I am confused by the second line and the third line of the autocovariance calculation. Like how the var() and the 0 come, and why there is t*sigma^2 at the end.
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$\begingroup$ $var(Y)=E[(Y -E[Y])^2]=E[Y^2]-(E[Y])^2$ may help $\endgroup$– HenryCommented Feb 14, 2023 at 1:12
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$\begingroup$ How the 0 term come? $\endgroup$– VivienCommented Feb 14, 2023 at 1:31
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$\begingroup$ It comes from $E[X_t]=0$ so $(E[X_t])^2=0$ and $E[X_t]E[X_s]=0$ $\endgroup$– HenryCommented Feb 14, 2023 at 9:24
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1 Answer
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The second term is $\mathbb E\left[\sum_{1\leq j\leq k}Z_j\sum_{1\leq j'\leq k}Z_{t+j'}\right].$ Now $\langle Z_j\rangle$ is a white noise process. That means $\mathbb E[Z_tZ_\tau] =0$ when $t\ne \tau.$ So, the concerned term yields zero.
answered Feb 14, 2023 at 7:53