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

  1. Sep 15, 2012 #1
    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

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


    3. 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: Sep 15, 2012
  2. jcsd
  3. Sep 15, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    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
     
  4. Sep 15, 2012 #3
    Feels stupid

    The rate was λ, the copypaste ate it.

    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]


    Thanks
     
    Last edited: Sep 16, 2012
  5. Sep 16, 2012 #4
    I've solved it btw, so no one wastes any time here.
     
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