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

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The discussion focuses on deriving the marginal distribution of the number of women customers (X) given that the total number of customers (Y) follows a Poisson distribution. It is established that the conditional distribution of X given Y is binomial due to the independence of customer entries and the fixed probability of a customer being a woman. The user initially struggles with the integration approach to find the marginal distribution of X but receives clarification that Poisson variables are discrete and should be summed rather than integrated. The conversation also touches on using the binomial approximation to the Poisson distribution for large values of n. Ultimately, the user confirms they have resolved their issue.
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
 
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I've solved it btw, so no one wastes any time here.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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