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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.

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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

\begin{align} & P(X_1 = y , X_2 = u_1 + y, X_3 = u_2 - 2y) \\ = & \begin{cases} \frac{e^{-(\lambda_1 + \lambda_2 + \lambda_3)}}{y!(u_1 + y)!(u_2 - 2y)!}\lambda_1^y\lambda_2^{u_1 + y}\lambda_3^{u_2 - 2y} & y = 0, \ldots, \lfloor u_2/2 \rfloor \\ 0 & \text{otherwise.} \end{cases} \tag{1} \end{align}

The above argument may shed some light on how to deal with the denominator. To begin with, apply the law of total probability: \begin{align} & P(X_2 = u_1 + X_1, X_3 = u_2 - 2X_1) \\ = & \sum_{z = 0}^{\lfloor u_2/2 \rfloor}P(X_2 = u_1 + X_1, X_3 = u_2 - 2X_1, X_1 = z) \\ = & \sum_{z = 0}^{\lfloor u_2/2 \rfloor}P(X_2 = u_1 + z, X_3 = u_2 - 2z, X_1 = z) \\ = & \sum_{z = 0}^{\lfloor u_2/2 \rfloor} \frac{e^{-(\lambda_1 + \lambda_2 + \lambda_3)}}{z!(u_1 + z)!(u_2 - 2z)!}\lambda_1^z\lambda_2^{u_1 + z}\lambda_3^{u_2 - 2z} \tag{2}. \end{align}

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.

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