Poisson distribution where each event can lead to different outcomes

Question

I'm trying to tackle the following problem.

Suppose that customers arriving at a bank follow a Poisson distribution with rate λ=5 every 15 minutes. For every customer, the probability that he performs one transaction is 0.4, two transactions 0.3, three transactions 0.2 and four transactions 0.1. Find the expected value and the variance of the number of transactions that will be performed in a 3 hour interval.

My starting point was to calculate that, per customer the expected value of transactions is $0.4\cdot1+0.3\cdot2+0.2\cdot3+0.1\cdot4=2$

Additionally, if X is the number of customers in a 3 hour interval, $X \sim \operatorname{Pois} \left({5\cdot12}\right)$. So the expected value is 60 customers in 3 hours, in other words 120 transactions in 3 hours. However, I wasn't able to figure out how to calculate the variance of the number of transactions. Does the number of transactions in 3 hours also follow the Poisson distribution or am I missing something?

Answer 1

Let us define $X$ as the random number of customers arriving in a 3 hour interval, so $X \sim \operatorname{Poisson}(\lambda = 60)$ as you noted. Now, for each $i = 1, 2, \ldots, X$, let $T_i$ be the number of transactions that the $i^{\rm th}$ customer to arrive makes, so that $$\Pr[T_i = k] = 0.1 (5 - k), \quad k = 1, 2, 3, 4.$$ We can now calculate $$\operatorname{E}[T_i] = \sum_{k=1}^4 \frac{k(5-k)}{10} = 2, \quad \operatorname{E}[T_i^2] = \sum_{k=1}^4 \frac{k^2(5-k)}{10} = 5,$$ so that $\operatorname{Var}[T_i] = 5-2^2 = 1$. Now we let $$S = \sum_{i=1}^X T_i$$ be the total number of transactions observed in the 3 hour interval. We wish to compute $\operatorname{E}[S]$ and $\operatorname{Var}[S]$. To this end, observe that

$$\begin{align*} \operatorname{E}[S] &= \operatorname{E}_X[\operatorname{E}[S \mid X]] \\[8pt] &= \operatorname{E}_X\left[\sum_{i=1}^X \operatorname{E}[T_i]\right] \\[8pt] &= \operatorname{E}[X \operatorname{E}[T_i]] \\[8pt] &= \operatorname{E}[X]\operatorname{E}[T_i] \\[8pt] &= 2\lambda = 120. \end{align*}$$

This is the law of total expectation. Next, we use the law of total variance:

$$\begin{align*} \operatorname{Var}[S] &= \operatorname{E}[\operatorname{Var}[S \mid X]] + \operatorname{Var}[\operatorname{E}[S \mid X]] \\[8pt] &\overset{\text{ind}}{=} \operatorname{E}[X \operatorname{Var}[T_i]] + \operatorname{Var}[X \operatorname{E}[T_i]] \\[8pt] &= \operatorname{E}[X]\operatorname{Var}[T_i] + \operatorname{E}[T_i]^2 \operatorname{Var}[X]. \end{align*}$$

Now it is a simple matter to recall that $\operatorname{E}[X] = \operatorname{Var}[X] = \lambda$ for a Poisson distribution, and substitute the expectation and variance for $T_i$ that we computed above. Note that this last formula $$\operatorname{Var}[S] = \operatorname{E}[X]\operatorname{Var}[T_i] + \operatorname{E}[T_i]^2 \operatorname{Var}[X]$$ applies to any random variable $S = T_1 + T_2 + \cdots + T_X$ where all the $T_i$s are iid, and $X$ is a nonnegative integer-valued random variable. This formula is frequently found in actuarial contexts.

Answer 2

Let $T$ be the number of transactions in a 3 hour interval, $X$ the number of customers in a 3 hour interval (which is $\sim poisson(60)$ as you found), and $N$ the number of transactions per customer (which has expectation 2 as you say and variance 5 as you can verify). Then we have that $T=\sum_{i=1}^X N$. I personally used the conditional variance formula from here, which is a straightforward calculation after this setup.