[Time series models] covariance in MA(1) process

_sebastian_

New Member
Hello,

the following question is about a stationary MA(1) process (Moving Average of Order 1) - see attached image. Epsilon represents white noise, i.i.d. E(epsilon) = 0.

Can someone please help me understand the last part of the bottom equation? I do understand the term inside the expected value brackets, however, the last part of the equation is a mystery to me…

Sebastian

Dason

Are you saying you see why you have $$Cov(y_t, y_{t-1}) = E[(\epsilon_t - \theta_1\epsilon_{t-1})(\epsilon_{t-1} - \theta_1\epsilon_{t-2})]$$ but not why the very last equality holds?

Dason

Just expand it out: (a+b)(c+d) = ac + ad + bc + bd

Then recognize that the epsilon terms are independent of each other - each having a mean of 0 (which means the expected value of the square of one of the epsilon terms is $$\sigma_e^2$$

_sebastian_

New Member
Just expand it out: (a+b)(c+d) = ac + ad + bc + bd

Then recognize that the epsilon terms are independent of each other - each having a mean of 0 (which means the expected value of the square of one of the epsilon terms is $$\sigma_e^2$$
Thanks for your quick response, Dason. I tried following your steps, but unfortunately that doesn't really work for me (see image). What am I doing wrong here?
Sorry being a total noob…

BGM

TS Contributor
Then recognize that the epsilon terms are independent of each other
Dason means

$$\epsilon_i, \epsilon_j$$ are independent when $$i \neq j$$

and therefore $$E[\epsilon_i\epsilon_j] = E[\epsilon_i]E[\epsilon_j]$$

- each having a mean of 0
so those cross moments terms vanish as calculate above, only one second moment term left.

Dason

Note that $$E[\epsilon_t^2] = \sigma_e^2$$ for any value of t but that doesn't mean that the expected value $$\epsilon_t\epsilon_{t+k}$$ is $$\sigma_e^2$$ when k isn't 0.

Recall that the epsilon terms are *independent* with a mean of 0.

_sebastian_

New Member
Dason means

$$\epsilon_i, \epsilon_j$$ are independent when $$i \neq j$$

and therefore $$E[\epsilon_i\epsilon_j] = E[\epsilon_i]E[\epsilon_j]$$

so those cross moments terms vanish as calculate above, only one second moment term left.
Now I got it, thanks a lot guys!