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

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.

• OGC: Notice that in the other question you pointed to ( which I think is correct. I just looked at it quickly ) , the error term is an AR(1). In the case, here, the error term is an MA(1). So, although the solution givenbelow is beautiful, a more compact answer is that the covariance is $\theta$ at lag 1 and zero everywhere else. ( because it's an MA(1) whose auto-covariance is well known ). Note that the auto-covariance is always symmetric by definition so lag[-1] autocovariance is the same as lag[1] autocovariance. Oct 28, 2019 at 1:47
• @mlofton Thanks for your explanation. However I was wondering if the covariance at lag 1 is $-\theta$ or just $\theta$.
– OGC
Oct 28, 2019 at 3:23
• @mlofton I think the covariance at lag 1 is $-\theta \sigma^{2}$. It would have been $-\theta$ if $\upsilon _{it}$ had a standard normal distribution with variance being equal to 1.
– OGC
Oct 28, 2019 at 4:33

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, $$$$cov(u_{it},u_{i(t+k)}) = cov(v_{it} - \theta v_{i(t-1)},v_{i(t+k)} - \theta v_{i(t+k-1)})$$$$ $$$$= cov(v_{it},v_{i(t+k)}) - \theta cov(v_{it},v_{i(t+k-1)}) - \theta cov(v_{i(t-1)},v_{i(t+k)}) + \theta^2 cov(v_{i(t-1)},v_{i(t+k-1)})$$$$ $$$$= \sigma^2(\delta[k] - \theta\delta[k-1] - \theta\delta[k+1] + \theta^2\delta[k])$$$$ $$$$= \sigma^2((1 + \theta^2)\delta[k] - \theta\delta[k-1] - \theta\delta[k+1])$$$$ where $$\delta$$ is the Kronecker delta function, defined as $$$$\delta[n] = \begin{cases} 1, & \text{if } n = 0\\ 0, & \text{otherwise} \end{cases}$$$$
• Thanks for your answer. I understand your explanation but I am not so clear how $\delta$ is there. Could you please clarify? Is it just that the covariance is $-\theta$ at lag 1 and zeros elsewhere?
• OGC: Yes, it's $-\theta \sigma^2$. My mistake and apologies. Oct 28, 2019 at 18:13
• OGC: Note though that if $v_t$ was normal, iid, then the covariance would be zero at all non-zero lags. Oct 28, 2019 at 18:15