Exercises

Example 5.7 (link)

Problem

Suppose that the number of customers visiting a fast food restaurant in a given day follows a Poisson distribution with parameter \(\lambda\). We further note that each customer purchases a drink with probability \(p\), independently from other customers and independently from Poisson distribution. Then, let \(X\) be the number of customers who purchase drinks, find \(\mathbb{E}[X]\).

Let \(Y\) be the number of customers visiting the restaurant in a given day. Then, we have

\[\begin{split} \begin{aligned} Y &\sim \text{Poisson}(\lambda) \\ \end{aligned} \end{split}\]

The problem mentioned that each customer purchases a drink with probability \(p\), independently from other customers \((\iid)\). This allows us to model each customer as a Bernoulli random variable with parameter \(p\). More concretely, let \(W\) be the indicator random variable of whether a customer purchases a drink or not, then we have \(W \sim \text{Bernoulli}(p)\).

We can then model \(X\) as the number of customers who purchase drinks in a day, which is a binomial random variable with parameters \(n\) and \(p\). More concretely, we have \(X\) as a sum of \(n\) independent Bernoulli random variables \(W\) with parameter \(p\), then we have \(X \sim \text{Binomial}(n, p)\).

The question asks us to find \(\mathbb{E}[X]\). We do know that \(\mathbb{E}[X] = np\), but we do not know what \(n\) is in a single day. Our intuition says that since the number of customers visiting the restaurant in a day follows a Poisson distribution, then the average number of customers visiting the restaurant in a day is \(\lambda\), by definition Property 15. Then we can instead say \(\mathbb{E}[X] = \lambda p\). It turns out our intuition is correct, and we can prove this result using the Law of Total Expectation.

The Law of Total Expectation states that

\[ \mathbb{E}[X] = \sum_{y} \mathbb{E}[X \mid Y=y] p_Y(y) \]

The tricky part is we need to sum all possible values of \(Y\), which is an infinite set \(\mathbb{N}\), let’s write it as follows:

\[ \mathbb{E}[X] = \sum_{n=0}^{\infty} \mathbb{E}[X \mid Y=n] p_Y(n) \]

For \(p_Y(n)\), we know that \(Y\) follows a Poisson distribution with parameter \(\lambda\), then we have

\[\begin{split} \begin{aligned} p_Y(n) &= \frac{\lambda^n e^{-\lambda}}{n!} \\ \end{aligned} \end{split}\]

For \(\mathbb{E}[X \mid Y=n]\), we need to find a general expression for this, since this should be a expression in terms of \(X\), it means that \(X \mid Y=n\) follows a Binomial distribution with parameters \(n\) and \(p\).

\[\begin{split} \begin{aligned} X \mid Y=n &\sim \text{Binomial}(n, p) \\ \end{aligned} \end{split}\]

where \(n\) is the number of customers visiting the restaurant in a day, and \(p\) is the probability of a customer purchasing a drink. Then

\[ \mathbb{E}[X \mid Y=n] = np \]

Putting everything together, we have

\[\begin{split} \begin{aligned} \mathbb{E}[X] &= \sum_{n=0}^{\infty} \mathbb{E}[X \mid Y=n] p_Y(n) \\ &= \sum_{n=0}^{\infty} np \frac{\lambda^n e^{-\lambda}}{n!} \\ &= p \sum_{n=0}^{\infty} n \frac{\lambda^n e^{-\lambda}}{n!} \\ &= p \sum_{n=0}^{\infty} n p_Y(n) \\ &= p \mathbb{E}[Y] \\ &= p \lambda \\ \end{aligned} \end{split}\]

where we used the fact that \(\sum_{n=0}^{\infty} n p_Y(n) = \mathbb{E}[Y]\).