The full question I'm working on:
Let $X_1$ and $X_2$ be two independent Poisson random variables with
mean parameter $\lambda > 0$. Let $T= T(X_1, X_2) = X_1 + X_2$. [Note:
You may use the fact that the sum of two independent Poisson random
variables has a Poisson distribution.]For $t \ge x_1 \ge 0$, find the conditional probability mass function
Pr$\{X_1 = x_1 | X_1 + X_2 = t\}$. Does this conditional distribution
depend on $\lambda$? Be sure to specify the support of this
conditional marginal distribution.
I know Pr$\{X_1 = x_1\} = \frac{\lambda^{x_1}}{x_1 !}e^{-\lambda} $ and that Pr$\{X_1 + X_2 = x\} = \frac{(2\lambda)^{x}}{x !}e^{-2\lambda}$. I imagine the answer is something like it, starting with $t$ and removing $X_2$. Maybe $\frac{(2\lambda)^{x}}{x !}e^{-2\lambda}-\frac{\lambda^{x_2}}{x_2 !}e^{-\lambda}$? My intuition is that there is a better answer. Any hints or suggestions are very appreciated.
Best Answer
As long as the event $\{X_1+X_2 = t\}$ has positive probability (which it does here) we can use $$ P(X_1 = x_1 | X_1+X_2 = t) = \frac{P(X_1 = x_1, X_1 + X_2 = t)}{P(X_1+X_2 = t)} =\frac{P(X_1 = x_1, X_2 = t-x_1)}{P(X_1+X_2 = t)} $$ which by independence of $X_1,X_2$ is $$ \frac{P(X_1 = x_1)P(X_2 = t-x_1)}{P(X_1+X_2 = t)}. $$ You have calculated everything else you need to solve the problem.