Homework

HW 10

1. (5 pts) Assume that the claims process $S_t$ is represented by a compound Poisson process. The mean time between adjacent claims is 2 hours and the random value $X$ of a particular claim is uniformly distributed on $[0,10]$. The insurance company sets premiums according to $c_t = 1.2 \E[S_t]$.

   • Estimate the ruin probability if the company sets aside an initial surplus of $u = 100$.
   • For which value of the initial surplus is the ruin probability less than $0.1$. You do not need to solve for the initial surplus exactly, you may just use the estimate in part 1.

   Solution: may be found here

   Note that for the second part, I changed the ruin probability to be less that $0.1$ instead of $0.05$ in the solutions. So set

   $$e^{-\gamma u} = 0.1 \quad \Rightarrow u = 43.94$$

2. (5 pts) Assume that $S_t$ is a compound Poisson process with rate $\lambda$ and $c_t = (1 + \theta) \E[S_t]$.

   • Find the adjustment coefficient when the process has exponential jumps with mean $1/a$.

   • Find the adjustment coefficient when the process has jumps that are $\Gamma(2,a)$ distributed. Hint I want you to compute this exactly like we did in class for the exponential case. Recall that $\Gamma(2,a)$ has MGF

     $$M(t) = \frac{1}{(1 - t/a)^{2}} \quad t < \beta$$

   If you flipped the roles of $2$ and $a$, that’s fine too. But you probably won’t be able to solve it explicitly.

   Solution: Again found in the solutions above for problem 3.

HW 9

1. Customers arrive at a service facility according to a Poisson process with an average rate of 5 per hour. Find

   • the prob that during 6 hours no customer will arrive
   • the prob that during 7 hours at most 30 customers will arrive
   • the prob that the waiting time between the third and fourth customer will be greater that 30 minutes.
   • the probability that the waiting time between the seventh and 8th customer will be less than 45 minutes,
   • The mean and variance of the time it takes between the arrival of the first customer and the fifth customer.

   Solutions Can be found here (problem 1).

2. The number of claims coming into a company is a Poisson process that has mean time between claims equal to 1/2 hours. With probability $p$, a claim is made by a woman over 60 years old. What kind of process is the ${N_t}_{t \in \R^+}$, the number of claims coming from women over 60 years old? Prove your answer.

   Solution The time between claims is of course $\tau \sim \Exponential(\lambda)$ is a $\Poisson(\lambda)$ process. This means $\E[\tau] = \lambda^{-1}$. Let

   $S_t = \sum_{i=1}^{N_t} X_i$ where $X_i$ are $\Bernoulli(p)$. $X_i = 1$ if the $i^{th}$ customer is over 60 years old. $N_t$ is the $\Poisson$ process. $S_t$ represents the total number of senior customers who come in by time $t$.

   We know that $S_t$ is compound Poisson and has MGF $\begin{align} M_t(x) & = e^{\lambda t (M_{X_i}(x) - 1)} \\ & = e^{ \lambda t( 1 - p + p e^x - 1)}\\ & = e^{ \lambda p t( e^x - 1) } \end{align}$ This is the MGF of a Poisson process for all $t$.

3. Customers arrive at a service facility according to a non-homogeneous Poisson process with a rate of 3 customers per hour between 9 and 11 am. After 11 am the rate is decreasing linearly from 3 at 11 am to 0 at 5pm. What is the probability that no more than 15 customers arrive between 10am and 4pm?

   Solution Recall that $\overline{\lambda}(s,t) = \int_s^t \lambda(r) dr$. In the 10am to 4pm(=16:00) period, we have to look at

   $$\begin{align} \overline{\lambda}(10,16) & = \int_{10}^{11} 3 dr + \int_{11}^{16} 3 - (t-11)\frac{3}{17 - 11} dr\\ & = 18 - \left. \frac{(t-11)^2}{4} \right|_11^{16} \\ & = 18 - \frac{25}{4} = 11.75 \end{align}$$

   Let $N_t$ be the number of customers. Then $\Prob(N_{16} - N_{11} \leq 15) = \sum_{i=0}^{15} e^{-11.75} \frac{11.75^i}{i!} \approx 0.862 ~ .$

HW 8

1. Assume the portfolio of a company consists of two homogeneous groups of risks. For the first group, the number of clients is $n_1 = 2000$ and the probability of loss event for each client is $q_1 = 0.1$. The payment, if a loss event occurs, is a non-random amount of $z_1 = 10$ units of money. For the second group, the corresponding quantities are $n_2 = 400$, $q_2 = 0.05$ and $z_2 = 30$. Let $n = n_1 + n_2$, and $S_n = S_{n_1} + S_{n_2}$.

   Find

   • $E[S_{n_i}]$ and $\Var(S_{n_i})$ for $i = 1,2$.
   • What is a good approximation to the distribution of $S_{n_i}$ for $i = 1,2$.
   • Suppose you have two normal random variables $X_1 \sim N(\mu_1,\sigma_1)$ and $X_2 \sim N(\mu_2,\sigma_2)$ that are independent, find the distribution of $X_1 + X_2$. You must prove this (you’ve also proved this earlier).
   • In view of the previous part, what is a good approximation for $S_n = \sum_{i=1}^2 S_{n_i}$?
   • Define $c = (1+\theta) \E[S_n]$ to be the total premium. Suppose $\beta = 0.99$, that is, the insurance company wants to be able to pay back their clients $99\%$ of the time. Find an expression for $\theta$ as we did in class by using the normal approxmation for $S_n$. Also calculate the value of $\theta$.
   • Once you’ve determined $\theta$, tell me how much a client in group $1$ and client in group $2$ should pay.

