(a) The event being conditioned on is \(\{X>c\}\). If \(x\le c\), then the events \(\{X\le x\}\) and \(\{X>c\}\) cannot occur together, so the conditional probability is 0. If \(x>c\), then \[ P(X\le x\mid X>c) = \frac {P(c<X\le x)}{P(X>c)} = \frac {F(x)-F(c)}{1-F(c)}. \] Thus the conditional CDF is \[ F_{X\mid X>c}(x)= \begin {cases} 0, & x\le c,\\[1mm] \dfrac {F(x)-F(c)}{1-F(c)}, & x>c. \end {cases} \]

(b) Differentiating the conditional CDF on the region \(x>c\) gives the conditional density. Therefore \[ f_{X\mid X>c}(x)= \begin {cases} \dfrac {f(x)}{1-F(c)}, & x>c,\\[1mm] 0, & x\le c. \end {cases} \]

(c) Since \(f(x)\ge 0\) and \(1-F(c)=P(X>c)>0\), the conditional density is nonnegative. Also, \[ \int _{-\infty }^{\infty } f_{X\mid X>c}(x)\,dx = \int _c^\infty \frac {f(x)}{1-F(c)}\,dx = \frac {1-F(c)}{1-F(c)} =1. \] So the function is indeed a valid PDF.

The area is \[ A=\pi R^2. \]

(a) Since \(R\sim \operatorname {Unif}(0,1)\), we have \(E(R^2)=1/3\) and \(E(R^4)=1/5\). Hence \[ E(A)=\pi E(R^2)=\frac {\pi }{3}. \] Also, \[ \operatorname {Var}(A) =\pi ^2\operatorname {Var}(R^2) =\pi ^2\left (E(R^4)-[E(R^2)]^2\right ) =\pi ^2\left (\frac 15-\frac 19\right ) =\frac {4\pi ^2}{45}. \]

(b) The possible values of \(A\) are from 0 to \(\pi \). For \(0\le a\le \pi \), \[ F_A(a)=P(A\le a) =P(\pi R^2\le a) =P\left (R\le \sqrt {\frac {a}{\pi }}\right ) = \sqrt {\frac {a}{\pi }}. \] Thus \[ F_A(a)= \begin {cases} 0, & a<0,\\[1mm] \sqrt {a/\pi }, & 0\le a\le \pi ,\\[1mm] 1, & a>\pi . \end {cases} \] Differentiating gives \[ f_A(a)= \begin {cases} \dfrac {1}{2\sqrt {\pi a}}, & 0<a<\pi ,\\[1mm] 0, & \text {otherwise}. \end {cases} \]

Assume the stick has length 1, and let \(U\sim \operatorname {Unif}(0,1)\) be the breakpoint. The two pieces have lengths \(U\) and \(1-U\), so the longer piece has length \[ L=\max (U,1-U). \] The possible values of \(L\) lie between \(1/2\) and \(1\).

If \(\ell <1/2\), then \(P(L\le \ell )=0\). If \(1/2\le \ell \le 1\), then \[ L\le \ell \quad \Longleftrightarrow \quad U\le \ell \text { and } 1-U\le \ell \quad \Longleftrightarrow \quad 1-\ell \le U\le \ell . \] The length of this interval is \(\ell -(1-\ell )=2\ell -1\). Thus \[ F_L(\ell )= \begin {cases} 0, & \ell <1/2,\\[1mm] 2\ell -1, & 1/2\le \ell \le 1,\\[1mm] 1, & \ell >1. \end {cases} \] This means that \(L\) is uniform on \([1/2,1]\), so its average is the midpoint of that interval, \[ E(L)=\frac {1/2+1}{2}=\frac 34. \]

For \(0\le x\le 1\), \[ F_X(x)=P(X\le x)=P(U_1\le x,\dots ,U_n\le x)=x^n, \] since the \(U_j\) are independent. Therefore \[ F_X(x)= \begin {cases} 0, & x<0,\\ x^n, & 0\le x\le 1,\\ 1, & x>1. \end {cases} \] Differentiating on \((0,1)\) gives \[ f_X(x)=nx^{n-1}, \qquad 0<x<1. \] The expectation is \[ E(X)=\int _0^1 x\,nx^{n-1}\,dx =n\int _0^1 x^n\,dx =\frac {n}{n+1}. \]

The exponential CDF is \[ F(x)=1-e^{-\lambda x}, \qquad x\ge 0. \] To use inverse transform sampling, set \[ U=1-e^{-\lambda X}. \] Solving for \(X\) gives \[ X=-\frac {1}{\lambda }\log (1-U). \] Since \(1-U\) is also uniform on \((0,1)\), the equivalent construction \[ X=-\frac {1}{\lambda }\log U \] is also valid.

