Conditional Expectation Given an Event with Poisson Distribution 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.
 A: 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 $E[X| N  = 30] = 6$.
A: 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?

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.
