# Find $Cov(e_n(1),e_n(2))$ in $AR(1)$ model

Let $n=144$ observations from an $AR(1)$ model $$y_t=\phi y_{t-1}+\epsilon_t$$ where $\epsilon_t$ is White Noise with mean zero and variance $\sigma^2$. If $y_1=-1.7$, $y_{144}=-2.1$, $\sum_{t=2}^n y_t y_{t-1}=-128.6$ and $\sum_{t=1}^n y_t^2=246.4$.

Suppose $\sigma^2$ know, find $Cov(e_n(1),e_n(2))$, where $e_n(j)$ is the forecast error $j$ steps ahead.

This question is related with another that I posted Estimation of $\phi$ in $AR(1)$ process , but I preferred to separate them.

First I found the forecasts $y_t(1)$ and $y_t(2)$ with $t=144$.

$$y_t(1)=E[y_{t+1}|y_1,\dots,y_t]=E[\phi y_t|y_1,\dots,y_t]=\phi y_t$$ $$y_t(2)=E[y_{t+2}|y_1,\dots,y_t]=E[\phi y_{t+1}+\epsilon_{t+2}|y_1,\dots,y_t]=\phi^2 y_t$$

Then the forecast errors are $$e_n(1)=y_{t+1}-y_t(1)=\epsilon_{t+1}$$ $$e_n(2)=y_{t+2}-y_t(2)=\phi y_{t+1}+\epsilon_{t+2}-\phi^2 y_t$$

Finally the covariance is $$Cov(e_n(1),e_n(2))=Cov(\epsilon_{t+1},\phi y_{t+1}+\epsilon_{t+2}-\phi^2 y_t)$$ $$=\phi Cov(\epsilon_{t+1},\phi y_t+\epsilon_{t+1})=\phi Var(\epsilon_{t+1})=\phi \sigma^2$$

Is it right?

NOTE: Sorry if I poste multiple questions about this topic, but I'm learning alone and I don't have solutions manual.

• The relation to the previous post is very similar. It is the exact same model with the exact same information on the data. The only difference I think is that you are trying to estimate a different parameter. Also you are formulating these problems in terms of checking your work. I don't think that is what this site is for. – Michael Chernick Dec 30 '16 at 19:18
• @MichaelChernick has a point there: where self-study questions are concerned I think we need to draw a line between discussing specific doubts you may have about your approach & providing a general answer-checking service. There's some discussion at A difficulty with self-study-like questions of the form “Is this correct?”. – Scortchi Dec 31 '16 at 22:53