# Time series expectation

#### Prometheus

##### Member
Maybe it's because i've been at it for 8 hours straight today, but i'm stumped trying to show this is true:

$$E[YtEt]=\sigma^2$$

Any help might spare my sanity.

#### JesperHP

##### TS Contributor
When you ask other people a question it is a good idea to provide them with the information needed for the question at hand to make sense....

So one guess is that you have and AR(q) model...for simplicity assume it is AR(1) no constantterm:
$$y_t = a_1 y_{t-1} + e_t$$ and by definition $$\sigma^2 = var(e_t)$$
then using the fact that $$E[y_t]=0$$ it follows that $$E[y_te_t]=cov(y_t,e_t)$$
then $$E[y_te_t]=cov(y_t,e_t)=cov(a_1 y_{t-1} + e_t,e_t)$$

$$=cov(a_1 y_{t-1} , e_t) + cov(e_t,e_t) = 0 + var(e_t) = \sigma^2$$