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Let say that $X_1,\dots ,X_m$ are independent random variables following Poisson law of parameter $λ_1,\dots, λ_m$.

I'm looking for the the conditional law of $X_1$ conditionated by $\{X_1+X_2=k\}$ which is said to be a Binomial law $B(k,\frac{\lambda_1}{\lambda_1+\lambda_2})$.

Yet, $\sum\limits_{k\in X_2(\Omega)}{P(X_2=k-j)}$ has sense while $k-j\ge 0$.

Thus \begin{align*} \sum_{k\in X_2(\Omega)}{P(X_2=k-j)}&=\sum\limits_{k-j>=0}{\frac{\lambda_2^{k-j}e^{-\lambda_2}}{(k-j)!}}\\[10pt] &= e^{-2\lambda_2} \end{align*}

Therefore, we can calculate $(PX_1=j\mid X_1+X_2=k)$

\begin{align*} (PX_1=j\mid X_1+X_2=k)&=\frac{\frac{\lambda_1e^{-\lambda_1-2\lambda_2}}{j!}}{\frac{(\lambda_1+\lambda_2)^ke^{-\lambda_1+\lambda_2}}{k!}} \end{align*}

Here I'm stuck, I don't think I can find a Binomial law from here... Have you any hint? Did I di a mistake?

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\begin{align} & \Pr(X_1 = j\mid X_1+X_2=k) = \frac{\Pr(X_1=j\ \&\ X_1+X_2=k)}{\Pr(X_1+X_2=k)} \\[12pt] = {} & \frac{\Pr(X_1=j\ \&\ X_2=k-j)}{\Pr(X_1+X_2=k)} = \frac{\Pr(X_1=j)\Pr(X_2 = k-j)}{\Pr(X_1+X_2=k)} \\[12pt] = {} & \frac{\left( \dfrac{\lambda_1^j e^{-\lambda_1}}{j!} \cdot \dfrac{\lambda_2^{k-j} e^{-\lambda_2}}{(k-j)!} \right)}{\left( \dfrac{(\lambda_1+\lambda_2)^k e^{-(\lambda_1+\lambda_2)}}{k!} \right)} = \frac{k!}{j!(k-j)!} \cdot \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^j \cdot \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^{k-j} \\[15pt] = {} & \binom k j \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^j \cdot \left( 1 - \frac{\lambda_1}{\lambda_1+\lambda_2}\right)^{k-j} \end{align}

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  • $\begingroup$ I missed the last step, that's why... $\endgroup$ – ThePassenger Nov 2 '15 at 18:20
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Hint: For the top, you want $\Pr(X_1=j \cap X_2=k-j)$. No summation, $k$ is fixed.

This is $$e^{-\lambda_1}\frac{\lambda_1^j}{j!}e^{-\lambda_2}\frac{\lambda_2^{k-j}}{(k-j)!}.$$ Now there will be nice cancellation. If trouble persists, I can finish the calculation.

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