# Method of moment through covariance derivation

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 $$\phi$$.

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