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

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Homework Help Overview

The problem involves a Poisson random variable representing the number of customers entering a store, with a focus on determining the marginal distribution of the number of women among these customers. The original poster attempts to explore the relationship between the conditional distribution of women given the total number of customers and the marginal distribution of women.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of probability relations and integration to find the marginal distribution. There are attempts to clarify the distinction between probability mass functions and probability density functions in the context of Poisson random variables. Questions arise about the solvability of integrals and the appropriateness of using summation versus integration.

Discussion Status

Some participants have provided guidance on the nature of Poisson distributions and the implications of using binomial distributions for conditional probabilities. There is an ongoing exploration of how to express the distribution of women given a fixed number of customers, with various interpretations being considered.

Contextual Notes

There is mention of missing information regarding the entry rate of customers, which may affect the analysis. The discussion also reflects on the assumptions made about the independence of customer entries and the probability of a customer being a woman.

Gullik
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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



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


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:
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Gullik said:

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



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


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.

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
 
Ray Vickson said:
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
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)


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

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