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

#1
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…

Many thanks in advance!
Sebastian
 

Dason

Ambassador to the humans
#2
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

Ambassador to the humans
#4
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 \(\sigma_e^2\)
 
#5
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
#6
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

Ambassador to the humans
#7
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.
 
#8
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!