   Solutions

   • So clearly, we have $\begin{align} \E[S_{n_1}] & = n_1 \E[X_1] = 2000 0.1 10 = 2000\\ \E[S_{n_2}] & = n_2 \E[X_2] = 400 0.05 30 = 600\\ \Var(S_{n_1}) & = n_1 \Var(X_1) = 2000 0.1 (1-0.1)10^2 = 17100\\ \Var(S_{n_2}) & = n_2 \Var(X_2) = 400 0.05 (1-0.05) 30^2 = 18000 \end{align}$
   • We know from the central limit theorem that $S_{n_i}$ are both approximately normally distributed; i.e., $\Prob( \frac{S_{n_i} - \E[S_{n_i}]}{\sqrt{\Var(S_{n_i})}} \leq x) \approx \Phi(x)$
   • Suppose $X_1$ and $X_2$ are normally distributed, then they have mgfs $M_{X_i} = e^{\mu_i t + \frac12 \sigma_i^2 t^2}$ We know the MGF of an independent sum is a product of MGFs. So $\begin{align} M_{X_1+X_2} & = e^{\mu_1 t + \frac12 \sigma_1^2 t^2} e^{\mu_2 t + \frac12 \sigma_2^2 t^2}\\ & = e^{(\mu_1 + \mu_2) t + \frac12 (\sigma_1^2 + \sigma_2^2) t^2} \end{align}$ which means that $X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$.

   • Since $S_{n_2}$ and $S_{n_1}$ are approximately normal and independent, a good approximation to their sum is the normal distribution! Clearly, $\begin{align} \E[S_n] & = \E[S_{n_2}] + \E[S_{n_1}] = 2600\\ \Var(S_n) & = \Var(S_{n_2}) + \Var(S_{n_1}) = 35100 \end{align}$

   • Then, we’re looking for the probability $\Prob(S_n \leq (1+\theta)\E[S_n]) = \Prob\left( \frac{S_n - \E[S_n]}{\sqrt{\Var(S_n)}} \leq \frac{\theta \E[S_n]}{\sqrt{\Var(S_n)}} \right) \approx \Phi\left( \frac{\theta \E[S_n]}{\sqrt{\Var(S_n)}} \right) = 0.99$

     How do you solve this?

     • Go to wolframalpha.com Type in “inverse standard normal cdf”.
     • It will show something like $-\sqrt{2} erfc^{-1}(2 x)$
     • Click on that function and it will show up in the search bar.
     • Now you need to evaluate this function at 0.99, our $\beta$ value. This gives $\frac{\theta \E[S_n]}{\sqrt{\Var(S_n)}} = 2.326$ which gives $\theta = 0.167$, which is something like $17\%$, which seems to make sense.

2. Assume that the occurrence of traffic accidents is well described by a Poisson process with rate $\lambda = 30$ per day. The probability that an accident causes serious injuries is $p = 0.1$, and the outcomes of different accidents are independent. Write down a formula for the probability that the number of accidents with serious injuries during a month (of $30$ days) exceeds $100$. Estimate this probability using normal approximation.

   Solution: may be found here

   You should also find the exact formula for the probability of having more than 100 serious accidents. Let $S_N = \sum_{i=1}^N X_i$ be the number of accidents with serious injuries. $N$ is the total number of accidents in a month. Also

   $$X_i = \begin{cases} 1 & \Prob = 0.1 \\ 0 & \Prob = 0.9 \end{cases}$$ $$\begin{align} \sum_{i=101}^{\infty} \Prob( S_i > 100, N = i) & = \sum_{i=101}^{\infty} \Prob( S_i > 100 | N = i) \Prob(N = i)\\ & = \sum_{i=101}^{\infty} \Prob( S_i > 100 ) \Prob(N = i)\\ & = \sum_{i=101}^{\infty} \sum_{j=101}^i \binom{i}{j} 0.1^j 0.9^{i-j} e^{-\lambda} \frac{\lambda^i}{i!} \end{align}$$

HW 7

1. Suppose that an insurance company has $d$ groups of clients, and that the clients within each group are homogeneous. That is, in the $i$th group there is a probability $r_i$ that any given client makes a claim. Also assume that the $i$th group has $n_i$ clients. Let $N$ be the total number of clients that make claims over all of the different groups, by Poisson approximation we can assume that $N$ will be Poisson distributed. What will be the parameter $\lambda$ that describes this random variable? Give mathematical justification.

   Solution: May be found here

2. Take a sequence of random variables $X_n$ that are $\Binomial(n,\lambda/n)$ distributed. Do they have a limit in distribution: that is, is there a random variable $Y$ such that $\lim_{n \to \infty} \Prob( X_n = k) = \Prob(Y = k)$

   • Find the limit (if it exists) and identify the random variable $Y$ (if it exists). Hint: This is something we discussed in class.

