MTH 202 Homework

HW 4 Due Fri Feb 20

Problem 1

1. Suppose $N$ is a geometric random variable taking values in $0,1,\ldots$ with parameter $p$. (By inspecting the range of $N$, we see that $N$ is in fact, a shifted geometric random variable that represents the number of failures before the first success.) Let

   $$X = \sum_{i=1}^N \xi_i$$

   where ${\xi_i}$ are iid random variables independent of $N$, and $X = 0$ if $N = 0$. Find the cdf of $X$. Hint be mindful of the $N=0$ case. Assume that $\xi_i$ are iid $\Exponential(1)$. Simplify your answers as much as possible.

   Show solution.

   Solution:

   Let $f$ be the density function of $\xi_{i}$. Then

   $$f(z)= \begin{cases} e^{-z} & z\geq 0\\ 0 & z<0 \end{cases}$$

   We know that a sum of independent exponentials produces a Gamma distribution (see page 30, section 1.4.4 in Karlin and Pinsky 4th edition). Then, the density of

   $$X_k = \sum_{i=1}^k \xi_i$$

   is

   $$f_k(z)= \begin{cases} \frac{1}{(k-1)!} z^{k-1} e^{-z} & z\geq 0\\ 0 & z<0 \end{cases}$$

   Let us find the cdf of $X$.

   $$\begin{align*} P(X \leq t) & = \sum_{k=0}^{\infty} P(X \leq t| N = k) \Prob(N = k)\\ & = \sum_{k=0}^{\infty} P\left( \sum_{i=1}^k \xi_i \leq t| N = k \right) p(1-p)^{k}\\ & = P\left( 0 \leq t | N = 0 \right) p + \sum_{k=1}^{\infty} P\left( X_k \leq t | N = k \right) p(1-p)^{k} \end{align*}$$

   Now we have to divide the cases up into $t \geq 0$ and $t < 0$ since when $N = 0$, we have $X = 0$.

   $$P\left( X \leq t | N = 0 \right) = \begin{cases} 0 & t < 0 \\ 1 & t \geq 0 \\ \end{cases}$$

   Thus, the only interesting case is when $t \geq 0$; we have

   $$\begin{align*} F_X(t) & = p + \sum_{k=1}^{\infty} \int_0^{t} \frac{1}{(k-1)!} z^{k-1} e^{-z} dz \quad p(1-p)^{k} \\ & = p + p (1-p) \int_0^{t} \sum_{k=1}^{\infty} \frac{1}{(k-1)!} (z(1-p))^{k-1} e^{-z} dz \\ & = p + p (1-p) \int_0^{t} e^{z(1-p)} e^{-z} dz \\ & = p + p (1-p) \int_0^{t} e^{-p z)} dz \\ & = p + (1-p)(1 - e^{-pt }) \\ & = 1 - (1-p) e^{-pt } \end{align*}$$

   Putting this together, we get

   $F_X(t) = \begin{cases} 0 & t < 0 \\ 1 - (1-p) e^{-pt } & t \geq 0 \\ \end{cases}$

2. Suppose that three contestants on a quiz show are each given the same question and that each answers it correctly, independently of the others with probability $p$. But the difficulty of the question is itself a random variable. So suppose that $p$ is $\Uniform(0,1)$. What is the probability that exactly two of the contestants answer correctly?

   Show solution.

   Solution:

   Let $X$ be the number of contestants answering question correctly.

   $\begin{equation} \begin{split} P(X=2) &=\int_{0}^{1}P(X=2\mid p=u)f_p(u) du\\ &=\int_{0}^{1}\binom{3}{2}u^{2}(1-u)du\\ &=\frac{1}{4} \end{split} \end{equation}$

3. Let $X_0,X_1,\ldots$ be iid nonnegative random variables having a pdf. Let $N$ be the first index $k$ for which $X_k > X_0$. $N = 1$ if $X_1 > X_0$, $N=2$ if $X_1 \leq X_0, X_2 > X_0$ and so on. Find the pmf for $N$ and the mean $\E[N]$. (Interpretation: $X_0$, $X_1$ are bids on the car you are trying to sell. $N$ is the index of the first bid that is better than the initial bid.)

