Dear all, I have a problem in solving covariance of Bivariate Poisson Distribution(adsbygoogle = window.adsbygoogle || []).push({});

Let X_i \sim POI (\theta_i) , i = 1,2,3

Consider

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

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# How to Get Covariance of Bivariate Poisson Distribution

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