Poisson Distribution with Condition

Question

Suppose that $X$ and $Y$ are independent Poisson random variables with parameter 1 and 2, respectively. Find $P(X=1|X+Y=4)$.

$X$ is $Poisson(1)$

$Y$ is $Poisson(2)$

$X+Y$ is $Poisson(3)$

If $X+Y=4$ is true, then the only way for $X=1$ is if $Y=3$.

Thus,

$P(X=1|X+Y=4)$

$=P(Y=3)$

$=\frac{e^{-2}2^3}{3!}$

$=0.180447044$

Textbook Answer:

What!

Answer 1

Note that

$$P(X=1|X+Y=4)=P(Y=3|X+Y=4)$$

We can't drop the condition.

dropping the condition means $X$ can take arbitrary value.

\begin{align}P(X=1|X+Y=4)&=P(X=1, Y=3|X+Y=4) \\&= \frac{\exp(-1)\frac{1^1}{1!}\exp(-2)\frac{2^3}{3!}}{\exp(-3)\frac{3^4}{4!}} \\ &=\frac{4!}{1!3!}\left(\frac13 \right)^1\left(\frac{2}{3}\right)^3\end{align}

Remark: what your textbook did is something called Poisson splitting.

Answer 2

In short, your answer is incorrect because the unconditional probability that $Y = 3$ says nothing about the value of $X$. You need to be more careful:

$$\Pr[X = 1 \mid X+Y = 4] = \frac{\Pr[(X = 1) \cap (X + Y = 4)]}{\Pr[X+Y = 4]} = \frac{\Pr[(X = 1) \cap (Y = 3)]}{\Pr[X + Y = 4]},$$ where the first equality is the definition of conditional probability; the second is true because of the reasoning you already provided. Then since $X$ and $Y$ are independent, we can write $$\Pr[X = 1 \mid X+Y = 4] \overset{\text{ind}}{=} \frac{\Pr[X = 1]\Pr[Y = 3]}{\Pr[X+Y = 4]}.$$ But notice how we can't simply get rid of the term in the denominator, and also notice how $X$ remains in the expression--as it clearly must.

We could now condition the denominator on the possible outcomes of $X$ and $Y$; e.g., $$\Pr[X + Y = 4] = \sum_{x=0}^4 \Pr[(X = x) \cap (Y = 4-x)] \overset{\text{ind}}{=} \sum_{x=0}^4 \Pr[X = x]\Pr[Y = 4-x].$$ Now that we have separated all probabilities in terms of the marginal (unconditional) probabilities of $X$ and $Y$, we may now apply the Poisson probability mass function and finish the computation.

However, the textbook answer also reveals a fact that you can prove as an exercise: that $$X \mid X+Y = n \sim \operatorname{Binomial}\left(n, \lambda_x/(\lambda_x + \lambda_y) \right),$$ where $X$ and $Y$ are independent Poisson random variables with rates $\lambda_x$ and $\lambda_y$, respectively.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

There are five possible pairs $\ds{\pars{x,y}}$. Namely, $\ds{\braces{\pars{0,4},\color{red}{\pars{1,3}},\pars{2,2}\pars{3,1}\pars{4,0}}}$.

\begin{align} {\pars{\expo{-1}/1!}\pars{\expo{-2}2^{3}/3!} \over \sum_{n = 0}^{4}\pars{\expo{-1}/n!}\bracks{\expo{-2}2^{4 - n}/\pars{4 - n}!}} & = {1 \over 3!}\, {1 \over (2/4!)\sum_{n = 0}^{4}{4 \choose n}\pars{1/2}^{n}} \\[5mm] & = 2\,{1 \over \pars{1 + 1/2}^{4}} = \bbx{32 \over 81} \approx 0.3951 \end{align}