# Finding joint and conditional distribution of a Poisson random variable and the sum of Poisson random variables

Let $$X_1,\dots,X_n$$ be independent Poisson random variables where $$X_j$$ has mean $$p_j\lambda_j$$ and let $$Y = \sum_{j=1}^n X_j$$.

1. How is the joint distribution between $$X_j$$ and $$Y$$ (or $$P(X_j=x_j,Y=y)$$) defined?

2. Is the conditional distribution $$P(X_j = x_j | Y = y)$$ simply the joint distribution divided by a Poisson distribution with mean $$\sum_{j=1}^n p_j \lambda_j$$?

• The very closely related thread at stats.stackexchange.com/questions/429564 essentially answers this question. Indeed, once you observe that $X_1+\cdots+X_{j-1}+X_{j+1}+\cdots+X_n$ is a Poisson variable independent of $X_j,$ that thread fully answers both your questions. In the language of my answer, all you need do is color events in the process associated with $X_j$ with one color and color all other events with another color.
– whuber
Commented Jan 11, 2021 at 17:02

The support of the random vector $$(X_j, Y)$$ is the set $$\mathcal A = \left \{ (p,q) \in \mathbb N^2 \mid p \leq q \right \}$$.

The distribution of $$(X_j, Y)$$ is given by the probabilities this vector takes for each elements of $$\mathcal A$$. Thus let $$p \leq q$$, then:

\begin{align*} \mathbb P\left (X_j = p, Y= q \right ) &= \mathbb P \left (Y=q \mid X_j = p \right ) \mathbb P \left ( X_j = p \right ) \\ &= \mathbb P \left (X_j + \sum_{i \neq j} X_i = q \mid X_j = p \right ) P \left ( X_j = p \right ) \end{align*}

Since $$X_j \sim \mathcal P(p_j \lambda_j)$$, $$\mathbb P \left ( X_j = p \right ) = e^{-p_j\lambda_j}\frac{(p_j\lambda_j)^p}{p!}$$.

For the remaining probability we can use the simplification \begin{align*} \mathbb P \left (X_j + \sum_{i \neq j} X_i = q \mid X_j = p \right ) &= \mathbb P \left ( \sum_{i \neq j} X_i = q-p \right ) \end{align*}

and the fact that the sum of independent Poisson random variables is a Poisson random variable whose rate is the sum of the individual rates. Thus let $$\rho_j = \sum_{i\neq j} p_i\lambda_i$$, we have:

\begin{align*} \mathbb P \left ( \sum_{i \neq j} X_i = q-p \right ) = e^{-\rho_j}\frac{\rho_j^{q-p}}{(q-p)!} & & (1) \end{align*}

and finally,

\begin{align*} \mathbb P\left (X_j = p, Y= q \right ) &= e^{-\rho_j}\frac{\rho_j^{q-p}}{(q-p)!} e^{-p_j\lambda_j}\frac{(p_j\lambda_j)^p}{p!} \\ &= e^{-\sum p_j\lambda_j} \frac{\rho_j^{q-p}(p_j\lambda_j)^p}{(q-p)!p!} \end{align*}

The conditonal distribution of $$X_j \mid Y$$ is a Binomial distribution.

First, since $$X_j \leq Y$$, the support of $$X_j \mid Y=y$$ is $$\{0,\dots,y \}$$.

Let $$p \in \{0,\dots,y \}$$, the Bayes formula gives us:

\begin{align*} \mathbb P(X_j = p \mid Y = y) = \frac{\mathbb P(Y = y \mid X_j = p)\mathbb P(X_j = p)}{\mathbb P(Y = y)} & & (2) \end{align*}

From what $$(1)$$ we have:

\begin{align*} \mathbb P(Y = y \mid X_j = p) = e^{-\rho_j}\frac{\rho_j^{y-p}}{(y-p)!} & & (3) \end{align*} Moreover, \begin{align*} \mathbb P(Y=y) = \exp \left(-\sum_j p_j \lambda_j \right)\frac{(\sum_j p_j\lambda_j)^y}{y!} & & (4) \end{align*} and

\begin{align*} \mathbb P(X_j=p) = \exp \left(- p_j \lambda_j \right)\frac{(p_j\lambda_j)^p}{p!} & & (5) \end{align*}

Combining $$(3)$$, $$(4)$$ and $$(5)$$ into $$(2)$$ we get:

\begin{align*} \mathbb P(X_j = p \mid Y = y) &= \frac{e^{-\rho_j}\frac{ \rho_j^{y-p}}{(y-p)!} e^{-p_j \lambda_j}\frac{(p_j\lambda_j)^p}{p!}}{e^{-\sum_j p_j \lambda_j}\frac{(\sum_j p_j\lambda_j)^y}{y!}} \\ &= \binom{y}{p}\frac{e^{-\rho_j + p_j\lambda_j}}{e^{-\sum p_j\lambda_j}} \times\frac{\rho_j^{y-p}(p_j\lambda_j)^p}{(\sum p_j\lambda_j)^y}\\ \end{align*}

Since $$\rho_j + p_j = \sum p_j\lambda_j$$ the last line can be simplified into:

$$\mathbb P(X_j = p \mid Y = y) =\binom{y}{p} \left( \frac{\rho_j}{\sum p_j\lambda_j}\right)^{y-p}\left(1- \frac{\rho_j}{\sum p_j\lambda_j}\right)^p$$

which the probability mass function of a Binomial distribution of parameters $$\left(y,1- \frac{\rho_j}{\sum p_j \lambda_j}\right).$$