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I am currently working on this problem that asked me to calculate the ACF of

$Y_{t}=x_{t}x_{t-1}, x \sim IID{\left\lbrace0, \sigma^{2}\right\rbrace}$

what I've done so far is, I have proved that $E[y_{t}]=0$, so I'll take it as given in the following calculation.

here I am calculating the ACF of $y_{t}$: $$ \begin{aligned} COV(y_{t+h}, y_{t}) &= E[(y_{t+h} - E[y_{t+h}])(y_{t} - E[y_{t})]\\ &= E[y_{t+h}y_{t}]\\ &= E[x_{t+h}x_{t+h-1}x_{t}x_{t-1}]\\ \end{aligned} $$

given $x \sim IID{\left\lbrace0, \sigma^{2}\right\rbrace}$, therefore we can separate the expectation above into: $$ E[x_{t+h}x_{t+h-1}x_{t}x_{t-1}] = E[x_{t+h}]E[x_{t+h-1}]E[x_{t}]E[x_{t-1}] \qquad(1) $$

if $h \neq 0$, then $(1)=0$ because we have $E[x_{t}]=0$

if $h = 0$, then (1) is

$$ E[x_{t}]E[x_{t-1}]E[x_{t}]E[x_{t-1}] = 0 $$

which we could then conclude that ACF of $y_{t}$ is 0.

it all looks correct me but the solution gives different answer when $h =0$, could anyone spot what I've done wrong?

thanks!

EDIT:

given the comments, there's one thing I've done wrong above; when $h = 0$, the independence cannot be applied because two or more indices are the same. this implies dependence (I think)

I therefore did,

$$ \begin{aligned} E[x_{t+h}x_{t+h-1}x_{t}x_{t-1}] &= E[x_{t}x_{t-1}x_{t}x_{t-1}]\\ &= E[x_{t}^{2}x_{t-1}^{2}] \qquad (2) \end{aligned} $$

I am stuck again because I don't know how to prove $$ (2) = E[x_{t}^{2}]E[x_{t-1}^{2}]\qquad (3) $$

one more hint, I could solve it!

stucash
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    The independence you used in (1) depends on which value $h$ takes – Taylor Jan 28 '18 at 19:01
  • @Taylor I guess you meant when h = 0, we can't separate it like we have done when $h \neq 0$ right? – stucash Jan 28 '18 at 19:03
  • this looks like a home work – Aksakal Jan 28 '18 at 19:05
  • @Aksakal I have done the work and checked against solution which baffles me. probably some hint could help understand what I've done wrong? – stucash Jan 28 '18 at 19:06
  • (1) is incorrect when two or more of the indexes $\{t+h,t+h-1,t,t-1\}$ coincide. – whuber Jan 28 '18 at 19:14
  • Haven't you assumed $x_t$ and $x_{t-1}$ are independent? That implies any functions of them (separately) are independent, too: see https://stats.stackexchange.com/questions/94872. – whuber Jan 28 '18 at 21:19
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    @whuber thanks very much for the link, that is exactly the question that I have, i remember that is the case, but I couldn't recall how to prove it. now I know the answer! – stucash Jan 28 '18 at 21:47
  • Thank you for confirming that. I'll close this question as a duplicate. That means your question will remain available for searching and reading, but viewers will be given a link to the other one for the answer. – whuber Jan 28 '18 at 21:47

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