I need to show that if Xi ∼ Poisson(λi), i = 1, 2, . . . , k are independent then the conditional distribution of X1 given X1 + X2 + . . .Xk is Binomial and determines the parameters of this Binomial distribution. I tried:
$$ P(X1 = x| X1+...+Xk = y)$$
$$=\frac{P(X1 = x)P(X1+...+Xk = y|X1= x)}{P(X1 +...+Xk = y)}$$
$$=\frac{P(X1=x, X1+...+Xk = y)}{P(X1+...+Xk = y)}$$
$$=\frac{P(X1 = x)P(X2+...+Xk = y-x)}{\sum_{m=0}^yP(X1=m)P(X2+...+Xk = y-m) }$$
$$=\frac{λ1^xexp(-λ1)/x! *\frac{(λ2 + ...+λk)^{y-x}exp(-[λ2+...+λk])}{(y-x)!} }{\sum_{m=0}^y\frac{λ1^mexp(-λ1)}{m!} * \frac{(λ2 + ...+λk)^{y-m}exp(-[λ2+...+λk])}{(y-m)!} }$$
Then I am completely stuck. Am I on the right track? How do I get the $\frac{n!}{k!(n-k)!}$ for the $nCk$ in the Binomial distribution? Any help is greatly appreciated.
-- Edit: I have managed to get to the following:
$$=\frac{\frac{λ1^xexp(-λ1)}{x!} *\frac{(λ2 + ...+λk)^{y-x}exp(-[λ2+...+λk])}{(y-x)!} } { \frac{(λ1 + ...+λk)^{y}exp(-[λ1+...+λk])} { y!}}$$ $$= \frac{y!}{x!{y-x}!}*\frac{λ1^x (λ2+...+λk)^{y-x}}{(λ1+...+λk)^y}$$
where I think I got the $nCk$ but not the $p^k(1-p)^{n-k}$ bit. How can I simplify this step further? Sorry if this is really straightforward. What is $λ1+...+λk$? Can they sum up to 1?
Your numerator is correct at $$P(X_1=x, X_2+\cdots+X_k=y-x)=\frac{e^{-\lambda_1}\lambda_1^x}{x!}\cdot \frac{e^{-(\lambda_2+\cdots+\lambda_k)}(\lambda_2+\cdots+\lambda_k)^{y-x}}{(y-x)!} $$ but your denominator is needlessly complex. Recall that, by independence, $X_1+\cdots+X_k$ has Poisson($\lambda_1+\cdots+\lambda_k$) distribution, so the denominator is simply $$ P(X_1+\cdots+X_k=y)=\frac{e^{-(\lambda_1+\cdots+\lambda_k)}(\lambda_1+\cdots+\lambda_k)^y}{y!}. $$ Computing the ratio of numerator over denominator, you should get some cancellation, and ${}_yC_x$ should pop out.
As a hint to uncover the binomial distribution, it will help to write $(\lambda_1+\cdots+\lambda_k)^y$ in the form $(\lambda_1+\cdots+\lambda_k)^x(\lambda_1+\cdots+\lambda_k)^{y-x}$.