Exercise 1 (Mean and variance of a Poisson) Let $$X$$ be a Poisson random variable with parameter $$\lambda >0$$, so that $\mathbb{P}( X = k) = e^{- \lambda} \frac{\lambda^k}{k!}$ for $$k =0,1,2, \ldots$$. Show that $$\mathbb{E} X = \lambda = var(X)$$.

Solution. First we show that $$\mathbb{E} X = \lambda$$.
$\begin{eqnarray*} \mathbb{E} X &=& \sum_{k=0} ^{\infty} k \mathbb{P}(X =k) = e^{- \lambda}\sum_{k=0} ^{\infty} k \frac{\lambda^k}{k!} \\ &=& e^{- \lambda}\lambda\sum_{k=1} ^{\infty} k \frac{\lambda^{k-1}}{k!} = e^{- \lambda}\lambda\sum_{k=1} ^{\infty} \frac{\lambda^{k-1}}{(k-1)!} = \lambda e^{- \lambda} \sum_{k=0} ^{\infty} \frac{\lambda^{k}}{k!} \\ &=& \lambda \end{eqnarray*}$
With the short-cut formula, it remains to compute $$\mathbb{E} (X^2)$$: $\begin{eqnarray*} \mathbb{E} (X^2) &=& e^{-\lambda} \sum_{k=0} ^ {\infty} k^2 \frac{\lambda^k}{k!} = e^{-\lambda} \sum_{k=1} ^ {\infty} k^2 \frac{\lambda^k}{k!} \\ &=& e^{-\lambda} \lambda \sum_{k=1} ^ {\infty} k \frac{\lambda^{k-1}}{(k-1)!} = e^{-\lambda} \lambda \sum_{k=0} ^ {\infty} (k+1) \frac{\lambda^{k}}{k!} \\ &=& e^{-\lambda} \lambda \Big( \sum_{k=0} ^ {\infty} k \frac{\lambda^{k}}{k!} + \sum_{k=0} ^ {\infty} \frac{\lambda^{k}}{k!} \Big) \\ &=& \lambda ( \mathbb{E} X + 1) = \lambda^2 + \lambda. \end{eqnarray*}$ Thus $$var(X) = \lambda^2 + \lambda - (\lambda)^2 = \lambda$$.

Exercise 2 (Memoryless property) Prove that exponential random variables have the memoryless property.

Solution. Let $$X$$ be a exponential random variable with rate $$\lambda >0$$.
The memoryless property states that for $$t,s \geq 0$$, we have $\mathbb{P}(X >t+s \ | \ X>s) = \mathbb{P}(X >t).$

We have $\begin{eqnarray*} \mathbb{P}(X >t+s \ | \ X>s) &=& \frac{\mathbb{P}\big( \{X >t+s\} \cap \{X >s\} \big)}{\mathbb{P}(X >s)}\\ &=& \frac{\mathbb{P}( X >t+s)}{\mathbb{P}(X >s)} \\ &=& \frac{e^{-\lambda (t+s)}}{e^{-\lambda s}} \\ &=& e^{-\lambda t} \\ &=& \mathbb{P}(X >t). \end{eqnarray*}$

Exercise 3 (Some of independent Poissons) Show that if $$X$$ and $$Y$$ are independent Poisson random variables with parameters $$\lambda$$ and $$\mu$$, respectively, then $$Z := X+Y$$ is a Poisson random variable with parameter $$\lambda + \mu$$.

Solution. Let $$p_{\lambda}$$ and $$p_{\mu}$$ be the pmfs for $$X$$ and $$Y$$. Since $$X$$ and $$Y$$ are independent, we know that $\begin{eqnarray*} \mathbb{P}(Z=n) &=& (p_{\lambda} \star p_{\mu})(n) \\ &=& \sum_{i \in \mathbb{Z}} p_{\lambda}(i) p_{\mu}(n-i). \end{eqnarray*}$ We know that $$\mathbb{P}(Z = n)=0$$ if $$n <0$$, since $$X$$ and $$Y$$ are nonnegative. Let $$n \geq 0$$. Note that $$p_{\lambda}(i)=0$$ for all $$i <0$$, and $$p_{\mu}(n-i) =0$$ for all $$i >n$$. Thus $\begin{eqnarray*} \sum_{i \in \mathbb{Z}} p_{\lambda}(i) p_{\mu}(n-i) &=& \sum_{i =0}^n p_{\lambda}(i) p_{\mu}(n-i) \\ &=& \sum_{i =0}^n\Big(\frac{e^{-\lambda} \lambda^i}{i!} \Big)\frac{e^{-\mu} \mu^{n-i}}{(n-i)!} \end{eqnarray*}$ Now recall that ${n \choose i} = \frac{n!}{i!(n-i)!}.$ So $\begin{eqnarray*} \mathbb{P}(Z=n) &=& \frac{e^{-(\lambda + \mu)}}{n!} \sum_{i=0}^n {n \choose i}\lambda^{i} \mu ^{n-i}. \end{eqnarray*}$ Now the binomial formula, with $$x= \lambda$$ and $$y= \mu$$ gives that $\mathbb{P}(Z=n) = ({e^{-(\lambda + \mu)}} ) \frac{(\lambda + \mu)^n}{n!}.$ So we obtain that $$Z$$ is a Poisson random variable with parameter $$\lambda + \mu$$.