Being within \(0.4\) meters of the true value means \[ |\epsilon |\le 0.4. \] Since \(\epsilon \sim N(0,0.04)\), its standard deviation is \(0.2\). Thus \[ P(|\epsilon |\le 0.4) =P\left (-2\le \frac {\epsilon }{0.2}\le 2\right ) =\Phi (2)-\Phi (-2). \] By symmetry, this is \[ 2\Phi (2)-1. \] Using \(\Phi (2)\approx 0.97725\), the probability is approximately \[ 2(0.97725)-1=0.9545. \]

(a) Since \(Y=|X|\), the random variable \(Y\) is always nonnegative. Thus \(F_Y(y)=0\) for \(y<0\). For \(y\ge 0\), \[ F_Y(y)=P(|X|\le y)=P(-y\le X\le y). \] Standardizing gives \[ F_Y(y) = \Phi \left (\frac {y-\mu }{\sigma }\right ) - \Phi \left (\frac {-y-\mu }{\sigma }\right ). \]

(b) Differentiating for \(y>0\) gives \[ f_Y(y) = \frac {1}{\sigma }\phi \left (\frac {y-\mu }{\sigma }\right ) + \frac {1}{\sigma }\phi \left (\frac {-y-\mu }{\sigma }\right ), \qquad y>0, \] and \(f_Y(y)=0\) for \(y<0\).

(c) At \(0\), the density has a jump if we define it to be 0 to the left and by the right-hand formula to the right. This is not a problem. A PDF only needs to integrate correctly, and changing the value of a density at a single point does not change any probability.

(a) By the memoryless property of the exponential distribution, once one of Bob or Claire finishes, Alice begins service while the other remaining original customer still has an exponential remaining service time with rate \(\lambda \). At that moment Alice and the remaining customer are in a perfectly symmetric race, with independent \(\operatorname {Expo}(\lambda )\) service times. Therefore Alice is last exactly if her own service time exceeds the other remaining service time, which has probability \(1/2\).

(b) Alice first waits until one of the two clerks becomes free. The minimum of two independent \(\operatorname {Expo}(\lambda )\) random variables is \(\operatorname {Expo}(2\lambda )\), so her expected waiting time is \[ \frac {1}{2\lambda }. \] After that, Alice receives service, whose expected duration is \[ \frac {1}{\lambda }. \] Hence her expected total time in the post office is \[ \frac {1}{2\lambda }+\frac {1}{\lambda }=\frac {3}{2\lambda }. \]

(a) If there are \(G\) failures before the first success, then the first success occurs at the next trial time. Since trial times are spaced by \(\Delta t\), we have \[ T=G\Delta t. \] This convention regards the trial at time 0 as the first possible trial, so \(G=0\) corresponds to success at time 0.

(b) For a fixed \(t\ge 0\), the event \(T>t\) means that all trials up to time \(t\) have failed. The number of such trial opportunities is naturally written using \(\lfloor t/\Delta t\rfloor +1\) under the convention that trials begin at time 0. Thus \[ P(T>t)=\left (1-\lambda \Delta t\right )^{\lfloor t/\Delta t\rfloor +1}. \] Therefore \[ F_T(t)=P(T\le t)=1-\left (1-\lambda \Delta t\right )^{\lfloor t/\Delta t\rfloor +1}, \qquad t\ge 0. \] For \(t<0\), \(F_T(t)=0\).

(c) For fixed \(t\ge 0\), as \(\Delta t\to 0\), \[ \left (1-\lambda \Delta t\right )^{\lfloor t/\Delta t\rfloor +1} \longrightarrow e^{-\lambda t}. \] Hence \[ F_T(t) \longrightarrow 1-e^{-\lambda t}, \] which is exactly the CDF of an \(\operatorname {Expo}(\lambda )\) random variable.

(a) The uniform distribution on \([a,b]\) is symmetric, and its CDF equals \(1/2\) at the midpoint. Hence the median is \[ \frac {a+b}{2}. \] The density is constant on the entire interval \([a,b]\), so every point in \([a,b]\) is a mode.

(b) For \(X\sim \operatorname {Expo}(\lambda )\), the median \(m\) solves \[ 1-e^{-\lambda m}=\frac 12. \] Thus \[ e^{-\lambda m}=\frac 12, \qquad m=\frac {\log 2}{\lambda }. \] The exponential density \(\lambda e^{-\lambda x}\) is decreasing on \([0,\infty )\), so its maximum occurs at \(x=0\).