   Show solution.

   Solution:

   Let $f$ and $F$ be pdf and cdf of $X_{i}$ resp.

   $$\begin{equation} \begin{split} P(N & = 1) = P(X_{i}\leq X_{0}) \\ &=\int_{0}^{\infty}dx_{i}\int_{x_{i}}^{\infty}dx_{0}f(x_{i})f(x_{0})\\ &=\int_{0}^{\infty}dx_{i}f(x_{i})(1-F(x_{i}))\\ &=1-\frac{1}{2}F(x_{i})^{2}\biggr\rvert_{0}^{\infty}\\ &=\frac{1}{2} \end{split} \end{equation}$$

   Before we tackle the general $k \geq 2$ case, we define the tail of $X_0$ as $T(t) = 1 - F(t)$, and hence $dT = T’(t)dt = -f(t)dt$.

   $$\begin{equation} \begin{split} P(N & = k) = P(X_{1}\leq X_{0}, \cdots, X_{k-1} \leq X_0, X_k > X_0) \\ &=\int_{0}^{\infty} P(X_{1}\leq t , \cdots, X_{k-1} \leq t , X_k > t|X_0 = t) f(t) dt\\ &=\int_{0}^{\infty} F^{k-1}(t) T(t) f(t) dt\\ & = \int_0^{\infty} F^{k-1} (1 - F) f dt \\ & = \int_0^{\infty} F^{k-1} f - \int_0^{\infty} F^{k} f dt \\ & := I_k - I_{k-1} \\ \end{split} \end{equation}$$

   where we have used the last line to define the integral $I_k$. Now we evaluate the integral $I_k$ for $k \geq 2$ using integration by parts (when in doubt…).

   $$\begin{align*} I_k & = \int_0^{\infty} F^{k-1} f dt \\ & = - \int_0^{\infty} F^{k-1} T'(t) dt \\ & = - F^{k-1} T(t) \big|_{0}^{\infty} + \int_0^{\infty} (k-1)F^{k-2} T dt \\ & = \int_0^{\infty} (k-1)F^{k-2} (1-F) dt \\ & = (k-1)\int_0^{\infty} F^{k-2} dt - (k-1)\int_0^{\infty} F^{k-1} dt \\ & = (k-1) I_{k-1} - (k-1)I_{k} dt \end{align*}$$

   and therefore moving the $(k-1)T_k$ to the LHS of the above equation

   $$\begin{align*} I_k & = \frac{(k-1)}{k} I_{k-1} \\ & = \frac{(k-1)}{k} \frac{k-2}{k-1} I_{k-2} \\ & = \frac{(k-1)}{k} \frac{k-2}{k-1} \cdots I_{1} \\ & = \frac{1}{k} \\ \end{align*}$$

   where we have used $I_1 = 1$. Thus

   $$\begin{align*} \Prob(N = k) & = \frac{1}{k} - \frac{1}{(k+1)}\\ & = \frac{1}{k(k+1)} \quad k \geq 2 \end{align*}$$

   Then,

   $\begin{equation} \begin{split} E[N] &=\sum_{k=1}^{\infty}kP(N=k)\\ &= \frac12 + \sum_{k=2}^{\infty} \frac{1}{k+1} &= \infty \end{split} \end{equation}$

Problem 2

1. Suppose $X_1,X_2,X_3$ are discrete random variables. Show that

   $$\E[ X_1 + X_2 + X_3 | X_2, X_3 ] = X_2 + X_3 + \E[X_1 | X_2,X_3]$$
   Show solution.

   Solution:

   $$\begin{equation} \begin{split} E[X_{1}+X_{2}+X_{3}\mid X_{2}, X_{3}] &=E[X_{1}\mid X_{2},X_{3}]+E[X_{2}\mid X_{2},X_{3}]+E[X_{3}\mid X_{2},X_{3}]\\ &=X_{2}+X_{3}+E[X_{1}\mid X_{2},X_{3}] \end{split} \end{equation}$$