   Solution We proved this in class:

   $\begin{align} \lim_{n \to \infty} \Prob(X_n = k) & = \lim_{n \to \infty} \binom{n}{k} \left( \frac{\lambda}{n} \right)^k \left( 1 - \frac{\lambda}{n} \right)^{n-k}\\ & = \frac{\lambda^k}{k!} \lim_{n \to \infty} \frac{n!}{(n-k)! n^k} \left( 1 - \frac{\lambda}{n} \right)^{n-k}\\ & = \frac{\lambda^k}{k!} \left( \lim_{n \to \infty} \frac{n!}{(n-k)! n^k} \right) \lim_{n \to \infty} \left( 1 - \frac{\lambda}{n} \right)^{n-k} \\ & = \frac{\lambda^k}{k!} 1 \cdot e^{- \lambda} \end{align}$ using taking the logarithm and using L’hospital’s rule to evaluate the limits. That is, you ought to show

   $\begin{align} \left( \lim_{n \to \infty} \frac{n!}{(n-k)! n^k} \right) = 1 \\ \lim_{n \to \infty} \left( 1 - \frac{\lambda}{n} \right)^{n-k} = e^{-\lambda} \end{align}$ to get full credit.

   Clearly this means $Y = \Poisson(\lambda)$.

   • If $X_n \sim \Binomial(n,\lambda/n^2)$, is there a random variable $Y$ such that $\lim_{n \to \infty} \Prob( X_n = k) = \Prob(Y = k)$

   Solution: The same calculation works $\begin{align} \lim_{n \to \infty} \Prob(X_n = k) & = \lim_{n \to \infty} \binom{n}{k} \left( \frac{\lambda}{n^2} \right)^k \left( 1 - \frac{\lambda}{n^2} \right)^{n-k}\\ & = \frac{\lambda^k}{k!} \lim_{n \to \infty} \frac{n!}{(n-k)! n^k} \frac{1}{n^k} \left( 1 - \frac{\lambda}{n^2} \right)^{(n-k)}\\ & = 0 \end{align}$

   To show that the above limit is $0$, we see that $\begin{align} \lim_{n \to \infty} (n-k) \log \left( 1 - \frac{\lambda}{n^2} \right) & = \lim_{n \to \infty} \frac{ 2 \lambda n^{-3} \left( 1 - \frac{\lambda}{n^2} \right)^{-1} }{ -(n-k)^{-2} }\\ & = 0 \end{align}$ and $\lim_{n \to \infty} \frac{1}{n^k} = 0 .$ Putting this and the previous calculation together shows that $\Prob(Y = k) = 0$ for all $k$. This means that there is no random variable that is the limit of $X_n$.

3. Suppose that 10,000 cars are traveling across a city. One out of 4 cards is gray. Suppose that the probability that any single car has an accident on a particular day is 0.002.

• Let $X$ be the random variable describing the number of gray cars that have an accident. What kind of random variable is it? What are the parameters of the distribution?
• What’s the probability that exactly 3 grey cars have an accident on this day?
• Using the approximation of a Binomial by a Poisson, compute the probability that exactly 3 gray cars have an accident on this day.

• Why can you approximate a Binomial by a Poisson in this case?

  Solution

  • $X$ is $\Binomial(10000/4,0.002)$
  • $\binom{10000/4}{3} 0.002^3 (1-0.002){10000/4-3}$
  • $\Binomial(10000/4,0.002) \approx \Poisson(10000/4*0.002)$ from the previous. So just write down the Poisson probability with the correct rate to get points.
  • Here $n = 10000/4$, which is pretty large, so this is “good enough” to use the Poisson approximation. In a more advanced course, you will learn how to estimate the error of this approximation.

HW 6, Due Mar 11

1. Well-Traveled insurance company sells a travel insurance policy that reimburses travelers for any expenses incurred for a planned vacation that is canceled because of airline bankruptcies. Individual claims follow Pareto with $\theta = 500$ and $\alpha = 2$.

   Because of financial difficulties in the airline industry, Well-Traveled imposes a limit of $$1000$ on each claim. If a policyholder’s planned vacation is canceled due to airline bankruptcies and he or she has incurred more than $$1000$ in expenses, what is the expected non-reimbursed amount of the claim? Recall that $\operatorname{Pareto}(\theta,\alpha)$ in its alternate form has the following tail

   $$T(x) = \begin{cases} \left( \frac{\theta}{\theta + x} \right)^{\alpha} & x \geq 0\\ 1 & x < 0 \end{cases}$$

   Solution: This is on page 153, example 3 (2nd edition). Let’s choose $$500$ (since this the Pareto parameter is $$500$) as a unit of money. We consider the distribution of the loss $\xi$ given that $\xi > 2$. The tail of the conditional is

   $$\Prob(\xi > x | \xi > 2) = \frac{9}{1+x^2} \quad x \geq 2$$

   Clearly, $\Prob( \xi > x

   \xi > 2) = 1$ if $x < 2$. Then,

   $$\begin{align} \E[ \xi | \xi > 2] & = \int_{0}^{\infty} \Prob(\xi > x | \xi > 2)dx \\ & = \int_{0}^{2} dx + \int_{2}^{\infty} \frac{9}{(1+x)^2} dx\\ = 5 \end{align}$$

   So what we’ve computed is the expected loss given that the loss is greater than $$1000$. This value is $5 \times 500 = $2500$.