Since the two die rolls are independent, \[ M_{X+Y}(t)=M_X(t)M_Y(t). \] The two rolls have the same distribution, so \[ M_X(t)=M_Y(t)=E(e^{tX})=\frac 16\sum _{j=1}^{6} e^{tj}. \] Therefore \[ M_{X+Y}(t)=\left (\frac 16\sum _{j=1}^{6}e^{tj}\right )^2. \] Equivalently, \[ M_{X+Y}(t)=\frac {1}{36}\sum _{i=1}^{6}\sum _{j=1}^{6} e^{t(i+j)}. \]

(a) Since \(X\) and \(Y\) are independent, \(X^2\) and \(Y^2\) are independent. Therefore \[ M_W(t)=M_{X^2}(t)M_{Y^2}(t) =(1-2t)^{-1/2}(1-2t)^{-1/2} =(1-2t)^{-1}, \] for \(t<1/2\).

(b) The MGF \[ (1-2t)^{-1} \] is the MGF of a \(\operatorname {Gamma}(1,1/2)\) distribution if we use the shape–rate convention, or equivalently an \(\operatorname {Expo}(1/2)\) distribution. It is also the chi-square distribution with 2 degrees of freedom.

(a) Since \(E(\bar X_n)=\mu \), we have \[ E(Z_n)=\frac {\sqrt n}{\sigma }(E(\bar X_n)-\mu )=0. \] Also, \[ \operatorname {Var}(\bar X_n)=\frac {\sigma ^2}{n}. \] Therefore \[ \operatorname {Var}(Z_n) =\frac {n}{\sigma ^2}\operatorname {Var}(\bar X_n) =\frac {n}{\sigma ^2}\cdot \frac {\sigma ^2}{n} =1. \]

(b) Write \[ Z_n=\frac {1}{\sigma \sqrt n}\sum _{j=1}^{n}(X_j-\mu ). \] Then \[ M_{Z_n}(t) =E\left [\exp \left (\frac {t}{\sigma \sqrt n}\sum _{j=1}^{n}(X_j-\mu )\right )\right ]. \] Using independence, this becomes \[ \prod _{j=1}^{n} E\left [\exp \left (\frac {t}{\sigma \sqrt n}(X_j-\mu )\right )\right ] = \left (e^{-\mu t/(\sigma \sqrt n)}M\left (\frac {t}{\sigma \sqrt n}\right )\right )^n. \]

The MGF of a \(\operatorname {Bin}(n,p_n)\) random variable is \[ M_{X_n}(t)=\left (1-p_n+p_ne^t\right )^n. \] Since \(p_n=\lambda /n\), \[ M_{X_n}(t) = \left (1+\frac {\lambda }{n}(e^t-1)\right )^n. \] As \(n\to \infty \), we use the standard limit \[ \left (1+\frac {a}{n}\right )^n\to e^a. \] With \(a=\lambda (e^t-1)\), this gives \[ M_{X_n}(t) \to \exp \{\lambda (e^t-1)\}. \] But \[ M_X(t)=\exp \{\lambda (e^t-1)\} \] is the MGF of \(X\sim \operatorname {Pois}(\lambda )\).

For \(U\sim \operatorname {Unif}(0,1)\), \[ M_U(s)=E(e^{sU})=\int _0^1 e^{su}\,du=\frac {e^s-1}{s}, \] with the value at \(s=0\) understood by continuity. Since \(U_1\) and \(U_2\) are independent, \[ M_{U_1+U_2}(s)=\left (\frac {e^s-1}{s}\right )^2. \]

Now compute the MGF of the triangular density. It is \[ M(s)=\int _0^1 e^{st}t\,dt+\int _1^2 e^{st}(2-t)\,dt. \] A direct integration gives \[ \int _0^1 e^{st}t\,dt =\frac {e^s(s-1)+1}{s^2}, \] and \[ \int _1^2 e^{st}(2-t)\,dt =\frac {e^{2s}-e^s(s+1)}{s^2}. \] Adding these two expressions, \[ M(s)=\frac {e^{2s}-2e^s+1}{s^2} =\left (\frac {e^s-1}{s}\right )^2. \] This is exactly the MGF of \(U_1+U_2\). Since MGFs determine distributions when they exist in a neighborhood of 0, \(U_1+U_2\) has the stated triangular distribution.