- #1

Gullik

- 62

- 6

## Homework Statement

Problem 2

Assume that the number Y of customers entering a store is a Poisson random variable with rate λ

. Let X denote the number of these customers being a woman. The probability that a customer

is a woman is denoted by p. Also, assume that all customers enter the store independently.

a) Explain why the conditional distribution of X given Y = y is binomial.

b) Show that the marginal distribution of X is Poisson with rate pλ

## Homework Equations

[itex]\Gamma(\alpha)=∫^\infty_0 t^{\alpha-1}e^{-t} dt[/itex]

## The Attempt at a Solution

I'm having troubles with b.

I think I should use the relation [itex]P(X=x|Y=y)=\frac{P(X=x,Y=x)}{P(Y=y)}[/itex] and [itex]f(x)=\int^\infty_0 f(x,y)dy[/itex] to get [itex]f(x)=\int^\infty_0 f(x|y)f(y)dy[/itex].

When I then put in probability distributions I get [itex]f(x)=∫^\infty_0 C_x^y*p^x(1-p)^{y-x}*\frac{\lambda^y*e^{\lambda}}{x!}dy=\frac{p^x}{x!*(1-p)^x}\int^\infty_0 \frac{((1-p)\lambda)^ye^{-\lambda}}{(y-x)!}[/itex]

Is that a solvable integral or is there an easier method? I'm thinking of trying to make a substitution so a get a gamma function out of it, but I can't get it right.

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