   We could also do this the following way for discrete $X_1,X_2$ and $X_3$. The conditional expectation is a FUNCTION of the values of $X_2,X_3$. So

   $$\begin{align*} \E[ X_1 + X_2 + X_3 | X_2 = a_2,X_3 = a_3] & = \sum_{a_1} (a_1 + a_2 + a_3) p_{X_1|X_2,X_3}(a_1 | a_2,a_3)\\ & = \sum_{a_1} (a_1 + a_2 + a_3) \frac{p_{X_1,X_2,X_3}(a_1 , a_2,a_3)}{p_{X_2,X_3}(a_2,a_3)}\\ & = \sum_{a_1} a_1 \frac{p_{X_1,X_2,X_3}(a_1 , a_2,a_3)}{p_{X_2,X_3}(a_2,a_3)} \\ & \qquad + \sum_{a_1} (a_2 + a_3) \frac{p_{X_1,X_2,X_3}(a_1 , a_2,a_3)}{p_{X_2,X_3}(a_2,a_3)}\\ & = \E[ X_1 | X_2 = a_2,X_3 = a_3] + a_2 + a_3 \end{align*}$$

   using the law of total probability in the last equation.

Problem 3

Conditional dependence and independence.

1. Draw $P$ uniformly randomly from $[0,1]$ and then draw two independent samples $X_1$ and $X_2$ with $\Bernoulli(P)$ distribution. Compute the (unconditional) correlation coefficient of $X_1$ and $X_2$.

2. Let $N$ have a Poisson distribution with parameter $\lambda > 0$. Suppose that, conditioned on $N$, we draw a random variable $X \sim \Binomial(N,p)$. Let $Y = N - X$. Show that $X$ and $Y$ have Poisson distributions with parameters $\lambda p$ and $(1 - p) \lambda$ and that $X$ and $Y$ are independent.

3. Let $X$ and $Y$ be jointly distributed random variables whose joint probability mass function is given in the following table:

Show that the covariance between $X$ and $Y$ is zero even though $X$ and $Y$ are not independent.

HW 3 Due Fri Feb 13

Problem 1: Conditional probability redux.

Let $A_0, A_1, \ldots, A_n$ be events of non-zero probability.

Show that $\Prob\left(\bigcap_{i=0}^n A_i \right) = \Prob(A_0) \Prob(A_1 | A_0) \cdots \Prob(A_n | A_{n-1} \ldots A_0 )$

Problem 2

1. Suppose $U$ has uniform distribution in $[0,1]$. Derive the density function for the random variables

   1. $Y = \log(1-U)$
   2. $W_n = U^n$ for $n \geq 1$

   Refer to section 1.2.6 in your textbook.

   Show solution.

   Solution: 1. As $U\sim \Uniform[0,1]$, range of $Y$ is [-$\infty$,0]. Given $t\leq 0$,

   $$\begin{equation} \begin{split} F_{Y}(t) &=P(Y\leq t)=P(\log(1-U)\leq t)\\ &=P(1-U\leq e^{t})=P(1-e^{t}\leq U)\\ &=e^{t} \end{split} \end{equation}$$

   So $f_{Y}(t)=e^{t}$ for $t\in[-\infty,0]$ and zero elsewhere.

   2. Range of $W_{n}$ is $[0,1]$. Given $t\in[0,1]$, \begin{equation} \begin{split} P(W_{n}\leq t) &=P(U^{n}\leq t)
   &=P(U\leq t^{1/n})
   &=t^{\frac{1}{n}} \end{split} \end{equation} So $f_{W_{n}}(t)=\frac{1}{n}t^{\frac{1}{n}-1}$ for $t\in[0,1]$ and zero elsewhere.

2. If $X \sim \Exponential(2)$ and $Y \sim \Exponential(3)$ are independent, find $\Prob(X < Y)$.

   Show solution.

   Solution: $\begin{equation} \begin{split} P(X<Y) &=\int_{0}^{\infty}\int_{x}^{\infty}2e^{-2x}3e^{-3y}dydx &=\int_{0}^{\infty}2e^{-2x}\int_{x}^{\infty}3e^{-3y}dydx &=\frac{2}{5} \end{split} \end{equation}$

Problem 3

1. A quarter is tossed repeatedly until a head appears. Let $N$ be the trial number when this head appears. Then, a dollar is tossed $N$ times. Let $X$ count the number of times the dollar comes up tails. Determine $P(X = 0)$, $P(X = 1)$ and $\E[X]$.

