Autocovariance of moving average process

Question

Let $\epsilon_t\text{ ~ i.i.d.}(0,1)$, and $X_t=\epsilon_{t}+0.5\epsilon_{t-1}$. I need to find its autocovariance function.

I know that $E(X_t)=0$, $E(\epsilon_{t})=0$. Let's say, that $s=t+1$:

$Cov(X_t,X_s)=E[(\epsilon_t+0.5\epsilon_{t-1})(\epsilon_{t+1}+0.5\epsilon_{t})]=E[\epsilon_{t}\epsilon_{t+1}+0.5\epsilon_{t}^2+0.5\epsilon_{t-1}\epsilon_{t+1}+0.25\epsilon_{t-1}\epsilon_{t}]=0+0.5E(\epsilon_{t}^2)+0.5E(\epsilon_{t-1}\epsilon_{t+1})+0=0.5E(\epsilon_{t}^2)+0.5E(\epsilon_{t-1}\epsilon_{t+1})=\ldots\text{?}$

Can I presume, that

$E(\epsilon_{t}^2)=E(\epsilon_{t}\epsilon_{t})=E(\epsilon_{t})E(\epsilon_{t})=0$ ?

Avitus · Accepted Answer · 2013-06-11T13:31:48.700

1

If $\epsilon_t$ is i.i.d $(0,1)$, then $E[\epsilon_t]=0$ and $E[(\epsilon_t-E[\epsilon_t])^2]=E[(\epsilon_t-0)^2]=1$ for any $t$. So the answer to your question is "no".

EDIT. A short add-on. It follows from the formulae above that

$Cov(X_t,X_s)=0.5*1+0.5*Cov(e_t,e_{t+1})$,

where

$Cov(\epsilon_t,\epsilon_{t+1})=E[(\epsilon_t-E[\epsilon_t])(\epsilon_{t+1}-E[\epsilon_{t+1}])]=E[\epsilon_t\epsilon_{t+1}]$.

edited Jun 11 '13 at 13:31

answered Jun 11 '13 at 13:14

Avitus

14,018

Why is $E[(\epsilon_t)^2]=1$ (by which property)? – atomoutside Jun 11 '13 at 13:20
@atomoutside: What does "i.i.d.(0,1)" mean? – Raskolnikov Jun 11 '13 at 13:21
@Raskolnikov: "i.i.d.(0,1)" = independent and identically distributed random variables, where $0$ is expected value, and $1$ variance. – atomoutside Jun 11 '13 at 13:23
@atomoutside: yes, but what about the (0,1) ? – Raskolnikov Jun 11 '13 at 13:24
@ Raskolnikov $(0,1)$ means "with mean=0 and variance =1$ – Avitus Jun 11 '13 at 13:27
Stupid me... $Var(\epsilon_{t})=E(\epsilon_{t})^2-(E(\epsilon_{t}))^2$. Thanks for help : ) – atomoutside Jun 11 '13 at 13:33
@ atomoutside not stupid! ARMAs are mysterious stuff :-) – Avitus Jun 11 '13 at 13:43

Autocovariance of moving average process

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