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How to Get Covariance of Bivariate Poisson Distribution

  1. Feb 22, 2012 #1
    Dear all, I have a problem in solving covariance of Bivariate Poisson Distribution

    Let X_i \sim POI (\theta_i) , i = 1,2,3
    X = X_1 + X_3
    Y = X_2 + X_3

    Then the joint probability function given :
    P(X = x, Y = y) = e^{\theta_1+\theta_2+\theta_3} \frac {\theta_1^x}{x!} \frac {\theta_2^y}{y!} \sum {k = 0}{min(x,y)} \left( \begin{array}{c} x \\ k \end{array} \right) \left( \begin{array}{c} y \\ k \end{array} \right) k! \left( \frac{\theta_3}{\theta_1 \theta_2} \right)^k

    Marginally, we get :
    X \sim POI (\theta_1+\theta_3)
    Y \sim POI (\theta_2+\theta_3)
    \theta_1, \theta_2, \theta_3 ≥ 0
    Then, the cov(X,Y) = \theta_3

    That's all information I have, but I have no idea how to get \theta_3 as the covariance of (X,Y). Please share anything you know about the way to get that value !
    Thanks a lot
  2. jcsd
  3. Feb 22, 2012 #2


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    Your notation is screwed up.
  4. Feb 26, 2012 #3
    I'm sorry, I'm a very new member here so I still confuse how to make equation correctly, I think it's just like writing equation in LaTEX..
    anyway, thanks for your reply.. I'll try to rewrite it correctly
  5. Feb 26, 2012 #4
    Let Xi ~ Poisson (θi) , i = 1,2,3
    X = X1 + X3
    Y = X2 + X3

    this two random variables X and Y follow the bivariate poisson distribution so that
    X ~ Poisson (θ1 + θ3)
    Y ~ Poisson (θ2 + θ3)

    and then the covariance of the bivariate poisson distribution is
    Cov(X,Y) = θ3

    I just don't know how to get θ3 as the covariance of this distribution.. please share me the way to get that Cov(X,Y) = θ3
  6. Feb 27, 2012 #5


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    I am not familiar with your notation. However, assuming Xi are independent, then the covariance between X and Y involves only X3.
  7. Feb 27, 2012 #6
    Oh, I am sorry.. maybe it's because the notations we usually used are different..

    Oke, I get it.. but I am confused how to explain it in mathematics equation
    Like we know,
    Cov(XY) = Cov(X1+X3,X2+X3)
    then, how must I explain about the assumption of independency between X1 and X2 in mathematics form?
  8. Feb 27, 2012 #7
    Oh, I get it..

    Correct me if I'm wrong

    Cov(X,Y) = Cov(X1+X3,X2+X3)
    = Cov(X1,X2+X3) + Cov(X3,X2+X3)
    = Cov(X1,X2) + Cov(X1,X3) + Cov(X3,X2) + Cov(X3,X3)

    but because there's assumption that Xi independent, so
    Cov(X1,X2) = 0
    Cov(X1,X3) = 0
    Cov(X3,X2) = 0

    and then I get

    Cov(X,Y) = Cov(X3,X3)
    = Var(X3)
    = θ3

    Am I wrong?
  9. Feb 28, 2012 #8


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    Your derivation is absolutely correct!
  10. Feb 29, 2012 #9
    wow, okay.. thanks a lot for your help..
    :) :) :)
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