   This is the total amount of the claim, given that the traveler occurred more than $$1000$ in expenses. The amount that the company reimburses is $$1000$. So the non-reimbursed amount is $$1500$.

2. Convolutions. Let $X$ and $Y$ be independent random variables. Compute the convolution of $X$ and $Y$ when

   • $X \sim \operatorname{Bernouilli}(p)$ and $Y \sim \operatorname{Bernouilli}(q)$.
   • $X \sim N(0,1)$ and $Y \sim \operatorname{Bernouilli}(p)$.
   • $X \sim \operatorname{Exp}(\alpha)$ and $Y \sim \operatorname{Poisson}(\beta)$.

   Solution:

   Let $S = X + Y$.

   • $X \sim \operatorname{Bernouilli}(p)$ and $Y \sim \operatorname{Bernouilli}(q)$. $\Prob(S = i) = \sum_{j = -\infty}^{\infty} \Prob(X = j) \Prob(Y = i - j)$

     Then, $\Prob(S = i) = \begin{cases} \Prob(X = 0) \Prob(Y = 0) = p q & i = 0\\ \Prob(X = 0) \Prob(Y = 1) + \Prob(X = 1) \Prob(Y = 0) = p(1-q) + q(1-p) & i=1 \\ \Prob(X = 1) \Prob(Y = 1) = (1-p)(1-q) & i = 2\\ 0 & \text{otherwise} \end{cases}$

   • $X \sim N(0,1)$ and $Y \sim \operatorname{Bernouilli}(p)$. $\begin{align} \Prob(S \leq t) & = \Prob(X + 0 \leq t | Y = 0) \Prob(Y = 0) + \Prob(X + 1 \leq t | Y = 1) \Prob(Y = 1)\\ & = p \int_{-\infty}^t \exp(-s^2/2) \frac{1}{\sqrt{2 \pi}}\, ds + (1-p) \int_{-\infty}^{t-1} \exp(-s^2/2) \frac{1}{\sqrt{2 \pi}}\, ds \end{align}$

   • $X \sim \operatorname{Exp}(\alpha)$ and $Y \sim \operatorname{Poisson}(\beta)$. For $t > 0$ $\begin{align} \Prob(S \leq t) & = \sum_{i=0}^{\infty} \Prob(X + i \leq t | Y = i) \Prob(Y = i) \\ & = \sum_{i=0}^{\lfloor t \rfloor } e^{-\beta} \frac{ \beta^i}{ i! } \int_{0}^{t-i} e^{-\alpha y} \alpha \, dy\\ & = \sum_{i=0}^{\lfloor t \rfloor } e^{-\beta} \frac{ \beta^i}{ i! } (1- e^{-\alpha (t-i)} \end{align}$

     where $\lfloor t \rfloor$ is the largest integer smaller than $t$.

3. Let $X \sim \operatorname{Gamma}(\alpha,\beta)$ so that its pdf is

   $$f_X(x) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}, x \geq 0$$
   • Find the MGF of $X$.
   • Show that if $X_1 \sim \operatorname{Gamma}(\alpha_1, \beta)$ and $X_2 \sim \operatorname{Gamma}(\alpha_2, \beta)$ are independent then $X_1 + X_2 \sim \operatorname{Gamma}(\alpha_1 + \alpha_2, \beta)$.
   • Assume that $X_i \sim \operatorname{Gamma}(2^{-i}, 1)$ for $i = 1, 2, 3, 4, \ldots$, and that the $X_i$ are all independent. Then use moment generating functions to show that $X_1 + X_2 + X_3 + X_4 + \ldots$ has an $\operatorname{Exp}(1)$ distribution.

   Solution:

   Recall $\Gamma(\alpha) = \int_0^{\infty} x^{\alpha - 1} e^{-x} dx$

   The MGF of the gamma function may be found as follows. Let $t < \beta$. Then, $\begin{align} M_X(t) & = \int_0^{\infty} e^{t x} \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}\\ & = \int_0^{\infty} \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-(\beta - t) x } dx\\ & = \int_0^{\infty} \frac{\beta^{\alpha}}{\Gamma(\alpha)} \frac{u^{\alpha-1}}{(\beta -t)^{\alpha}} e^{-u } du & = \left(\frac{\beta}{(\beta -t)} \right)^{\alpha} \end{align}$ where we’ve used the change of variables $(\beta - t)x = u$.

