# How to Get Covariance of Bivariate Poisson Distribution

1. Feb 22, 2012

### ahdika

Dear all, I have a problem in solving covariance of Bivariate Poisson Distribution

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

2. Feb 22, 2012

### mathman

3. Feb 26, 2012

### ahdika

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..

4. Feb 26, 2012

### ahdika

Let Xi ~ Poisson (θi) , i = 1,2,3
consider
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
Thanks

5. Feb 27, 2012

### mathman

I am not familiar with your notation. However, assuming Xi are independent, then the covariance between X and Y involves only X3.

6. Feb 27, 2012

### ahdika

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?

7. Feb 27, 2012

### ahdika

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?

8. Feb 28, 2012