ f_S(s)= \frac {\lambda ^{a+b}e^{-\lambda s}}{\Gamma (a)\Gamma (b)} \int _0^s x^{a-1}(s-x)^{b-1}\dd x. \] Using the beta integral, \[ \int _0^s x^{a-1}(s-x)^{b-1}\dd x = s^{a+b-1}\frac {\Gamma (a)\Gamma (b)}{\Gamma (a+b)}. \] Thus \[ f_S(s)=\frac {\lambda ^{a+b}}{\Gamma (a+b)}s^{a+b-1}e^{-\lambda s}, \qquad s>0, \] which is the \(\operatorname {Gamma}(a+b,\lambda )\) density.

(b) The MGF of \(\operatorname {Gamma}(a,\lambda )\), using the rate parameter \(\lambda \), is \[ M_X(t)=\left (\frac {\lambda }{\lambda -t}\right )^a, \qquad t<\lambda . \] By independence, \[ M_{X+Y}(t)=M_X(t)M_Y(t) = \left (\frac {\lambda }{\lambda -t}\right )^a \left (\frac {\lambda }{\lambda -t}\right )^b = \left (\frac {\lambda }{\lambda -t}\right )^{a+b}. \] This is the MGF of \(\operatorname {Gamma}(a+b,\lambda )\).

(c) When \(a\) and \(b\) are positive integers, \(\operatorname {Gamma}(a,\lambda )\) is the waiting time for the \(a\)th arrival in a Poisson process of rate \(\lambda \). Likewise, \(\operatorname {Gamma}(b,\lambda )\) is the additional waiting time for the next \(b\) arrivals. By the independent increments property of the Poisson process, the total time to wait for \(a+b\) arrivals is the sum of these two independent waiting times. Hence the total waiting time is \(\operatorname {Gamma}(a+b,\lambda )\).

The sample mean has \[ E(\bar X_n)=\mu , \qquad \Var (\bar X_n)=\frac {\sigma ^2}{n}. \] Chebyshev’s inequality gives \[ P(|\bar X_n-\mu |\ge 2\sigma ) \le \frac {\Var (\bar X_n)}{(2\sigma )^2} = \frac {1}{4n}. \] To make the probability of being outside the interval at most \(0.01\), it is enough to require \[ \frac {1}{4n}\le 0.01. \] Thus \(n\ge 25\) is sufficient. With \(n=25\), Chebyshev gives \[ P(|\bar X_n-\mu |<2\sigma )\ge 0.99. \]

Let \(X\) take the values \(a_1,a_2,\dots ,a_n\) with equal probabilities: \[ P(X=a_j)=\frac 1n, \qquad j=1,\dots ,n. \] Then \[ E(X)=\frac {a_1+\cdots +a_n}{n} \] and \[ E(\log X)=\frac 1n\sum _{j=1}^n \log a_j = \log \left ((a_1a_2\cdots a_n)^{1/n}\right ). \] Since \(\log x\) is concave, Jensen’s inequality gives \[ E(\log X)\le \log (E X). \] Therefore \[ \log \left ((a_1a_2\cdots a_n)^{1/n}\right ) \le \log \left (\frac {a_1+\cdots +a_n}{n}\right ). \] Because the logarithm is increasing, exponentiating both sides gives \[ (a_1a_2\cdots a_n)^{1/n} \le \frac {a_1+a_2+\cdots +a_n}{n}. \]

The answers are as follows. \[ \begin {array}{c|c|l} \text {Part} & \text {Symbol} & \text {Reason}\\ \hline (a) & \le & \log x\text { is concave, by Jensen.}\\ (b) & \le & (E X)^2\le E(X^2).\\ (c) & = & \sin ^2 X+\cos ^2 X=1\text { pointwise.}\\ (d) & \le & (E|X|)^2\le E(X^2).\\ (e) & \le & \text {Markov applied to }X^3.\\ (f) & = & \text {Symmetry of i.i.d. }X,Y.\\ (g) & \le & \text {Cauchy--Schwarz.}\\ (h) & \le & X+Y>10\text { implies }X>5\text { or }Y>5.\\ (i) & \le & \min (X,Y)\le X\text { and }\min (X,Y)\le Y.\\ (j) & ? & \text {No fixed relation holds in general.}\\ (k) & \ge & E(X^4)\ge (E X^2)^2.\\ (l) & = & \text {Symmetry, and the two fractions exchange when }X,Y\text { are swapped.} \end {array} \]

