Finding the covariance of $u_{it} = \nu _{it} - \theta \nu _{i\left ( t-1 \right )}$ for $t>1$

Question

Given $u_{it} = \nu _{it} - \theta \nu _{i\left ( t-1 \right )}$ for $t>1$

$u_{i1} = \nu _{i1}$

and the $\nu _{it}$ are white noise with variance equal to $\sigma^{2}$.

I can find the expected value and the variance.

Since $u_{it} = -\sum_{j=0}^{\infty }\theta^{j} \nu_{i\left ( t-j \right )}$ and $$E\left ( u_{it} \right ) = E\left [ -\sum_{j=0}^{\infty }\theta^{j} \nu_{i\left ( t-j \right )} \right ] = 0.$$

Also $$Var\left ( u_{it} \right ) = Var\left [ -\sum_{j=0}^{\infty }\theta^{j} \nu_{i\left ( t-j \right )} \right ] = \sigma^{2}\frac{1}{1 - \theta^{2}}$$

But I am not sure about the covariance.

Is it the one I derived below?

$$Cov\left ( u_{it},u_{i\left ( t+k \right )} \right ) = \theta^{k} Var(u_{it}) = \theta^{k}\frac{\sigma^{2}}{1 - \theta^{2}}$$

This relates to the question I asked here which didn't get too many views and hence no response.

Answer 1

I'm assuming that the first equation was supposed to be $ u_{it} = v_{it} - \theta v_{i(t-1)}$. It seems like you are looking at this as a recurrence relation and solving from there, but this equation has no recurrence, so I'm not sure where you are getting that from. Following from this equation, \begin{equation} cov(u_{it},u_{i(t+k)}) = cov(v_{it} - \theta v_{i(t-1)},v_{i(t+k)} - \theta v_{i(t+k-1)}) \end{equation} \begin{equation} = cov(v_{it},v_{i(t+k)}) - \theta cov(v_{it},v_{i(t+k-1)}) - \theta cov(v_{i(t-1)},v_{i(t+k)}) + \theta^2 cov(v_{i(t-1)},v_{i(t+k-1)}) \end{equation} \begin{equation} = \sigma^2(\delta[k] - \theta\delta[k-1] - \theta\delta[k+1] + \theta^2\delta[k]) \end{equation} \begin{equation} = \sigma^2((1 + \theta^2)\delta[k] - \theta\delta[k-1] - \theta\delta[k+1]) \end{equation} where $\delta$ is the Kronecker delta function, defined as \begin{equation} \delta[n] = \begin{cases} 1, & \text{if } n = 0\\ 0, & \text{otherwise} \end{cases} \end{equation}