Adding independent Poisson variables

8 minA Level6.2.1

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can add independent Poisson variables, combine this with changing the length of an interval, and explain why $X - Y$ and $2X$ are not Poisson.

Key to the symbols

Symbol	Meaning
$X \sim \text{Po}(\lambda)$	$X$ has a Poisson distribution with mean $\lambda$ ("lambda")
$\lambda_1$ , $\lambda_2$	the means of two different Poisson variables
$P(X = r)$	$\dfrac{e^{-\lambda}\lambda^r}{r!}$ , the Poisson probability of exactly $r$ events

Remember that a Poisson variable counts random events (calls, customers, flaws) that happen singly, independently and at a constant average rate.

Two streams of events

A help desk has two phone lines. Calls on line A arrive at random at an average of 2 per hour, and calls on line B at an average of 3 per hour, independently.

Look at all the calls together. They are still random, single, independent events at a constant rate, just a busier stream: on average $2 + 3 = 5$ per hour.

So the total number of calls in an hour is Poisson with mean 5.

Definition

Sum of independent Poisson variables

If $X \sim \text{Po}(\lambda_1)$ and $Y \sim \text{Po}(\lambda_2)$ are independent, then

$X + Y \sim \text{Po}(\lambda_1 + \lambda_2)$

This extends to three or more independent Poisson variables: add all their means.

You may quote this result without proof. A quick consistency check: $X + Y$ has mean $\lambda_1 + \lambda_2$ and, by the variance rule, variance $\lambda_1 + \lambda_2$ . Mean and variance are equal, as they must be for a Poisson variable.

Worked example: Example 1: adding two Poisson variables

$X \sim \text{Po}(2.5)$ and $Y \sim \text{Po}(1.5)$ are independent. Find $P(X + Y = 3)$ .

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