Conditional Expectation Given an Event with Poisson Distribution

Question

I am studying for my final exam and am attempting to solve a problem from Blitzstein and Hwang's Introduction to Probability without much success.

The problem is;

While Fred is sleeping one night, X legitimate emails and Y spam emails are sent to him. Suppose that X and Y are independent, with X ~ Pois(10) and Y ~ Pois(40). When he wakes up, he observes that he has 30 new emails in his inbox. Given this information, what is the expected value of how many new emails he has?

I know that we are trying to find E[X] conditioning on N = X + Y = 30, and that N ~ Pois(50) as X and Y are independent, but im just conceptually stuck on how to apply what I know about conditional expectation to this problem. Any advice would be much appreciated.

Answer 1

Here is a hint: the random variable $X+Y$ is Poisson with rate $10+40 = 50$, so in an intuitive sense, the number of legitimate emails that he would expect out of $30$ is $$\operatorname{E}[X \mid X+Y = 30] = 30(10/50) = 6.$$ But you need to justify this formally. How would you go about it?

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If you know (or are allowed to assume) that the sum of independent Poisson random variables is Poisson, i.e., $X_i \overset{\text{ind}}{\sim} \operatorname{Poisson}(\lambda_i)$ for $i = 1, 2, \ldots, n$ implies $\sum_i X_i \sim \operatorname{Poisson}(\sum_i \lambda_i)$, then the conditional distribution of $X \mid X+Y = n$ is trivial via the definition of conditional probability: $$\Pr[X = x \mid X+Y = n] = \frac{\Pr[(X = x) \cap (X+Y = n)]}{\Pr[X+Y = n]} = \frac{\Pr[X = x]\Pr[Y = n-x]}{\Pr[X+Y = n]},$$ which after substituting the Poisson PMFs, you will obtain a binomial PMF with a suitable choice of parameters.

Answer 2

On page 166 of Blitzsteins book you find the following theorem:

• If $X\sim Pois(\lambda_x)$ and $Y\sim Pois(\lambda_y)$ are independent, then $X| N = n \sim Bin(n, \frac{\lambda_x}{\lambda_x + \lambda _y})$, where $N = X+ Y$.

Using this as starting point, we get $$E[X| N = n] = n \cdot p$$ directly from the binomial distribution. Here $n=30$ and $p = \frac{\lambda_x}{\lambda_x + \lambda _y} = \frac{10}{50}$. Plugging these numbers into the formula, we get $E[X| N = 30] = 6$.