The number of people who visit the Leftorium store in a day is $\operatorname{Pois}(\lambda)$. Suppose that 10% of customers are sinister (left-handed), and 90% are dexterous (right-handed). Half of the sinister customers make purchases, but only a third of the dexterous customers make purchases. The characteristics and behavior of people are independent, with probabilities as described in the previous two sentences. On a certain day, there are 42 people who arrive at the store but leave without making a purchase. Given this information,what is the conditional PMF of the number of customers on that day who make a purchase?
Let $N$ be the total number of customers coming at that particular day, let $S_i$ be the event that the $j^\text{th}$ customer is sinister. Then we have
\begin{align}
N \sim \operatorname{Pois}(\lambda) \\
S_j \sim \operatorname{Bern}(0.1) \\
\end{align}
Let $Y$ be the number of customers who make a purchase and $X$ the number who don't make a purchase, and let $I_j$ be an indicator random variable taking on value $1$ if person $i$ does NOT make purchase and $0$ otherwise. Then $N=X+Y$ and $X=\sum_j I_j$.
As
$$
P(I_j=1) = 1-P(I_j=0) = 1- \left(P(I_j=1 \mid S_j) P(S_j) + P(I_j=1 \mid S_j^c) P(S_j^c) \right) = 0.65,
$$
$X$ is the sum of i.i.d. Bernoulli random variables, and hence we have $X \mid N=n \sim \operatorname{Bin}(n,0.1)$. Marginally, $X$ follows a Poisson distribution with rate $0.65\lambda$ (I do not show this for the sake of brevity). The marginal distribution of the number of customers who do actually buy something can the be computes with
\begin{align}
P(Y=y \mid X=x) &= P(N=x+y, X=x) \\
&= \frac{P(X=x \mid N=x+y) P(N=x+y)}{P(x=x)}\\
&= \frac{(0.35\lambda)^y e^{-0.35\lambda}}{y!}
\end{align}
Such that $P(Y=y \mid X=x) = P(Y=y)$ and $Y \sim \operatorname{Pois}(0.35\lambda)$ Suprisingly, It seems like the number of customers who didn't buy gives us no information about the number of customers who made a purchase. Is this actually right?
Best Answer
It is correct.
You have $N\sim\operatorname{Poisson}(\lambda)$ and $X\mid N\sim\operatorname{Binomial}(N,p)$ and $Y=N-X$.
A standard result says that in that case
The first and second bullet points in a sense both say the same thing.
Notice that if you assume these three bullet points as hypotheses, you can conclude that