   Using the theorem about moment generating functions and convolutions $\begin{align} M_{X_1+X_2}(t) & = M_{X_1}(t)M_{X_2}(t)\\ & = \left(\frac{\beta}{\beta-t} \right)^{\alpha_1}\left(\frac{\beta}{\beta-t} \right)^{\alpha_2}\\ & = \left(\frac{\beta}{\beta-t} \right)^{\alpha_1 + \alpha_2} \end{align}$

   Similarly let $S = \sum_{i=1}^{\infty} X_i$. Recall that $\sum_{i=0}^{\infty} x^i = \frac{1}{1 - x} \quad |x| < 1.$ In particular, this gives $\sum_{i=1}^{\infty} 2^{-i} = 1.$

   Then $\begin{align} M_S(t) & = \prod_{i=1}^{\infty} M_{X_i}(t)\\ & = \prod_{i=1}^{\infty} \left(\frac{1}{(1-t)} \right)^{\alpha_i}\\ & = \left(\frac{1}{1-t} \right)^{\sum_{i=1} 2^{-i} }\\ & = \left(\frac{1}{1 -t} \right) \end{align}$

   The earlier definition of the density of $\operatorname{Gamma}(1,1)$ gives $f_X(x) = e^{-x} \quad x \geq 0$ with MGF $1/(1-t)$. Hence, $M_S(t)$ is the MGF of $\operatorname{Exp}(1)$.

HW 5, Due Mar 4

1. Fun with MGFs. Compute the moment generating functions of

   • The Poisson random variable $X \sim Poisson(\lambda)$.
   • A uniform random variable on the interval $[0,1]$.

   • A Gaussian random variable $X \sim N(0,\sigma^2)$.

     Hint:

     $$\begin{align} M_X(z) & = \int_{-\infty}^{\infty} e^{z x} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx\\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-(x-z)^2/2 + z^2/2} \, dx \end{align}$$

     Can you make a change of variable and complete the proof? IMPORTANT. The computation I’ve shown is you for the $N(0,1)$ case; i.e., the variance of the Gaussian random variable was assumed to be $1$. I want you do it for $N(0,\sigma^2)$, which is slightly different.

2. Let $X_1 \sim \operatorname{Poisson}(\lambda_1)$ and $X_2 \sim \operatorname{Poisson}(\lambda_2)$ be independent. Show that $X_1 + X_2 \sim \operatorname{Poisson}(\lambda_1 + \lambda_2)$ in two distinct ways:

   • By direct computation, i.e. compute $\Prob(X_1 + X_2 = k)$ for $k \in {0, 1, 2, 3, \ldots }$. Hint: for a fixed $k$ what are all the possible combinations of values $X_1$ and $X_2$ such that $X_1 + X_2 = k$

   • By using moment generating functions.

     Hint: When you have two independent random variables, $X_1$ and $X_2$, then $M_{X_1 + X_2}(z) = M_{X_1}(z) M_{X_2}(z)$ Using this formula, show that the $X_1 + X_2$ has the moment generating function of $\operatorname{Poisson}(\lambda_1 + \lambda_2)$.

3. Find the moment generating function of a Pareto random variable with tail ($\alpha > 1$)

   $$T(x) = \overline{F}(x) = \begin{cases} \frac{1}{x^{\alpha}} & x \geq 1\\ 1 & x < 1 \end{cases}$$

   Is the Pareto random variable heavy or light-tailed? State precisely why or why not.

   Hint: It turns out that Pareto is heavy tailed (it’s merely Polynomial). Start your proof like this. Suppose there are constants $B,c > 0$ such that $\begin{align} T(x) & \leq B e^{- c x} \quad \forall x\\ \frac{1}{x^{\alpha}} & \leq B e^{- c x}\\ \frac{B e^{c x}}{x^{\alpha}} & \leq 1 \end{align}$

   However, we can prove (you must prove this) that $\lim_{x \to \infty} \frac{B e^{c x}}{x^{\alpha}} = + \infty$

   This is a contradiction! This means that the Pareto distribution cannot be light-tailed. Therefore the moment generating function of the Pareto distribution must be $M(z) = +\infty$ for all $z > 0$. (Why? Recall the theorem we proved in class)

   Alternate approach: You may directly show that $M(z) = +\infty$, of course. It follows that the Pareto random variable is heavy-tailed.

HW 4, Due Feb 26

Solutions: This week’s solutions may be found here

1. An EUM customer of an insurance company has a total wealth of 100 and is facing a arandom loss $\xi$ distributed as follows.

   $$\xi = \begin{cases} 0 & \Prob = 0.9\\ 50 & \Prob = 0.05\\ 100 & \Prob = 0.05 \end{cases}$$
   1. Let the utility function of the customer be $u(x) = x - 0.005x^2$ for $0 \leq x \leq 100$. Graph it. Is the customer a risk averter?
   2. Is the customer a risk averter if instead $u(x) = x + 0.005 x^2$ for $0 \leq x \leq 100$.
   3. Assume the customer uses the utility function from part $a$. What is the maximum premium the customer would be willing to pay in order to insure his wealth against the loss? Is this greater than or less than $\E[\xi]$?
   4. Find the minimum premium that an insurance company would accept to cover the risk $\xi$, assuming the company uses the utility function $u_1(x) = \sqrt{x}$ and has initial wealth 300. Is this greater than or less than $\E[\xi]$?

   *Hint:* Recall that for the insurance company, if they take premium $p$, their final wealth is
   $$
       X = w + p - \xi
   $$
   Use the EUM criterion for the company to determine the minimum premium. 1.   Repeat the previous question when $\xi$ is uniformly distributed on $[0,100]$ instead.
   

