# Statistics - Finding marginal distribution through integration (I think?)

1. Sep 15, 2012

### Gullik

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$\Gamma(\alpha)=∫^\infty_0 t^{\alpha-1}e^{-t} dt$

3. The attempt at a solution
I'm having troubles with b.
I think I should use the relation $P(X=x|Y=y)=\frac{P(X=x,Y=x)}{P(Y=y)}$ and $f(x)=\int^\infty_0 f(x,y)dy$ to get $f(x)=\int^\infty_0 f(x|y)f(y)dy$.

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

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.

Last edited: Sep 15, 2012
2. Sep 15, 2012

### Ray Vickson

Poisson random variables are discrete: they take only values 0,1,2,3,... so they do not have probability density functions and you cannot integrate. Instead, they have probability mass functions and you have to sum.

Instead of X and Y I will use C = number of customers and W = number of women, within some specified time interval of length t. So, C is Poisson with mean m = r*t (where r = entry rate: your post did not actually give a value here).

Now suppose C = 4 (4 customers enter); can you give the distribution of W? (That would be P{W = k|C = 4} for k = 0,1,2,3,4.) Suppose, instead, that 100 customers enter; can you now give the distribution of W? (That would be P{W = k|C = 100} for k = 0,1,2,...,100.) So, what is P{W = k|C = n} for k = 0,1,2,...,n?

RGV

3. Sep 15, 2012

### Gullik

Feels stupid

The rate was λ, the copypaste ate it.

With C=4 will it be
$P(W=k|C=4)=C^4_kp^x(1-p)^{4-k}$

So
$P(W=k|C=n)=C^n_kp^x(1-p)^{n-k}$?

And
$P(W=k)=\Sigma^n_{i=1}P(W=k|C=i)P(C=i)$?

Or can I say that $\lim_{n\to\infty}bin(n,p)\approx poiss(np)$

Thanks

Last edited: Sep 16, 2012
4. Sep 16, 2012

### Gullik

I've solved it btw, so no one wastes any time here.

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