White noise assumption in the autocorrelation proof I followed the proof presented in Quantitative Risk Management: Concepts, Techniques and Tools by D. Duffie, S. Schaefer (proposition 4.9, pages 128-129).
To arrive at the numerator for the autocorrelation expression they state that, since $\varepsilon_t$ is $WN(0,\sigma^2_\varepsilon)$ white noise, it follows that $\mathbb E(\varepsilon_{t-i}\varepsilon_{t + h - j}) \neq 0 \iff j = i + h$.
I don't get fully how they arrive at $j = i + h$. Could someone point me in the right direction?
 A: The formula $ ~ \mathbb E(\varepsilon_{t-i}\varepsilon_{t + h - j}) \neq 0 \iff j = i + h$ is derived from white noise assumptions:

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*Assumption: $Cov(\varepsilon_s, \varepsilon_r) = 0 $ for every $s\neq r$:

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*Any two different epsilons are uncorrelated, then they satisfy the condition: $Cov(\varepsilon_s, \varepsilon_r) = \mathbb E(\varepsilon_s \varepsilon_r) - \mathbb E(\varepsilon_s) \mathbb E(\varepsilon_r) = 0$ for every $s\neq r$.



*Assumption: $\mathbb E(\varepsilon_t) = 0$ for every $t$:

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*Assuming this, we can simplify above condition to: $\mathbb E(\varepsilon_s \varepsilon_r) = 0$ for every $s\neq r$.



*Assumption: $Var(\varepsilon_t) = \sigma^2 < \infty$ for every $t$:

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*Then we know, that $Var(\varepsilon_t) = \mathbb E(\varepsilon_t^2) - \mathbb E^2(\varepsilon_t) = \mathbb E(\varepsilon_t^2) \neq 0$.



So $E(\varepsilon_{t-i}\varepsilon_{t + h - j}) \neq 0$ can happen if and only if the two epsilons inside the expected value are the same random variable, and this happens only for $ - i = h-j $.
The same holds when the white noise is assumed an I.I.D. process.
