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)
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])}
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)