Interesting probability question involving a sum of Bernoulli

Question

I have an interesting question but I am a bit confused on a certain aspect of it and I am hoping I can get it cleared up. I am also not very confident in my answers so it is very possible I am making some big or small mistakes.

Suppose that in any 1 hour time period, either a person shows up to wait for a new doctor, or no one shows up, with probabilities $p$ and $(1-p)$ respectively.

Now suppose we check the waiting room after M hours, where $M$ is Poisson distributed with parameter $\lambda$.

I was asked,

What would

$P[S_{M}=k]$ be for $k=0,1,.... $and

$E[M|S_{M}=k]$

Il explain what I have and where I am confused now.

$P[S_{m}=0]=(1-p)^{m}$

$P[S_{m}=1]=(m | 1) (1-p)^{m-1}(p)$ . . .

$P[S_{m}=k]=(m|n)(1-p)^{m-k}(p)^{k}$

( where (m|k) stands for m choose k)

Now the issue here is that I am calculating the probability using little m, not big random variable M like in the question

But $P[S_{M}=k]= \sum_{m=k}^{\infty} P[S_{m}=k]P[M=m]$ moreover M is Poisson so we know its pdf.

So is that as simplified as I could make it?

Essentially, $$P[S_{M}=k]=P[S_{k}=k]P[M=k]+P[S_{k+1}=k]P[M=k+1]+...$$

For the expected value part I am more lost, so far I have used Bayes theorem to get that

$$P[M=m|S_{M}=k]= \frac{P[S_{m}=k]P[M=m]}{P[S_{M}=k]}$$

but this to isnt very neat, it consists of $(m|k)(1-p)^{m-k}(p)^{k} \lambda^{m}e^{- \lambda}/m!$ all divided by the original expression,

thus by best thought would be that using law of total expectation

$$E[M|S_{M}=k]=P(M=1|S_{M}=k)+2P(M=2)|S_{M}=k]+....$$

but again I am not sure this is clean or correct,

I am looking for some guidance and help on understanding either where I went wrong, or how to understand it better.

Also I believe I could represent M as a sum of the events that a person showed versus they did not. For example, could expressions for other such questions be asked such as the $E[M_{0}|S_{M}=k]$ where $M=M_{0}+M_{1}$ where these represent the total number of hours that a person did show up and the total hours that they did not, summed.

Thanks all in advance.

Answer 1

We have $S_M\mid M~\sim~ \mathcal{Bin}(M, p)$ and $M\sim\mathcal{Pois}(\lambda)$

That is $M$ is the count of poisson arrivals, and $S_M$ is the count of successes among them.   (A success is "no one shows up in that hour", and $M$ is the count of hours checked.)

So indeed $$\begin{align}\mathsf P(S_M{=}k) ~&=~ \sum_{m=k}^\infty \mathsf P(S_M{=}k\mid M{=}n)~\mathsf P(M{=}m) \\[1ex] &=~ \sum_{m=k}^\infty\dfrac{m!~p^k(1-p)^{m-k}}{k!(m-k)!}\cdot\dfrac{\lambda^m\mathsf e^{-\lambda}}{m!}\\[1ex] &=~ \dfrac{\lambda^k~p^k~e^{-\lambda}}{k!}\sum_{m-k=0}^\infty\dfrac{(1-p)^{m-k}\lambda^{m-k}}{(m-k)!}\\[1ex] &=~ \dfrac{(\lambda p)^k~e^{-\lambda}}{k!}\sum_{n=0}^\infty\dfrac{((1-p)\lambda)^{n}}{n!}\end{align}$$

Which you may simplify by recalling that $\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$

It should yeild that $S_M\sim\mathcal{Pois}(\lambda p)$ . Successes occur in a Poisson distribution with rate $\lambda p$.

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And $\mathsf E(M\mid S_M=k) = \sum_{m=k}^\infty m~\mathsf P(M=m\mid S_M=k)$

Where $\mathsf P(M=m\mid S_M=k) $ is the probabiluity for $m-k$ failures, given $k$ successes, among the possion arrivals ...

Then $\mathsf E(M\mid S_M=k)$ is the expected number of successes and failures when given there are $k$ successes...

Also I believe I could represent M as a sum of the events that a person showed versus they did not. For example, could expressions for other such questions be asked such as the $E[M_0|S_M=k]$ where $M=M_0 +M_1$ where these represent the total number of hours that a person did show up and the total hours that they did not, summed.

Indeed, and further that $M_1= S_M$; the count of hours no one showed.

So $\mathsf E(M\mid S_M=k) = k+\mathsf E(M_0\mid M_1=k)$.

Thus, $\mathsf E(M\mid S_M=k)= k+(1-p)\lambda$. ... why?