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Math Help - Poisson and Binomial Variables

  1. #1
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    Poisson and Binomial Variables

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  2. #2
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    For the Poisson random variables I get ((lamda)^x)/((e^2lambda)*x!) when y =1. Is this a Poisson Random variable? Also, when y = 0, I get ((lamda)^x)/((e^lambda)*x!) - ((lamda)^x)/((e^2lambda)*x!) I cannot get the binomial. I don't know how to show that this is Binomial as x is in the range from 1...infinity.

    For part 1 I used the fact that P(X=x and Y=y) = P(X|Y)P(Y)
    and P(Y=y) = sum(1 to x)P(X=x and Y=y) = 1, so P(X=x|Y=y)(P(y) = P(X=x and Y=y)
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  3. #3
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    Quote Originally Posted by lord12 View Post
    For the Poisson random variables I get ((lamda)^x)/((e^2lambda)*x!) when y =1. Is this a Poisson Random variable? Also, when y = 0, I get ((lamda)^x)/((e^lambda)*x!) - ((lamda)^x)/((e^2lambda)*x!) I cannot get the binomial. I don't know how to show that this is Binomial as x is in the range from 1...infinity.

    For part 1 I used the fact that P(X=x and Y=y) = P(X|Y)P(Y)
    and P(Y=y) = sum(1 to x)P(X=x and Y=y) = 1, so P(X=x|Y=y)(P(y) = P(X=x and Y=y)
    (a) You need to show that:

    1. \sum_{x = 0}^1 \sum_{y=0}^{1} p_{XY}(x,y) = 1. (Some clues to doing this are given in the solution to (b))

    2. p_{XY}(x,y) \geq 0 for x = 0, 1, 2, ..... and y = 0, 1.


    (b) p_X(x) = \sum_{y=0}^{1} p_{XY}(x,y) =\frac{\lambda^x(1 - e^{-\lambda})}{e^{\lambda} x!} + \frac{\lambda^x}{e^{2 \lambda} x!} = \frac{\lambda^x e^{-\lambda}}{x!}.


    p_Y(y) = \sum_{x=0}^{\infty} p_{XY}(x,y) = \left[ \frac{(1 - e^{-\lambda})^{1-y}}{e^{\lambda(y+1)}}\right] \sum_{x=0}^{\infty} \frac{\lambda^x}{x!} = \left[ \frac{(1 - e^{-\lambda})^{1-y}}{e^{\lambda(y+1)}}\right] e^{\lambda}  = (1 - e^{-y})^{1-y} (e^{-\lambda})^y

    which is clearly recognised as the pdf for a binomial random variable with n = 1 and p = e^{-\lambda}.


    (c) Does p_{XY}(x, y) = p_{X}(x) p_{Y}(y) .... ?
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  4. #4
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    Quote Originally Posted by lord12 View Post
    I need help
    This is the fourth infraction you have received for bumping, you are banned for a week.

    CB
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