2. Let $u(x) = x$ for $x \in [0,1]$ and $u(x) = 1/2 + x/2$ for $x \geq 1$. Is an EU maximizer with this utility function a risk averter (according to condition $Z$)? Why?

3. Let $d > 0$ and $r_d(x)$ be the function

   $$r_d(x) = \begin{cases} 0 & x \leq d \\ x-d & x > d \end{cases}$$
   • Let $X$ be exponential with mean $m$. Find the number $d$ such that $\E[r_d(x)] = m/2$.
   • Let $X$ be uniform on $[0,2m]$. Find the number $d$ such that $\E[r_d(x)] = m/2$.
   • Which is bigger: the deductible for the exponential $X$ or the uniform $X$? Could you have guessed this without doing computations?

HW 3, Due Feb 12

1. (8 pts) Let us construct an example where the mean-variance criterion is non monotone. Let $X$ be Pareto distributed. That is, let

   $$\Prob(X > x) = \frac{1}{x^{\alpha}}$$

   The Pareto distribution has $\alpha > 1$.

   Let $Y$ be uniformly distributed on $[0,1]$.

   • Find the mean and variance of $X$ and $Y$.

   Solution: Clearly the pdf is $f_X(t) = \frac{\alpha}{x^{\alpha + 1}}$

   This gives $\begin{align} \E[X] & = \int_1^{\infty} x \alpha x^{-(\alpha + 1)} dx\\ & = \int_1^{\infty} \alpha x^{-(\alpha )} \\ & = \left( \frac{1}{x^{\alpha - 1}} \frac{\alpha}{\alpha + 1} \right)_1^{\infty}\\ & = \frac{\alpha}{\alpha - 1} \end{align}$

   $$\begin{align} \E[X^2] & = \int_1^{\infty} x^2 \alpha x^{-(\alpha + 1)} dx\\ & = \int_1^{\infty} \alpha x^{-\alpha + 1)} dx\\ & = \frac{\alpha}{\alpha - 2} \left( x^{-\alpha + 2} \right)_1^{\infty}\\ & = \frac{\alpha}{\alpha - 2} \end{align}$$

   Which gives $\begin{align} \Var(X) & = \frac{\alpha}{\alpha - 2} - \left( \frac{\alpha}{\alpha - 1} \right)^2 \\ & = \frac{\alpha (\alpha^2 - 2\alpha + 1 - \alpha^2 + 2\alpha)}{(\alpha - 1)^2 (\alpha - 2) }\\ & = \frac{\alpha}{(\alpha - 1)^2 (\alpha - 2) } \end{align}$

   $$E[Y] = \frac{1}{2} \quad \Var(Y) = \frac{1}{12}$$
   • Is $X \geq Y$ almost surely or is $Y \geq X$? Does it make a difference whether $X$ and $Y$ are independent or not?

   Solution: Since $Y \in [0,1]$ and $X \in [1,\infty)$. Clearly $X \geq Y$, and it makes no difference whether or not they’re independent.

   The mean-variance criterion says that

   $$X \gtrsim Y \Leftrightarrow V(X) \geq V(Y)$$

   where

   $$V(X) = \E[X] - \tau \Var(X)$$
   • For what range of $\tau$ and $\alpha$ is $Y$ preferred over $X$ ($Y \gtrsim X$)?

   Solution: We need to compute the range of $\tau$ for which

   $$\frac{1}{2} - \tau \frac{1}{12} \geq \frac{\alpha}{\alpha - 1} - \tau \frac{\alpha}{(\alpha - 1)^2 (\alpha - 2) }$$

   This results in the solution

   $$\tau \geq \frac{-6(\alpha - 1)(\alpha - 2)}{\alpha^2 - 5 \alpha -2}$$

   You will find that we can show that the condition can be satisfied by choosing $\tau$ large enough when $\alpha \in (2,1/2(5 + \sqrt{33})]$.

   • Is the mean-variance criterion monotone? Explain.

   Solution: We have found an interval where $V(Y) \geq V(X)$ even though $Y \leq X$ with probability $1$! This means the condition cannot be monotone.

2. (5 pts) Which of the following functions are convex on their entire domain of definition? Justify your answer.

   • $u \colon \R \to \R$, $u(x) = x^2$

   • $u \colon \R \to \R$, $u(x) =

     x

     $

   • $u \colon [0,\infty) \to \R,~u(x) = x^3$
   • $u \colon (-\pi,\pi) \to \R,~u(x) = \cos(x)$
   • $u \colon (-\pi/2,\pi/2) \to \R,~u(x) = -\cos(x)$

3. (7 pts) Another gambler type: Let a gambler’s utility function be $u(x) = e^x$. Consider a game where the gambler wins $a$ dollars with probability $p$ and loses the same amount with probability $q=1-p$.

   • Based on the EUM criterion, determine the ranges of values for $a$ and $p$ for which the gambler should play the game.

   • Is the EUM criterion monotone for this particular utility function? Why?

   • Would you say the gambler likes to takes risks, or is he very cautious? Why or why not?

HW 2, Due Feb 5

1. Interest rates: There are two banks in the town of Bankopia, UT. J.P Young offers offers you a nominal interest rate of $4.5\%$ compounded every quarter on your savings. Goldman-Woodruff offers an effective rate of $5.3 \%$ per annum.

   a. What is the effective interest rate that J.P Young offers?

   b. If you started with $1000 in your account, how much would you have at the end of one year? Compare both banks.

