$$X_1 \overset{d}{\sim} \operatorname{Poisson}(\alpha_1)$$ $$X_2 \overset{d}{\sim} \operatorname{Poisson}(\alpha_2)$$ $$S = X_1 + X_2$$

Find $E[X_1 | S =n]$

My argument is that since $X_1 + X_2 = n$, $X_1$ range from $0$ to $n$ only, thus the expected value is

$$E[X_1 | S= n] = \sum_{k=1}^n k \Pr\{X_1 =k | S = n\}$$


$$\Pr\{X_1=k | S = n\} = \sum_{k=0}^{n} \frac{e^{-\alpha_1} \alpha_1^k}{k!}$$

Please argue for or against it?

Additionally, how would you evaluate such a series on the condition that it's correct?


1 Answer 1


Use the fact that $S=X_1+X_2 \sim \operatorname{Poisson}(a_1+a_2)$, then compute the conditional PMF:

\begin{align} P(X_1=x_1|X_1+X_2=n) &= \frac{P(X_1+X_2=n|X_1=x_1) P(X_1=x_1)}{P(X_1+X_2=n)}\\ &=\frac{P(X_2=n-x_1)P(X_1=x_1)}{P(X_1+X_2=n)}\\ &=\frac{\frac{1}{(n-x_1)!}a_2^{n-x_1} e^{}-a_2 \frac{1}{x_1!}a_1^{x_1}e^{-a_1}}{\frac{1}{n!}(a_1+a_2)^n e^{-(a_1+a_2)}}\\ &= \binom{n}{x_1} \frac{a_1^{x_1}a_2^{n-x_1}}{(a_1+a_2)^n} \end{align}

Having this, you can compute the expected value:

\begin{align} E(X_1|S=n) &= \sum_{x_1=0}^n x_1 \cdot P(X_1=x_1|X_1+X_2=n)\\ &= \frac{1}{(a_1+a_2)^n} \sum_{x_1=0}^n x_1\cdot \binom{n}{x_1} a_1^{x_1}a_2^{n-x_1} \\ &= \frac{n}{(a_1+a_2)^n} \sum_{x_1=0}^n \binom{n-1}{x_1-1} a_1^{x_1}a_2^{n-x_1} \\ &= \frac{na_1}{(a_1+a_2)^n} \sum_{x_1=0}^n \binom{n-1}{x_1-1} a_1^{x_1-1}a_2^{n-x_1}\\ &= \frac{na_1}{(a_1+a_2)^n} \sum_{j=0}^{n-1} \binom{n-1}{j} a_1^j a_2^{n-1-j}\\ &= \frac{na_1}{(a_1+a_2)^n} (a_1+a_2)^{n-1}\\ &=n \frac{a_1}{a_1+a_2} \end{align}

Alternatively you can see after the first step that $X_1|S=n \sim \operatorname{Bin}(n,\frac{a_1}{a_1+a_2})$.


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