What is the conditional probability $P(Y|X=x)$, where $Y$ is binomial with Poisson distributed $n=X$?

Question

Question

The number of patients visiting the dentist on a day follows a Poisson distribution with $\lambda= 20$. Patients can either have one or two issues, the probability of a patient having one issue is $0.7$, and two issues is $0.3$.

What is the expected total number of issues across all patients for a day?

I can figure this question out just using logic, the expected number of patients would be $20$ since the expected value of a Poisson variable is $\lambda$. And then $30\%$ will have two issues and $70\%$ will have one which equals $26$ total issues.

I am just not sure how to derive mathematically $P(Y|X=x)$ since we have to include that in our answer. I thought of treating the number of issues as binomial and then taking the total issues as $2np + nq$ and then using $E(Y) = E[E(Y|X)]$ but I am not sure that is along the lines of what the teacher asked, to derive the term $P(Y|X=x)$.

Answer 1

Let $X_1,X_2$ denote the number of patients who visited throughout the course of a day with $1$ and $2$ issues, respectively.

Then $X_1\sim \text{Poisson}(20\times 0.7)$, $X_2=\text{Poisson}(20\times 0.3)$, and $X_1+2X_2$ denotes the total number of issues treated that day.

The expected number of issues is $$\mathbb{E}(X_1+2X_2)=14+2\times 6=26$$

Answer 2

Let $X$ represent the random number of patients visiting the dentist in a day, and let $Y$ represent the number of issues that are seen by the dentist in a day. We are given that $$X \sim \operatorname{Poisson}(\lambda = 20).$$ But what is $Y$? To understand the relationship between $Y$ and $X$, we introduce a third variable, say $B$, that counts the random number of patients that visit the dentist in a day who also have two issues. Then $$B \mid X \sim \operatorname{Binomial}(n = X, p = 0.3).$$ This is because for each patient that visits, their probability of having two issues is $0.3$, independent of any other patient. So the total number of such patients, given that $X$ visited, is binomially distributed.

Therefore, $$Y \mid X = (X + B \mid X).$$ For instance, if $X = 18$, then $Y = 18 + B$ where $B$ is binomial with $n = 18$ and $p = 0.3$. This is because we count all the patients who visited for having at least one issue, then count a second time the patients who had two issues.

From this, we calculate

$$\begin{align} \operatorname{E}[Y] &= \operatorname{E}[\operatorname{E}[Y \mid X]] \\ &= \operatorname{E}[\operatorname{E}[X + B \mid X]] \\ &= \operatorname{E}[X + X p] \\ &= (1+p)\operatorname{E}[X] \\ &= (1+p)\lambda \\ &= (1.3)(20) \\ &= 26. \end{align}$$

Notice how the fact that $\operatorname{E}[Y] = (1+p)\operatorname{E}[X]$ does not depend on $X$ being Poisson distributed, only that the conditional distribution $B \mid X$ is binomial, thus $\operatorname{E}[B \mid X] = Xp$. While we can exploit the property of Poisson thinning to show that the unconditional distribution of $B$ is Poisson with rate $\lambda^* = \lambda p$, we do not need to do so.

Answer 3

What is the conditional probability $P(Y|X=x)$, where $Y$ is binomial with Poisson distributed $n=X$?

Clearly it is: $$\mathsf P(Y=y\mid X=x)~=~\dbinom xy p^y(1-p)^{x-y}~\mathbf 1_{x,y\in\Bbb N~:~0\leq y\leq x}$$

In this case, $p=0.30$ the probability that a patient will have a second issue. $Y$ being the count of patients with two issues.

$$\mathsf E(Y\mid X)~{=\sum_{y=0}^X\dbinom Xy p^y(1-p)^{X-y}\\=Xp}$$

So you may use this (a well known properties of a binomial distribution), and the Law of Total Expectation.

Thus the expected count of issues is :$\mathsf E(X+Y)~{=\mathsf E(X)+\mathsf E(\mathsf E(Y\mid X))\\=\mathsf E(X)+\mathsf E(Xp)\\=(1+p)~\lambda\\=(1.30)~ 20\\=26}$