   Show solution.

   Solution:

   $$\begin{equation} \begin{split} P(X=0) &=\sum_{n=1}^{\infty}P(N=n, \text{no tails}) \\ &=\sum_{n=1}^{\infty}\big(\frac{1}{2}\big)^{n}\binom{n}{0}\big(\frac{1}{2}\big)^{n} \\ &= \frac{1}{4} \sum_{n=0}^{\infty}\big(\frac{1}{4}\big)^{n} \\ &= \frac{1}{4} \frac{1}{1-1/4} \end{split} \end{equation}$$ $$\begin{equation} \begin{split} P(X=1) &=\sum_{n=1}^{\infty}P(N=n, \text{one tail})\\ &=\sum_{n=1}^{\infty}\big(\frac{1}{2}\big)^{n}\binom{n}{1}\big(\frac{1}{2}\big)^{n}\\ &=\sum_{n=0}^{\infty}\frac{n}{4^{n}} \end{split} \end{equation}$$

   Let $S=P(X=1)$, then

   $$\begin{equation} S-\frac{1}{4}S=\sum_{n=0}^{\infty}\big(\frac{1}{4}\big)^{n}=\frac{1}{3} \end{equation}$$

   So $S=P(X=1)=\frac{4}{9}$. The way to see this in general is by defining

   $$\begin{align} S(p) & = \sum_{n=0}^{\infty} p^n = \frac{1}{1-p} \\ S'(p) & = \sum_{n=0}^{\infty} n p^{n-1} = \frac{1}{(1-p)^2}\\ p S'(p) & = \sum_{n=0}^{\infty} n p^n = \frac{p}{(1-p)^2} \end{align}$$

   and then substituting $p=1/4$.

   $$\begin{equation} \begin{split} E[X] &=\sum_{k=0}^{\infty}kP(X=k)=\sum_{k=0}^{\infty}\sum_{n=1}^{\infty}P(X=k\mid N=n)P(N=n)\\ &=\sum_{n=1}^{\infty}\sum_{k=0}^{\infty}k\binom{n}{k}\big(\frac{1}{2}\big)^{n}P(N=n)\\ &=\sum_{n=1}^{\infty}P(N=n)\sum_{k=0}^{\infty}k\binom{n}{k}\big(\frac{1}{2}\big)^{n}\\ &=\sum_{n=1}^{\infty}\frac{n}{2}P(X=n)=\frac{1}{2}\sum_{n=1}^{\infty}nP(N=n)\\ &= \frac12 E[N] = 1 \end{split} \end{equation}$$

   since $N$ is geometric taking values in ${1,\ldots,}$ and wikipedia says $E[N] = 1/p$.

   In the above we used that if $Z \sim \Binomial(n,p)$ then

   $$\E[Z] = \sum_{k=0}^n k \binom{n}{k} p^{n-k} (1-p)^k = n p$$

   with $p = 1/2$.

2. Let $N$ have Poisson distribution with parameters $\lambda = 1$. Conditioned on $N = n$, let $X$ have a uniform distribution over the integers $0,1,\ldots,n+1$. What is the marginal distribution for $X$?

   Show solution.

   Solution

   $$\begin{equation} \begin{split} P(X=t) &=\sum_{n=0}^{\infty}P(X=t\mid N=n)P(N=n)\\ &=e^{-1}\sum_{n=0}^{\infty}\frac{1}{n+2}\frac{1}{n!}=e^{-1} \sum_{n=t-1}^{\infty}\frac{n+1}{(n+2)!}\\ &=e^{-1}\sum_{n=t+1}^{\infty}\frac{n-1}{n}=e^{-1}\sum_{n=t+1}^{\infty}\frac{n-1}{n!}\\ &=e^{-1}\sum_{n=t+1}^{\infty}\frac{1}{(n-1)!}-e^{-1}\sum_{n=t+1}^{\infty}\frac{1}{n!}\\ &=\frac{e^{-1}}{t!} \end{split} \end{equation}$$

Problem 4

Do men have more sisters than women have? In a certain society, all married couples use the following strategy to determine the number of children that they will have: If the first child is a girl, they have no more children. If the first child is a boy, they have a second child. If the second child is a girl, they have no more children. If the second child is a boy, they have exactly one additional child. (We ignore twins, assume sexes are equally likely, and the sex of distinct children are independent random variables, etc.)

