Recent content by ahdika

wow, okay.. thanks a lot for your help.. :) :) :)

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...

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...

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...

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

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...