The classic example… $X \sim Po\left (\lambda\right ), Y \sim Po\left (\mu\right)$, X and Y are independent. Show that the conditional distribution of X is binomially distributed. Or in other words, $P(X=k\mid X+Y = n) = P (\tilde{X} = k), \tilde{X} \sim B\left (n ,\frac{\lambda}{\lambda + \mu}\right )$.
I've so far managed to reach to this step, and have been stuck since. Just somehow gotta get a $\frac{1}{n!}$ in the denominator, that would then complete the proof..or at least I think..
$$P(X=k\mid X+Y=n) = \frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{P(X+Y = n)}=
\frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{\sum_{i=1}^{n} \frac{\lambda^{i}\mu^{n-i}}{i!(n-i)!}}$$
Thanks for the help!
Best Answer
HINT
Remember that $$\sum_{i=0}^{n} \binom{n}{i} \lambda^i \mu^{n-i} = \left( \lambda + \mu\right)^n$$ The above gives us that $$\sum_{i=0}^{n} \dfrac{n!}{i! (n-i)!} \lambda^i \mu^{n-i} = \left( \lambda + \mu\right)^n$$ which inturn gives us that $$\sum_{i=0}^{n} \dfrac{ \lambda^i \mu^{n-i}}{i! (n-i)!} = \dfrac{\left( \lambda + \mu\right)^n}{n!}$$