1. What is the probability distribution for the number of children in a family?
2. What is the probability distribution for the number of girl children in a family?
3. A male child is chosen at random from all of the male children in the population. What is the probability distribution for the number of sisters of this child?
4. What is the probability distribution for the number of brothers of the previously chosen male child?

HW 2 Due Fri Feb 6

1. Let $N$ cards carry the distinct numbers $x_1,\ldots,x_N$. If two cards are drawn at random without replacement, show that the correlation coefficient $\rho$ between the numbers appearing on the two cards in $-1/(N-1)$.

   Show solution.

   Solution

   Let $X,Y$ be the number on the first and second cards respectively.

   $$\begin{equation} \begin{split} P(X=x_{n}) &=\frac{1}{N}\\ P(Y=x_{n}) &=P(X=x_{n}, Y=x_{n})+P(X\neq x_{n}, Y=x_{n})=P(X\neq x_{n}, Y=x_{n})\\ & =P(Y=x_{n}\mid X\neq x_{n})\times P(X\neq x_{n}) =\frac{1}{N-1}\times \frac{N-1}{N}\\ &=\frac{1}{N} \end{split} \end{equation}$$

   So we know $E[X]=E[Y]$ and $E[X^{2}]=E[Y^{2}]$. Let

   $$\begin{equation} T=\sum_{n=1}^{N}x_{n} \quad \text{ and } \quad S=\sum_{n=1}^{N}x_{n}^{2} \end{equation}$$

   Then

   $$\begin{equation} \begin{split} E[X]&=E[Y]=\sum_{n=1}^{N}x_{n}P(X=x_{n})=\frac{T}{N}\\ E[X^{2}]&=E[Y^{2}]=\sum_{n=1}^{N}x_{n}^{2}P(X=x_{n})=\frac{S}{N} \end{split} \end{equation}$$

   We know $\rho$ is the correlation coefficient of $X,Y$, the formula of $\rho$ is

   $$\begin{equation} \rho=\frac{\Cov(X,Y)}{\sigma_{X}\sigma_{Y}} =\frac{\E[XY]-\E[X]\E[Y]}{\sigma_{X}\sigma_{Y}} \end{equation}$$

   Also we know

   $$\begin{equation} \sigma_{X}=(\E[X^{2}]-(\E[X])^{2})^{\frac{1}{2}}=(\frac{S}{N}-\frac{T^{2}}{N^{2}})^{\frac{1}{2}}=\sigma_{Y} \end{equation}$$

   There is only $\E[XY]$ left to compute.

   $$\begin{equation} \begin{split} \E[XY] &=\sum_{n=1}^{N}x_{n}\sum_{m\neq n}x_{m}P(X=x_{n}, Y=x_{m})\\ &=\sum_{n=1}^{N}x_{n}(T-x_{n})\frac{1}{N-1}\frac{1}{N}\\ &=\frac{T^{2}}{N(N-1)}-\frac{S}{N(N-1)} \end{split} \end{equation}$$

   Bring those formulas back to $\rho$, we have

   $\begin{equation} \rho=-\frac{1}{N-1} \end{equation}$

2. Let $X$ and $Y$ be independent random variables having distribution $F_x$ and $F_y$ respectively.

   • Let $Z = \max(X,Y)$. Express $F_Z(t)$ in terms of $F_X(s)$ and $F_Y(u)$.
   • Let $W = \min(X,Y)$. Express $F_W(t)$ in terms of $F_X(s)$ and $F_Y(u)$.
   Show solution.

   Solution

   Since $X,Y$ are independent,

   $$\begin{equation} \begin{split} F_{Z}(t) &=P(Z\leq t)\\ &=P(\max(X,Y)\leq t)\\ &=P(X\leq t, Y\leq t)\\ &=P(X\leq t)P(Y\leq t)\\ &=F_{X}(t)F_{Y}(t)\\ \end{split} \end{equation}$$

   since if $\max(X,Y) \leq t$ then both $X \leq t$ and $Y \leq t$.

