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Problem setup:

$X_1=Y_1+Y_0,X_2=Y_2+Y_0$ where $Y_1, Y_2\text{ and }Y_0$ are independent Poisson random variables with parameters $θ_1, θ_2\text{ and }θ_0$, respectively.

I know that the joint probability function of bivariate Poisson distribution is given by:

$P(X_1 = x_1, X_2 = x_2) = e^{-(\theta_{1}+\theta_{2}+\theta_{0})} \displaystyle\frac{\theta_{1}^{x_1}}{x_1!}\frac{\theta_{2}^{x_2}}{x_2!} \sum_{i=0}^{min(x_1,x_2)}\binom{x_1}{i}\binom{x_2}{i}i!\left(\frac{\theta_{0}}{\theta_{1}\theta_{2}}\right)^{i}$

Also, the joint probability generating function is: $exp⁡{(θ_1 (t_1-1)+θ_2(t_2-1)+θ_0(t_1 t_2-1))}$

My question is: how can we derive this probability generating function from the joint probability function?

I tried but I couldn't find this result! If anyone knows, please give me a helping hand.

Also, how can I derive the probability generating function in general for the multivariate case? Thanks in advanced.

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  • $\begingroup$ In general, I believe you can calculate the generating function $P[X_1]$ from the joint generating function $P[X_1,X_2]$ by noting that $P[X_1] = \cfrac{\partial P[X_1,X_2]}{\partial X_1}$ evaluated at $(X_1,X_2)=(1,1)$. Unfortunately, the added $\theta$ and $Y$ variables make this more convoluted, so I'm not sure if this would help. $\endgroup$ Aug 17, 2015 at 15:36
  • $\begingroup$ I found in the book that the PGF of bivariate Poisson distribution is given by:$exp⁡{(θ_1 (t_1-1)+θ_2(t_2-1)+θ_0(t_1 t_2-1))}$, I asked my teacher and he said to me that I can derive it from the joint probability function (noting that $Y_i$ are independent Poisson random variables with the previous parameters. $\endgroup$
    – Lean Bader
    Aug 17, 2015 at 15:43

1 Answer 1

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Since $X_1=Y_1+Y_0,$ and $X_2=Y_2+Y_0$, the joint pgf $E\left[t_1^{X_1} t_2^{X_2}\right]$ can be neatly found as: $$E\left[t_1^{Y_1+Y_0} t_2^{Y_2+Y_0}\right]$$ The latter does not require the joint pmf of $(X_1,X_2)$ ... just the simple joint pmf of $(Y_0,Y_1,Y_2)$ [recall the $Y_i$ are independent Poisson random variables].

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  • $\begingroup$ That what I tried before. I used this rule but I confused about how can I take $x_i$ from the joint pmf and put instead $y_i$?, $\endgroup$
    – Lean Bader
    Aug 17, 2015 at 16:06
  • $\begingroup$ What is the joint pmf of $(Y_0,Y_1,Y_2)$? $\endgroup$
    – wolfies
    Aug 17, 2015 at 16:35
  • $\begingroup$ This is my confusing. I don't know! $\endgroup$
    – Lean Bader
    Aug 17, 2015 at 16:38
  • $\begingroup$ I begin solving this as follows: $$E\left[t_1^{Y_1+Y_0} t_2^{Y_2+Y_0}\right]=E\left[t_1^{Y_1}t_2^{Y_2}(t_1t_2)^{Y_0}\right]$$, then: $$= t_1^{Y_1}t_2^{Y_2}(t_1t_2)^{Y_0}e^{-(\theta_{1}+\theta_{2}+\theta_{0})} \displaystyle\frac{\theta_{1}^{x_1}}{x_1!}\frac{\theta_{2}^{x_2}}{x_2!} \sum_{i=0}^{min(x_1,x_2)}\binom{x_1}{i}\binom{x_2}{i}i!\left(\frac{\theta_{0}}{\theta_{1}\theta_{2}}\right)^{i}$$, then how can I change $x_i$ or $Y_i$?, I lost here. $\endgroup$
    – Lean Bader
    Aug 17, 2015 at 16:47
  • $\begingroup$ Thanks . I finally derive it. $\endgroup$
    – Lean Bader
    Aug 18, 2015 at 11:44

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