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Thread: Poisson random variable question

  1. #1
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    Poisson random variable question

    Let X and Y be independent Poisson random variables with means L and u respectively. Then it is known that X+Y is a Poisson with means L+u. Now let X1,...,X10 be independent Poisson random variables with the same mean, L=3. Find the probability that X1+...+X10=20 when you already know that X1<2.
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  2. #2
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    So this is asking to find

    $\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2)$

    So break it down. We try to work this stuff out into independent events so that we can calculate the pieces.

    Standard equality for conditional probability:
    $\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2) =P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 < 2) /P(X_1 < 2)$

    Break numerator into disjoint sets.
    $\displaystyle = (P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =1))/P(X_1 < 2)$

    Now that $\displaystyle X_1$ is fixed in the two summands, you can write the expressions as what happens to $\displaystyle X_2,\ldots,X_{10}$ and what happens to $\displaystyle X_1$ separately.
    $\displaystyle = (P(\sum_{i=2}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=2}^{10} X_i = 19 \textrm{ and } X_1 =1))/P(X_1 < 2)$

    Now we have "and"'s between independent events.
    $\displaystyle = (P(\sum_{i=2}^{10} X_i = 20)P(X_1 =0)+P(\sum_{i=2}^{10} X_i = 19)P(X_1 =1))/P(X_1 < 2)$

    Now everything is standard Poisson which you should be able to calculate.
    Last edited by meymathis; Mar 18th 2009 at 08:14 PM. Reason: Made the edit that matheagle pointed out.
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  3. #3
    MHF Contributor matheagle's Avatar
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    switched the /\

    Quote Originally Posted by meymathis View Post
    So this is asking to find

    $\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2)$

    So break it down. We try to work this stuff out into independent events so that we can calculate the pieces.

    Standard equality for conditional probability:
    $\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2) =P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 < 2) /P(X_1 < 2)$

    Break numerator into disjoint sets.
    $\displaystyle = (P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =1))/P(X_1 < 2)$

    Now that $\displaystyle X_1$ is fixed in the two summands, you can write the expressions as what happens to $\displaystyle X_2,\ldots,X_{10}$ and what happens to $\displaystyle X_1$ separately.
    $\displaystyle = (P(\sum_{i=2}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=2}^{10} X_i = 19 \textrm{ and } X_1 =1))/P(X_1 < 2)$

    Now we have "and"'s between independent events.
    $\displaystyle = (P(\sum_{i=2}^{10} X_i = 20)P(X_1 =0)+P(\sum_{i=2}^{10} X_i = 19)P(X_1 =1))/P(X_1 < 2)$

    Now everything is standard Poisson which you should be able to calculate.
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  4. #4
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    Could you please guide me to the solution?

    I'm having trouble doing the "standard poisson" part, and am getting a result of 1.

    Thank you!
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  5. #5
    MHF Contributor matheagle's Avatar
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    Your $\displaystyle \lambda $ seems to be 3.
    So the distribution is $\displaystyle P(X=x)= e^{-\lambda}\lambda^x/x! = e^{-3}3^x/x! $
    From that you can easily obtain $\displaystyle P(X=0)$ and $\displaystyle P(X=1)$
    and $\displaystyle P(X<2)=P(X=0)+P(X=1)$.

    Now for the rv $\displaystyle S=\sum_{i=2}^{10}X_i$ well thats a sum of nine Poisson's so its distribution is...

    $\displaystyle P(S=s)= e^{-9\lambda}(9\lambda)^s/s! = e^{-27}(27)^s/s! $

    Plug in 19 and 20 for s, and that's all folks.
    Last edited by matheagle; Mar 22nd 2009 at 07:49 PM.
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