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Given a Bivariate INAR(1) Poisson Process:

$Y_t^1 = \rho_1 * Y_{t-1}^1+R_t^1$

$Y_t^2 = \rho_2 * Y_{t-1}^1+R_t^2$

Where $R_t^1$ and $R_t^2$ are the innovation terms and follow the bivariate Poisson distribution such that marginally

$R_t^1\sim P_o (\mu_t^1-\rho_1\mu_{t-1}^1)$ and

$R_t^2\sim P_o (\mu_t^2-\rho_2\mu_{t-1}^2)$ and

$Cov(R_t^1,R_t^2)=\phi$

Derive $Cov(Y_t^1,Y_t^2)$ and hence use the method of moments to estimate $\phi$.


I started deriving:

$$Cov(Y_t^1,Y_t^2)=Cov(\rho_1 * Y_{t-1}^1+R_t^1,\rho_2 * Y_{t-1}^1+R_t^2)$$

$$Cov(Y_t^1,Y_t^2)=\rho_1 \rho_2Cov(Y_{t-1}^1,Y_{t-1}^2)+ Cov(R_t^1,R_t^2)$$ (Since $R_t$ is independent of $Y_{t-1}$, cancelation of some terms.)

Further,I assumed strict stationarity of $Y_t^1$ and $Y_t^2$ which was not mentioned in the question.

$$Cov(Y_y^1,Y_t^2)= \frac{\phi}{1-\rho_1 \rho_2}$$

I know that method of moment generating function:

$m(t)=E(e^{tX})$ Setting $t=0$ 1st derivative setting $t=0$,for mean, 2nd moment is the variance.

I cannot suss how to deal with this method of moment to get $\phi$.

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    $\begingroup$ It seems you mixed up the moment generating function and moment method in point estimate. $\endgroup$ – user158565 May 3 '19 at 3:05
  • $\begingroup$ I checked the point estimate on the net. I still do not understand it. $\endgroup$ – Tosh May 3 '19 at 5:37
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    $\begingroup$ Try to find a book to learn the method of moment. I do not think it is so simple that can be answered here. $\endgroup$ – user158565 May 4 '19 at 21:05

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