   $\begin{equation} \begin{split} F_{W}(t)&=P(W\leq t)\\ &=1-P(W>t)\\ &=1-P(\min(X,Y)>t)\\ &=1-P(X>t, Y>t)\\ &=1-P(X>t)P(Y>t)\\ &=1-(1-P(X\leq t))(1-P(Y\leq t))\\ &=1-(1-F_{X}(t))(1-F_{Y}(t))\\ &=F_{X}(t)+F_{Y}(t)-F_{X}(t)F_{Y}(t) \end{split} \end{equation}$

3. Let $U$ be Poisson distributed with parameter $\lambda$ and let $V = 1/(1+U)$. Find the expected value of $V$.

   Show solution.

   Solution

   $$U\sim\Poisson(\lambda), \qquad P(U=k)=\frac{\lambda^{k}e^{-\lambda}}{k!}$$

   $\begin{equation} \begin{split} \E[V] &=\sum_{k=0}^{\infty}\frac{1}{1+k} P(U=k) \\ &=\sum_{k=0}^{\infty}\frac{1}{1+k}\frac{\lambda^{k}e^{-\lambda}}{k!}\\ &=e^{-\lambda}\sum_{k=0}^{\infty}\frac{1}{1+k}\frac{\lambda^{k}}{k!}\\ &=e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^{k}}{(k+1)!}\\ &=\frac{e^{-\lambda}}{\lambda}\sum_{k=0}^{\infty}\frac{\lambda^{k+1}}{(k+1)!}\\ &=\frac{e^{-\lambda}}{\lambda}\sum_{n=1}^{\infty}\frac{\lambda^{k}}{k!}\\ &=\frac{e^{-\lambda}}{\lambda}(e^{\lambda}-1)\\ &=\frac{1-e^{-\lambda}}{\lambda} \end{split} \end{equation}$

4. Let $U,V,W$ be independent random variables with equal variances $\sigma^2$. Let $X = U + V$ and let $Y = V - W$. Find the covariance of $X$ and $Y$.

   Show solution.

   Solution

   As $U,V,W$ are independent,

   $$\begin{equation} \begin{split} \Cov(X,Y) &=\E[XY]-\E[X]\E[Y]\\ &=\E[(U+V)(V-W)]-\E[U+V]\E[V-W]\\ &=\E[UV+V^{2}-UW-VW]-(\E[U]\E[V]+(\E[V])^{2}-\E[U]\E[W]-\E[V]\E[W])\\ &=\E[V^{2}]-(\E[V])^{2}\\ &=\Var(V)\\ &=\sigma^{2} \end{split} \end{equation}$$

   Another way to do this is to write

   $\begin{align} \Cov(X,Y) & = \Cov(U,V) - \Cov(U,W) + \Cov(V,V) \\ & \qquad + \Cov(U,V) - \Cov(U,W) + \Cov(V,V) + \Cov(V,W) \\ & = 0 - 0 + \Var(V) + 0 \\ & = \sigma^2 \end{align}$

HW 1 Due Fri Jan 30

Homework will be a little long this week, since it is mostly review.

1. Which topics about stochastic processes most interest you? (E.g. Brownian motion, stock market modeling, statistical physics, etc.)

2. Given that you end up working as hard as you think you can work this semester, what grade do think you can achieve in this class? How important is the letter grade to you?

3. Would you prefer a challenging class at a good pace so you can learn a lot, or do you prefer an easy-paced class so you can learn the material well?

4. Of the three general principles for modeling probabilities stated in K&P chapter 1, in your opinion, which one applies best to the modeling of share prices in the stock market?

5. Plot the distribution function

   $$F(x) = \begin{cases} 0 & x \leq 0\\ x^3 & 0 < x < 1\\ 1 & x \geq 1 \end{cases}$$
   • Determine the corresponding density function $f(x)$ in the three regions.
   • What is the mean of the distribution?
   • If $X$ is a random variable with distribution $F$, then evaluate $\mathbb{P}(1/4 \leq X \leq 3/4)$.
   Show solution.

