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 \[ E(X)=\int _0^1 x\,nx^{n-1}\,dx =n\int _0^1 x^n\,dx =\frac {n}{n+1}. \]

(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 \(\operatorname {Exp}(\lambda )\) 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 \(\operatorname {Exp}(\lambda )\) random variables is \(\operatorname {Exp}(2\lambda )\), 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 \(\operatorname {Exp}(\lambda )\) 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 \(X\sim \operatorname {Exp}(\lambda )\), 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 \(x=0\).

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 \(X\sim \operatorname {Pois}(\lambda )\).

The random variable \(X\) has two components. If the shoe gets stuck in the tree, then \(X=0\). Otherwise, \(X\) has a \(\operatorname {Unif}(0,1)\) distribution. Such a random variable is called a mixed random variable.

(a) The CDF is \[ F_X(x)=P(X\leq x). \]

If \(x<0\), then \(X\geq 0\), so \[ F_X(x)=0. \]

Now suppose that \(0\leq x<1\). The event \(X\leq x\) occurs in either of two disjoint ways:

• the shoe gets stuck in the tree, in which case \(X=0\leq x\);
• the shoe does not get stuck in the tree and \(U\leq x\).

Therefore, \[ \begin {aligned} F_X(x) &=P(I=1)+P(I=0,U\leq x)\\ &=P(I=1)+P(I=0)P(U\leq x)\\ &=p+(1-p)x, \end {aligned} \] where independence was used in the second line.

If \(x\geq 1\), then \(X\leq x\) with probability \(1\). Hence \[ \boxed { F_X(x)= \begin {cases} 0, & x<0,\\[1mm] p+(1-p)x, & 0\leq x<1,\\[1mm] 1, & x\geq 1. \end {cases}} \]

(b) Immediately to the left of \(0\), \[ F_X(0^-) = \lim _{x\uparrow 0}F_X(x) = 0. \]

At \(x=0\), \[ F_X(0)=p. \]

Hence the CDF has a jump at \(0\) of size \[ F_X(0)-F_X(0^-)=p. \]

For any random variable, \[ P(X=a)=F_X(a)-F_X(a^-). \]

Therefore, \[ P(X=0)=F_X(0)-F_X(0^-)=p. \]

The jump in the CDF represents the probability mass concentrated at \(0\).

After the jump, the CDF increases linearly from \(p\) to \(1\) on the interval \((0,1)\). The slope of this part is \(1-p\), representing the continuous uniform component of the distribution.

(c) At \(x=0\), \[ P(X=0)=P(I=1)=p. \]

For every \(x\in (0,1)\), \[ \begin {aligned} P(X=x) &=P(I=0,U=x)\\ &=P(I=0)P(U=x)\\ &=(1-p)\cdot 0\\ &=0, \end {aligned} \] because a continuous uniform random variable assigns probability \(0\) to every individual point.

Also, \(P(X=x)=0\) for \(x\notin [0,1]\). Therefore, \[ \boxed { P(X=x)= \begin {cases} p, & x=0,\\[1mm] 0, & x\neq 0. \end {cases}} \]

These point probabilities do not account for the entire distribution, since \[ \sum _{x\in \mathbb R}P(X=x)=p<1. \]

The remaining probability \(1-p\) is spread continuously over \((0,1)\). Therefore, \(X\) is not a discrete random variable.

More formally, let \(C\subseteq \mathbb R\) be any countable set. The uniform component assigns probability \(0\) to \(C\). Therefore, \[ P(X\in C) = \begin {cases} p, & 0\in C,\\ 0, & 0\notin C. \end {cases} \]

In either case, \[ P(X\in C)\leq p<1. \]

Thus there is no countable set that contains all the probability, so \(X\) is not discrete.

(d) Suppose that \(X\) were an absolutely continuous random variable. Then there would exist an ordinary PDF \(f_X\) such that \[ P(X\in A)=\int _A f_X(x)\,dx \] for every suitable set \(A\).

In particular, \[ P(X=0) = \int _{\{0\}}f_X(x)\,dx = 0, \] because a single point has length \(0\).

However, \[ P(X=0)=p>0. \]

This is a contradiction. Therefore, \(X\) cannot be described by an ordinary PDF alone and is not an absolutely continuous random variable.

Hence \(X\) is neither discrete nor continuous.

(e) Using the CDF, \[ \begin {aligned} P\left (X\leq \frac 12\right ) &=F_X\left (\frac 12\right )\\ &=p+(1-p)\frac 12\\ &=\boxed {\frac {1+p}{2}}. \end {aligned} \]

The event \(0<X\leq 1/2\) requires that the shoe not get stuck in the tree and that \(U\in (0,1/2]\). Therefore, \[ \begin {aligned} P\left (0<X\leq \frac 12\right ) &=P(I=0)P\left (0<U\leq \frac 12\right )\\ &=(1-p)\frac 12\\ &=\boxed {\frac {1-p}{2}}. \end {aligned} \]

Finally, \(X>0\) whenever the shoe does not get stuck in the tree, except for the event \(U=0\), which has probability \(0\). Hence \[ \boxed {P(X>0)=1-p}. \]

Notice that \[ \begin {aligned} P\left (X\leq \frac 12\right ) &=P(X=0)+P\left (0<X\leq \frac 12\right )\\ &=p+\frac {1-p}{2}\\ &=\frac {1+p}{2}. \end {aligned} \]

(f) Let \(\delta _0\) denote the probability distribution that places all its probability mass at \(0\). The distribution of \(X\) can be written as \[ \boxed { P_X = p\delta _0 + (1-p)\operatorname {Unif}(0,1). } \]

The first component places probability \(p\) at the single point \(0\), while the second component distributes probability \(1-p\) continuously and uniformly over \((0,1)\).

Equivalently, for every measurable set \(A\subseteq \mathbb R\), \[ \boxed { P(X\in A) = p\mathbf {1}_{\{0\in A\}} + (1-p)\int _A\mathbf {1}_{(0,1)}(x)\,dx. } \]

Thus the distribution consists of \[ \underbrace {p\delta _0}_{\text {discrete component}} \qquad +\qquad \underbrace {(1-p)\operatorname {Unif}(0,1)} _{\text {continuous component}}. \]