Poisson Conditional Probability - Very lost!

Question

A discrete random variable is said to have a Poisson distribution if its possible values are the non-negative integers and if, for any non-negative integer $k$,

$$P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}$$

where $\lambda>0$. It turns out that $E(X)=\lambda$.

Minitab has a calculator for calculating Poisson probabilities, which is very similar to the calculator for Binomial probabilities.

The Poisson distribution model is widely used for modeling the number of "rare" events.

Suppose we have a Poisson random variable $X$ with mean (or expected value) equal to $2$ and another Poisson random variable $Y$ with mean $3$.

Suppose $X$ and $Y$ are independent random variables, in which case $W = X+Y$ will be a Poisson random variable with mean equal to $5 (= 2+3)$.

Find the conditional probability that $X = 5$ given that $W = 10$.

Answer 1

$$\frac{\Pr(X=5\ \&\ W=10)}{\Pr(W=10)} = \frac{\Pr(X=5\ \&\ Y=5)}{\Pr(W=10)}$$

The event $(X=5\ \&\ W=10)$ is the same as $(X=5\ \&\ Y=5)$. Then use independence of $X$ and $Y$.

Answer 2

The distribution of the conditional probability is $$f_{X\mid W}(x\mid w)f_W(w)=f_{X,W}(x,w).$$

The probability that $X=5$ and $W=10$ is the same as the probability that $X=5$ and $Y=5$, so the joint probability on the right is the same as $f_{X,Y}(x,y)$, where $Y=W-X$. (Explicitly, this is a change of coordinates, but it has unit Jacobian). Since $X$ and $Y$ are independent, the joint probability is easy to understand. Dividing it by the distribution of $W$, which we also know, gives us the conditional probability.

Answer 3

A Poisson distributed random variable measures a number of events occurring in a fixed interval if these events occur with a known average rate ($\lambda$) and independently of the time since the last event.   As you noted, the mean of a Poisson distribution is this rate, $\lambda$.

In this case, we have events occurring at rate 5, which are a combination X-type events occurring at rate 2 and Y-type events occurring at rate 3.   Hence, on average, when any event happens, 2/5 of the time it will be X-type and 3/5 it will be Y-type; that is a Bernoulli trial determines the type of each event that occurs.

So, for a given number of events, $W$, the count of X-type among them will be binomially distributed. $$X\mid W \sim \mathcal{Bin}(W, 2/5)$$

Hence, we can find : $\mathsf P(X=5\mid W=10)$