   Solution: Let $e$ be effective rate of interest offered by J.P. Young.

   $$\begin{align} (1+e) & = (1+0.045/4)^4\\ e & = (1+0.045/4)^4 - 1\\ & = 0.0457 \end{align}$$

   If you started with $1000 in your account, you’d get $(1+0.045/4)^4 1000$ with J.P Young and $(1+0.053)1000$ with Goldman-Woodruff.

2. Double integral: Let $X$ and $Y$ independent random variables. Suppose $X$ and $Y$ are standard normal. We will find the probability distribution of the following random variable $M$.

   $$M = \frac{1}{\sqrt{2}} X + \frac{1}{\sqrt{2}} Y.$$

   NOTE You must follow the steps I’ve written down here. You may not state the distribution of $M$ by citing a theorem.

   1. First compute the conditional distribution.

   $\begin{align} F_{M|Y=y}\,(t) & = \Prob( M \leq t | Y = y) \\ & = \Prob( \frac{1}{\sqrt{2}} X + \frac{1}{\sqrt{2}} y \leq t )\\ \end{align}$

   All I want you to do here is to write $F_{M

   Y=y}~(t)$ as an integral from $-\infty$ to $\infty$.

   Hint: The Gaussian pdf is given by $f_Y(y) = (1/\sqrt{2\pi}) e^{-y^2/2}$.

   Solution: We have $\begin{align} F_{M|Y=y}(t) & = \Prob\left(X \leq \sqrt{2} t - y \right) \\ & = \int_{-\infty}^{\sqrt{2} t - y} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx \end{align}$

   1. Now compute $F_M(t)$ by integrating over all the values $y$ that $Y$ can take. That is, compute the following double integral.
   $$\begin{align} F_M(t) & = \int_{-\infty}^{\infty} \Prob\left(X \leq \sqrt{2} t - y \right) \frac{1}{\sqrt{2\pi}} e^{-y^2/2} \, dy\\ & = \int_{-\infty}^{\infty} \int_{-\infty}^{\sqrt{2}t - y} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \frac{1}{\sqrt{2\pi}} e^{-y^2/2} \, dx \, dy\\ \end{align}$$

   Hint: Make the following change of variables

   $$\begin{align} u & = \frac{1}{\sqrt{2}} (x + y)\\ v & = \frac{1}{\sqrt{2}} (x - y) \end{align}$$

   Remember double integrals from Calc 3? First compute the limits of integration in the $u$ and $v$ variables. You will need to compute a Jacobian (a determinant) and plug it into the integral.

   Solution: First we make the change of variables to $(u,v)$. Notice that that $u^2 + v^2 = x^2 + y^2$ Another miracle of mathematics. Then compute the Jacobian of the transformation $det\left( \begin{array}{cc} \partial_u x & \partial_v x \\ \partial_u y & \partial_v y \end{array} \right) = det\left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right) = 1$

   Also notice that the condition on $(x + y)/\sqrt{2} \leq t$ translates to $u \leq t$. I suggest you draw a picture of the area to verify this.

   Therefore $\begin{align} F_M(t) & = \int_{-\infty}^{\infty} \int_{-\infty}^{\sqrt{2}t - y} \frac{1}{2\pi} e^{-(x^2+y^2)/2} \, dx \, dy\\ & = \int_{-\infty}^{\infty} \int_{-\infty}^{t} \frac{1}{2\pi} e^{-(u^2+v^2)/2} \, du \, dv\\ & = \int_{-\infty}^{t} \frac{1}{2\pi} e^{-u^2/2} \, du \\ & = \Phi(t) \end{align}$

   1. What is the distribution of $M$? What kind of random variable is it?

   Solution: From part $2$, it follows that $M$ is normally distributed with mean zero and variance $1$.

HW 1, Due Jan 22 (in class)

1. Conditional distributions. Let $Y$ be any random variable and let $X$ be discrete. The conditional distribution of $Y$ given $X$ is defined to be

The conditional expectation is a function of $X$, and is defined to be

Let $X$ be the number that shows up on the roll of a standard fair die. Let $Y$ be an exponential random variable with parameter $X$. That is $Y \overset{d}{=} \exp(X)$.

• Find $F_{Y

  X}(y)$.

  Hint: The definition may also be found on page 49 of your textbook under the section called conditional expectations. The notation used in your textbook is $F_Y(y

  X = x)$, which is the same thing as $F_{Y

  X}(y)$. Once the value of $X$ is fixed, the distribution of $Y$ is completely determined.

  To make things even more explicit, let’s think of the problem in the following way. Let there be 6 variables $Y_1,\ldots, Y_6$ all independent of each other and $X$. Let $Y_i$ be exponentially distributed with parameter $i$. Then,

  $$Y = Y_i \text{ when } X = i.$$

  For example, then $\begin{align} F_{Y|X=1} \,(y) & = \frac{ \Prob( Y \leq y , X = 1) }{\Prob(X = 1)}\\ & = \frac{ \Prob( Y_1 \leq y , X = 1) }{\Prob(X = 1)}\\ & = \frac{ \Prob( Y_1 \leq y | X = 1) \Prob(X = 1) }{\Prob(X = 1)}\\ & = \Prob( Y_1 \leq y ) \\ & = 1 - \exp(-y) \end{align}$

  Solution: Simply redo the above calculation to get

  $$F_{Y | X = i} \, (y) = 1 - \exp( - i y)$$

• Find $\E[Y

  X]$.

