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Poisson Distribution

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Let N,X1, X2, .... be independant random bariables where ?N has a poission Distribution with mean 3 while X1, X2.... each has a poisson distribution with mean 7

    Determine [tex]E[N \sum^N_{i=1} X_i][/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]E[N \sum^N_{i=1} X_i] = E[N] * E[\sum^N_{i=1} X_i] = (3)(7) = 21[/tex]

    but that can't be right.
     
  2. jcsd
  3. Mar 27, 2009 #2

    lanedance

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    Homework Helper

    Hi cse63146

    I would first look at for constant sum of m Xi's, with the Xi's representing the same distribution X independently, the sum gives:
    [tex]E[ \sum^m_{i=1} X_i] = *E[X_1 + X_2 + ... + X_m ] = m*E[X] [/tex]
     
  4. Mar 27, 2009 #3
    so the expectation would be E[N2]*E[X]

    E[N2] = Var(x) + E[N]2 - 3 + 32 = 12

    E[N2]*E[X] = (12)(7) = 84
     
  5. Mar 28, 2009 #4
    You may want to use the tower law
    [tex]\mathbb{E}[X]=\mathbb{E}[\mathbb{E}[X|Y]] [/tex]

    You need to condition on the N to pull out the sum.
     
  6. Mar 28, 2009 #5
    Let T = X1 + X2 + XN. E[T|N] = E[X1 + X2 + XN | N ]

    E[T|N] = E[X1 + X2 + XN] = N*E[X]

    E[T] = E[E[T|N]] = E[NE[X]] = E[N]*E[X]

    [tex]E[N \sum^N_{i=1} X_i] = E[N^2]E[X][/tex]

    E[N2] = Var(x) + E[N2] = 3 + 32 = 12

    E[N2]*E[X] = (12)(7) = 84.

    There's a second part of the question - Determine the variance of

    [tex]\sum^N_{i=1} X_i[/tex]

    so Var (T|N) = Var(X1+ ... + Var(XN). Since X1 X2 XN are independant: Var(T|N) = N*Var(X).

    Is that correct?
     
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