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 $X=\sum_j I_j$.


$$ P(I_j=1) = 1-P(I_j=0) = 1- \left(P(I_j=1 \mid S_j) P(S_j) + P(I_j=1 \mid S_j^c) P(S_j^c) \right) = 0.65, $$

$X$ 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?


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

  • $X\sim\operatorname{Poisson}(p\lambda)$,
  • $Y\sim\operatorname{Poisson}((1-p)\lambda)$,
  • $X,Y$ are independent.

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

  • $N\sim\operatorname{Poisson}(\lambda)$,
  • $X\mid N\sim\operatorname{Binomial}(N,p)$,
  • $Y\mid N\sim\operatorname{Binomial}(N,1-p)$.
  • $X,Y\mid N$ are perfectly negatively correlated, i.e. the conditional correlation between $X$ and $Y$ given $N$ is $-1$.

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