  Hint: This object is again a function of the value of $X$, but the dependence on $Y$ has been integrated out so to speak. Using integration by parts, we get

  $$\begin{align} \E[Y | X = 1] & = \int_0^{\infty} y e^{-y} dy\\ & = y (-e^{-y}) \Bigg|_{0}^{\infty} + \int_0^{\infty} e^{-y} \,dy\\ & = 1 \end{align}$$

  Now repeat this caclulation for general $X = i$.

  Question to ask yourself: Why did the integral change from $(-\infty,\infty)$ to $(0,\infty)$?

  Solution: $\E[Y | X = i] = \frac{1}{i}$

  The integral changed from $(-\infty,\infty)$ to $(0,\infty)$ because $Y$ is exponentially distributed; its density is zero on the negative reals.

• When $y,x \in \R$, find the joint distribution $\begin{equation} F_{Y,X}\,(y,x) = \Prob( Y \leq y , X \leq x) \end{equation}$

  Hint: Use $\Prob(A , B) = \Prob (A \cap B) = \Prob(A

  B) \Prob(B)$. Notice that the joint depends on $x$ and $y$. The events here are $A = { Y \leq y }$ and $B = { X \leq x }$.

  Solution:

  $$\begin{align} F_{Y,X}(y,x) & = \sum_{i=1}^x \Prob(Y \leq y | X = i) \Prob(X = i)\\ & = \sum_{i=1}^x (1 - \exp(-iy)) \frac{1}{6} \end{align}$$

  We’ve used the result of an earlier computation in the second step of the equation.

• Find the marginal distribution $F_{Y}$.

  Hint: The marginal distribution of $Y$ can be found the following way.

  $$F_Y(y) = \Prob(Y \leq y) = \sum_{i=1}^6 \Prob( Y \leq y, X = i)$$

  Solution: We may simply set $x = 6$ in the joint marginal distribution (Why?!). That is,

  $$F_Y(y) = \Prob(Y \leq y) = \Prob(Y \leq y, X \leq 6) = F_{Y,X}(y,6)$$

• Given that $3 \leq Y \leq 4$, calculate the probability that a three was shown on the dice.

  Hint: Bayes rule. So the two events here are $A = { 3 \leq Y \leq 4 }$ and $B = { X = 3}$. The problem asks you to compute $P(B

  A)$. It ought to be easy to compute $\Prob( A

  B)$. Can you use Bayes rule to compute $\Prob( B

  A)$?

  Solution: We have to compute $\begin{align} \Prob(B | A) & = \frac{\Prob(3 \leq Y \leq 4, X = 3)}{\Prob(3 \leq Y \leq 4 )}\\ & = \frac{\left( F_{Y|X=3}\,(4) \Prob(X = 3)- F_{Y|X=3}\,(3)\Prob(X = 3) \right) }{F_Y(4) - F_Y(3)} \end{align}$

  The second line in the above followed from the definition of the conditional marginal in the earlier part of the problem. It’s left to you to complete this computation.

2. Tower property

The conditional expectation satisfies a very important property called the tower property.

Consider the previously defined random variables $Y$ and $X$ again. Verify the tower property by

• Calculating $\E[Y]$ using the joint distribution $F_{Y,X}$.

  Hint: It’s better to compute this using the marginal distribution $F_Y$.

  Solution: $\begin{align} \E[Y] & = \int_0^{\infty} y \frac{d}{dy} F(y) \, dy\\ & = \int_0^{\infty} y \sum_{i=1}^6 i \exp(-iy) \frac{1}{6} \, dy\\ & = \frac{1}{6} \sum_{i=1}^6 \int_0^{\infty} y i \exp(-iy) \, dy\\ & = \frac{1}{6} \sum_{i=1}^6 \frac{1}{i} \end{align}$

• Calculating $\E[Y]$ using the tower property. That is, calculate

  $$\begin{align} \E[ \E[ Y | X ] ] & = \sum_{i=1}^6 \E[ Y | X = i] \Prob(X = i)\\ \end{align}$$

  Hint: We’ve computed $\E[ Y

  X = i]$ in a previous step. Just plug it in.

  Solution: $\begin{align} \E[ \E[ Y | X ] ] & = \sum_{i=1}^6 \E[ Y | X = i] \Prob(X = i)\\ & = \sum_{i=1}^6 \frac{1}{i} \frac{1}{6} \end{align}$

3. Independence and random variables We say that $Y$ does not depend on an event $A$ if $Y$ and $1_A$, the indicator of the event $A$ are independent. The indicator function $1_A = 1$ if $A$ occurs and is $0$ if $A$ does not occur. Show that when $Y$ is any random variable, and $A$ is any event independent of $Y$, we have

Solution:

We have used independence in the third step. This can be repeated on $1_A = 0$, and this proves the result.