For part (k), independence gives \[ E\!\left (X^2(Y^2+1)\right )=(E X^2)^2+E(X^2), \] while the left-hand side is \(E(X^4)+E(X^2)\), so the comparison reduces to \(E(X^4)\ge (E X^2)^2\).

By the Strong Law of Large Numbers, \[ \frac {X_1+\cdots +X_n}{n}\to 2 \qquad \text {with probability }1, \] and \[ \frac {Y_1+\cdots +Y_n}{n}\to 3 \qquad \text {with probability }1. \] On the event where both convergences occur, which has probability \(1\), we may divide the two limits since the denominator tends to the nonzero constant \(3\). Hence \[ \frac {X_1+\cdots +X_n}{Y_1+\cdots +Y_n} = \frac {(X_1+\cdots +X_n)/n}{(Y_1+\cdots +Y_n)/n} \to \frac {2}{3}. \] The independence between the \(X_i\)’s and the \(Y_j\)’s is not needed. What is needed is that each sequence separately satisfies the Strong Law. The two probability-one events can be intersected, and the intersection still has probability \(1\).

(a) For one \(\operatorname {Unif}(0,1)\) random variable, \[ E(U_j)=\frac 12, \qquad \Var (U_j)=\frac 1{12}. \] Therefore \[ E(X)=60\cdot \frac 12=30, \qquad \Var (X)=60\cdot \frac 1{12}=5. \] By the Central Limit Theorem, the distribution of \(X\) is very close to \[ N(30,5). \]

(b) Using the normal approximation, \[ P(X>17) \approx P\left (N(30,5)>17\right ) = P\left (Z>\frac {17-30}{\sqrt 5}\right ). \] The standardized value is \[ \frac {17-30}{\sqrt 5}\approx -5.81. \] Thus \[ P(X>17)\approx P(Z>-5.81)=\Phi (5.81), \] which is essentially \(1\).

(a) Since \(X\sim \operatorname {Expo}(3)\), its MGF is \[ M_X(t)=\frac {3}{3-t}, \qquad t<3. \] Because \(Y=e^X\), \[ E(Y)=E(e^X)=M_X(1)=\frac 32. \] Also \[ E(Y^2)=E(e^{2X})=M_X(2)=3. \] Therefore \[ \Var (Y)=E(Y^2)-[E(Y)]^2 =3-\frac 94 =\frac 34. \]

(b) By the Central Limit Theorem, when \(n\) is large, \[ \bar Y_n\approx N\left (E(Y),\frac {\Var (Y)}{n}\right ) = N\left (\frac 32,\frac {3}{4n}\right ). \]

(a) We compute \[ M_n(t)=E\left (e^{t\bar X_n}\right ) =E\left (e^{(t/n)(X_1+\cdots +X_n)}\right ). \] By independence, \[ M_n(t)=\prod _{j=1}^n E\left (e^{(t/n)X_j}\right ) =\left [M\left (\frac tn\right )\right ]^n. \] Using the Poisson MGF, \[ M_n(t)=\left [e^{\lambda (e^{t/n}-1)}\right ]^n = \exp \left (n\lambda (e^{t/n}-1)\right ). \]

(b) As \(n\to \infty \), \[ e^{t/n}-1\sim \frac tn. \] Therefore \[ n\lambda (e^{t/n}-1)\to \lambda t. \] It follows that \[ M_n(t)\to e^{\lambda t}. \] This is the MGF of the constant random variable equal to \(\lambda \), which agrees with the Law of Large Numbers.