   Solution: The density function is

   $$f(x)= \begin{cases} 3x^{2} & 0<x<1 \\ 0 & \text{otherwise} \end{cases}$$ $$\begin{equation*} E(X)=\int_{0}^{1}xf(x)dx=\frac{3}{4} \end{equation*}$$

   $\begin{equation*} P(\frac{1}{4}\leq X \leq \frac{3}{4})=F(\frac{3}{4})-F(\frac{1}{4})=\frac{13}{32} \end{equation*}$

6. Determine the distribution function, mean and variance corresponding to the triangular density

   $f(x) = \begin{cases} x & 0 \leq x \leq 1,\\ 2 - x & 1 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}$

   Show solution.

   Solution

   $$F(x)= \begin{cases} 0 & x<0 \\ \frac{1}{2}x^{2} & 0\leq x \leq1 \\ -\frac{1}{2}x^{2}+2x-1 & 1<x\leq 2 \\ 1 & x>2 \end{cases}$$ $$\begin{equation*} E[X]=1 \end{equation*}$$

   $\begin{equation*} \text{Var}(X)=\frac{1}{6} \end{equation*}$

7. Let $1_{A}$ be the indicator random variable associated with an event $A$, defined to be one if $A$ occurs, and zero otherwise. Show

   • $1_{A^c} = 1 - 1_{A}$
   • $1_{A \cap B} = 1_{A} 1_{B} = \min( 1_{A}, 1_{B} )$
   • $1_{A \cup B} = \max( 1_{A}, 1_{B} )$.
   Show solution.

   Solution:

   If $\omega \in A^{c}$, then

   $$\begin{equation*} 1_{A^{c}}(\omega)=1=1-1_{A}(\omega) \end{equation*}$$

   If $\omega \in A$, then

   $$\begin{equation*} 1_{A}(\omega)=1=1-1_{A^{c}}(\omega) \end{equation*}$$

   So $1_{A^{c}}=1-1_{A}$

   If $\omega\in A\cap B$, then

   $$\begin{equation*} 1_{A\cap B}(\omega)=1=1_{A}1_{B}(\omega) \end{equation*}$$

   If $\omega \notin A\cap B$, then $\omega$ is either in $X-(A\cup B)$, $A-B$ or $B-A$, we have

   $$\begin{equation*} 1_{A\cap B}(\omega)=0=1_{A}1_{B}(\omega) \end{equation*}$$

   So

   $$1\_{A\cap B}=1_{A}1_{B}$$

   Note that $\min(1_{A},1_{B})$ has only two possible values ${1,0}$.

   $$\begin{equation*} \min(1_{A},1_{B})(\omega)=1 \Longleftrightarrow 1_{A}(\omega)=1=1_{B}(\omega) \Longleftrightarrow \omega\in A\cap B \end{equation*}$$

   So

   $$\begin{equation*} 1_{A\cap B}=1_{A}1_{B}=\min(1_{A},1_{B}) \end{equation*}$$

   Note that $\max(1_{A},1_{B})$ has only two possible values ${1,0}$.

   $\begin{equation*} \omega\in A\cup B\Longleftrightarrow 1_{A}(\omega)=1\text{ or }1_{B}(\omega)=1 \Longleftrightarrow \max(1_{A},1_{B})(\omega)=1 \end{equation*}$

8. A pair of dice is tossed. If the two outcomes are equal, the dice are tossed again, and the process is repeated. If the dice is unequal, their sum is recorded. Determine the probability mass function for the sum.

   Show solution.

   Solution:

   There are only 30 possible outcomes. Among 30 outcomes, there are 2 outcomes of having 3, 4, 10 and 11 as their sum. 4 outcomes of having 5,6,8 and 9 as sum. And 6 outcomes of having 7 as sum.
   Define $X$= sum of two dice. Then sample space is ${3,4,5,6,7,8,9,10,11}$ And we have the following probability

   $\begin{equation*} \begin{split} P(X=3) &=P(X=4)=P(X=10)=P(X=11)=\frac{1}{15}\\ P(X=5) &=P(X=6)=P(X=8)=P(X=9)=\frac{2}{15}\\ P(X=7) &=\frac{3}{15} \end{split} \end{equation*}$