Exercise 4 (Shop keeper) Suppose we model the number of customers that arrive at a high street shop on at particular day by a Poisson process of intensity $$\lambda >0$$, where $$\lambda$$ is measured in customers per hour. We wish to estimate $$\lambda$$. The shop keeper has records of how many customers arrive each day for $$n$$ days given by $$x = (x_1, \ldots, x_n)$$ and opens everyday for $$6$$ hours. Find an estimate for $$\lambda$$. Carefully justify why this is a reasonable estimate.

Solution. Let $$X_1, \ldots, X_n$$ be random variables representing the arrivals each day. We will assume that they these random variables are independent. The model gives that $$X_i$$ is Poisson with mean $$6\lambda$$. The law of large numbers tells us that $\frac{1}{n} \big( X_1+ \cdots + X_n\big) \to \mathbb{E} X_1 = 6 \lambda$ as $$n \to \infty$$. Thus the estimator $\frac{1}{6n} \big(X_1 + \cdots + X_n \big)$ is consistent. It is also easy to see that it is unbiased. With this estimator and the sample data, we give the corresponding estimate $\frac{1}{6n}( x_1 + \cdots +x_n).$

Exercise 5 (Orderliness) Show that Poisson processes constructed as exponential inter-arrial times satisfy orderliness.

Solution. Let $$N$$ be a Poisson process of rate $$\lambda$$. We will consider the case $$\lambda = 1$$, as it will be become apparent that the value of $$\lambda$$ does not matter. Let $$X_1$$ and $$X_2$$ be exponential random variables with rate $$\lambda =1$$. Recall the sum of exponentials gives a gamma. Let $$h >0$$. We have $\begin{eqnarray*} \mathbb{P}( N(h) \geq 2) &=& \mathbb{P}( X_1 + X_2 \leq h)\\ &=& \int_0 ^h x e^{-x} dx \\ &=& -xe^{-x} \big|_0 ^h - e^{-x} \big|_0 ^h \\ &=& -h e^{-h} - e^h + 1. \end{eqnarray*}$ Hence an easy application of l’hopital’s rule gives the desired result.

Exercise 6 (Scaling) Prove the scaling property for Poisson point processes on $$[0, \infty)$$.

Solution. Let $$c >0$$. Observe that if $$X$$ is an exponential random variable with rate $$\lambda$$, then $$c X$$ is an exponential random variable with rate $$c^{-1} \lambda$$.
Let $$X_i$$ be independent exponential random variables with rate $$\lambda$$. Let $$T_n = X_1 + \cdots + X_n$$ be their partial sums. We know that $\Pi = \{ T_n : n \in \mathbb{Z}^{+} \}$ is a Poisson point process on $$[0, \infty)$$ with intensity $$\lambda$$. The upshot is that $c \Pi = \{ c T_n: n \in \mathbb{Z}^{+} \}$ and we know that $cT_n = cX_1 + \cdots +cX_n$ which is the sum of exponentials with rate $$c^{-1} \lambda$$. Hence we conclude that $$c \Pi$$ is a Poisson point process of intensity $$c^{-1} \lambda$$ as desired.

Exercise 7 (Markov Chains) Can you guess what $$mean(y)$$ will roughly be without implementing the code?

Solution. Yes! It is easy to see that the give transition matrix is aperiodic and irreducible and thus has a stationary distribution $$\pi = \pi p$$, which we can compute by finding the left eigenvalue of the matrix $$p$$ and normalizing. We know that regardless of the initial distribution the chain will tend towards stationary distribution and by a version of the law of large numbers for Markov chains the long term average of ones is just given by $$\pi(1)$$.

Exercise 8 (A simple queue) Suppose customers arrive according to a Poisson process of intensity $$\lambda$$, and then are served by a single person at an exponential rate of $$\mu$$ (service time). Only person can be served at a each time, and customers are served in order, and may have to wait in a queue (waiting time). Explore such queues by simulation of various intensities. In particular, plot a histogram of the total time a customer spends in the system (response time = service time + waiting time).

Solution. We will simulate in the case $$\lambda = 1$$ and $$\mu = 1.5$$, and plot a histogram of the response time. The histogram approximates the density of the exponential distribution at rate $$\mu - \lambda$$.
inter=rexp(100000, 1)
arr=cumsum(inter)
service = rexp(100000,1.5)

output <- arr[1] + service[1]
for (i in 1:99999){
if (arr[i+1] < output[i]){output <- c(output, output[i] + service[i+1])}
if (arr[i+1] > output[i]){output <- c(output, arr[i+1] + service[i+1])}
output
}
time = output - arr
x = seq(0,max(time)+1, by=0.01)
b = seq(0,max(time)+1, by=0.1)
hist(time, prob=T, breaks=b)
curve(dexp(x,0.5), add=T)