# Finding conditional probability of two independent poison rv

Let $X$ and $Y$ be identically independent Poisson r.v. with parameter $\lambda$. Find $P(X=k|X+Y=n )$.

### Attempt

By defitnion,

$$P(X=k|X+Y=n ) = P(X=k | Y=n-k) = \frac{ p_{XY}(k,n-k)}{p_Y(n-k)}$$

now since independence we have $p_X(k)p_Y(n-k)$ and so we have

$$P(X=k|X+Y=n ) = p_X(X=k)$$ but this is not the answer I should get. What am I doing wrong in this problem?

• Your "definition" is actually an incorrect statement. Maybe you switched with the correct $P(X+Y=n\mid X=k)=P(Y=n-k\mid X=k)$ (which however is not a definition). – drhab Apr 27 '18 at 11:19

It is not correct to state that $P(X=k\mid X+Y=n)=P(X=k\mid Y=n-k)$.

\begin{aligned}P\left(X=k\mid X+Y=n\right)P\left(X+Y=n\right) & =P\left(X=k\wedge X+Y=n\right)\\ & =P\left(X=k\wedge Y=n-k\right)\\ & =P\left(X=k\right)P\left(Y=n-k\right)\\ & =e^{-\lambda}\frac{\lambda^{k}}{k!}e^{-\lambda}\frac{\lambda^{n-k}}{\left(n-k\right)!}\\ & =e^{-2\lambda}\frac{\lambda^{n}}{n!}\binom{n}{k} \end{aligned} \tag1

Further if $X,Y\sim\mathsf{Poisson}\left(\lambda\right)$ are independent then $X+Y\sim\mathsf{Poisson}\left(2\lambda\right)$ leading to:$$P\left(X+Y=n\right)=e^{-2\lambda}\frac{\left(2\lambda\right)^{n}}{n!}\tag2$$

Based on $(1)$ and $(2)$ we find: $$P\left(X=k\mid X+Y=n\right)=\binom{n}{k}2^{-n}$$

• you are doing what I did in third equality. how is that version diferent then mine – James Apr 27 '18 at 11:20
• You are dividing by $p_Y(n-k)$ which is wrong. You should divide by $P(X+Y=n)$. – drhab Apr 27 '18 at 11:26
• im so confused right now – James Apr 27 '18 at 11:27
• What makes you think that $P(X=k\mid X+Y=n)=P(X=k\mid Y=n-k)$? That is your essential mistake. This wrong statement leads indeed to $P(X=k\mid X+Y=n)=P(X=k)$ which is evidently false. E.g. $P(X=n+1)>0$ but $P(X=n+1\mid X+Y=n)=0$. – drhab Apr 27 '18 at 11:30