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

Question

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 $\sum_{j=1}^n p_j \lambda_j$?

Answer 1

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*}

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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 $\left(y,1- \frac{\rho_j}{\sum p_j \lambda_j}\right).$