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

  • Thread starter Gullik
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  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



Problem 2
Assume that the number Y of customers entering a store is a Poisson random variable with rate
. 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 p



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.
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
62
6
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.
Feels stupid

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).
The rate was λ, the copypaste ate it.

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?
With C=4 will it be
[itex]P(W=k|C=4)=C^4_kp^x(1-p)^{4-k}[/itex]

So
[itex]P(W=k|C=n)=C^n_kp^x(1-p)^{n-k}[/itex]?

And
[itex]P(W=k)=\Sigma^n_{i=1}P(W=k|C=i)P(C=i)[/itex]?

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


RGV
Thanks
 
Last edited:
  • #4
62
6
I've solved it btw, so no one wastes any time here.
 

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