# Conditional probability of independent Poisson random variables

Let $X_{j}; j = 1,2,3$ be independent Poisson distribution with mean $\lambda_{j} ; j = 1,2,3$. I want to find the conditional distribution of

$P(X_{1} = y|X_{2}-X_{1}=u_{1}, X_{3}+2X_{1}=u_{2})$, where $u_{1}+u_{2}=U$.

My work is as follows.

$P(X_{1} = y|X_{2}-X_{1}=u_{1}, X_{3}+2X_{1}=u_{2}) = \frac{P(X_{1} = y, X_{2}-X_{1}=u_{1}, X_{3}+2X_{1}=U-u_{1})}{P(X_{2}-X_{1}=u_{1}, X_{3}+2X_{1}=U-u_{1})}$

Then, I can further rewrite the above as,

$\frac{P(X_{1} = y, X_{2}=u_{1}+y, X_{3}=U-u_{1}-2y)}{P(X_{2}-X_{1}=u_{1}, X_{3}+2X_{1}=U-u_{1})} = \frac{P(X_{1} = y) P(X_{2}=u_{1}+y)P(X_{3}=U-u_{1}-2y)}{P(X_{2}-X_{1}=u_{1}, X_{3}+2X_{1}=U-u_{1})}$

Would it be right? Also, I don't know how to deal with denominator. If you have any ideas please share with me.

There seems no need to introduce the additional symbol $u$ -- you can present your answer in terms of $y, u_1, u_2$.
Your treatment to numerator is correct. Just one more thing that you might want to pay attention to -- the domain of pmf. To be complete, always display the domain of $y$ explicitly as follows: given $u_2 \geq 0$ and $X_3$ is Poisson, $u_2 - 2y$ must be at least zero, which gives an upper bound of $y$, i.e., $y \leq \lfloor u_2/2 \rfloor$. Hence
You can get the final result by combining $(1)$ and $(2)$. Some common factors may be cancelled with each other, but further simplification seems difficult due to the range of $z$ in the summation.
In general, without special constraints such as $X_3 + 2X_1 = u_2$, the conditional distribution of $(X_1, X_2, X_3)$ given their total $U$ is multinomial with parameter $U$ and component probabilities $\lambda_j/(\lambda_1 + \lambda_2 + \lambda_3), j = 1, 2, 3$. The derivation is exactly the same as this exercise.