Poisson law link with binomial distribution in a problem

Question

I have the following problem :

Tom is calling people. He has $10\%$ chance of calling the wrong number. Let $Y$ be a random variable which count the number of calls Tom makes. $Y$ follow a Poisson law with parameter $\lambda = 20$. Let $X$ denote the number of bad calls Tom makes.

I would like to compute the law of $X$.

First I can see that : $$\mathbb{P}(X = k | Y = n) = \binom {n} {k} 0.1^k0.9^{n-k}$$

Yet from there I don't see how I can compute the law of $X$. Since the évènement $\{Y = n\}_n$ aren't independant I can't just do the following :

$$\mathbb{P}(X = k ) = \sum_n \mathbb{P}(X = k | Y = n) = \sum_n \binom {n} {k} 0.1^k0.9^{n-k}$$

Thus I don't know how to proceed. Thank you.

Answer 1

A Poisson variable with parameter $\lambda$ counts the events that occur in one time unit of a Poisson process with rate parameter $\lambda$. If we uniformly randomly select $10\%$ of the events, the result is a Poisson process with rate parameter $\frac{\lambda}{10}$, so the count $X$ of events in one time unit is a Poisson variable with parameter $\frac{\lambda}{10}$.

Answer 2

Perhaps something along the following lines is what you're aiming for?

First, follow StubbornAtom's comment; they are correct that the total probability is not the sum of conditional probabilities, but the sum of joint probabilities. Then

\begin{align} P(X = k) & = \sum_{n=k}^\infty P(X = k, Y = n) \\ & = \sum_{n=k}^\infty P(X = k \mid Y = n) P(Y = n) \\ & = \sum_{n=k}^\infty \binom{n}{k} p^k (1-p)^{n-k} e^{-\lambda} \frac{\lambda^n}{n!} \\ & = \sum_{n=k}^\infty \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} e^{-\lambda} \frac{\lambda^n}{n!} \\ & = e^{-\lambda} \frac{1}{k!} \sum_{n=k}^\infty \frac{1}{(n-k)!} p^k (1-p)^{n-k} \lambda^n \\ & = e^{-\lambda} \frac{1}{k!} \sum_{n=k}^\infty \frac{1}{(n-k)!} p^k (1-p)^{n-k} \lambda^k \lambda^{n-k} \\ & = e^{-\lambda}\frac{(\lambda p)^k}{k!} \sum_{n=k}^\infty \frac{[\lambda(1-p)]^{n-k}}{(n-k)!} \\ & = e^{-\lambda}\frac{(\lambda p)^k}{k!} \sum_{m=0}^\infty \frac{[\lambda(1-p)]^m}{m!} \qquad [m = n-k] \\ & = e^{-\lambda}\frac{(\lambda p)^k}{k!} e^{\lambda(1-p)} \\ & = e^{-\lambda p}\frac{(\lambda p)^k}{k!} \end{align}

which was to be shown.

Some useful ideas to be carried away from this example:

• Make sure you have the right probability set-up! :-)
• Expand the binomial coefficient into factorial terms.
• Break out whatever constant terms you can outside the summation—especially those terms you want to end up there.
• Re-index summations to begin at $0$ when possible.

Obviously, these tips can't be applied willy-nilly, but they are helpful to keep in mind in case you get stuck.