Autocorrelation function of MA(1)

Question

I'm very new to this subject. My teacher is following Walter Enders Applied Econometric Time Series, and Autocorrelation functions is something I have to know. However I can´t understand this step made in the book. Thanks in advance.

Considering the process $MA(1)$: $y_t=\varepsilon_t+\beta\varepsilon_{t-1}$, we can obtain the Yule–Walker equations by multiplying $y_t$ by each $y_{t-s}$ >and take expectations.

$\gamma_0=Ey_ty_t=E[(\varepsilon_t+\beta\varepsilon_{t-1})(\varepsilon_t+\beta\varepsilon_{t-1})]=(1+\beta^2)\delta^2$

But I don't understand why it's equal to $(1+\beta^2)\delta^2$

I think this should be in the Stat's stack exchange. – Chickenmancer May 03 '17 at 20:38 — Chickenmancer, May 03 '17 at 20:38
@Justin No. Fully on-topic here. – Did May 03 '17 at 20:48 — Did, May 03 '17 at 20:48

score 0 · Accepted Answer · answered May 03 '17 at 20:45

Out $\varepsilon$'s should be iid and follow $$ \varepsilon_t\sim\mathcal{N}(0,\delta^2) $$ (I am assuming giving the form) $$ \mathbb{E}[(\varepsilon_t+\beta\varepsilon_{t-1})^2] = \mathbb{E}[\varepsilon_t^2+2\beta\varepsilon_t\varepsilon_{t-1} + \beta^2\varepsilon_{t-1}^2] = \mathbb{E}[\varepsilon_t^2] + 2\beta\mathbb{E}[\varepsilon_t\varepsilon_{t-1}] + \beta^2\mathbb{E}[\varepsilon_{t-1}^2] $$ In theory you should have $$ \mathbb{V}[\varepsilon_t] = \mathbb{E}[\varepsilon_t^2] - \mathbb{E}[\varepsilon_t] = \mathbb{E}[\varepsilon_t^2] = \delta^2 $$ and since they should be independent $$ \mathbb{E}[\varepsilon_t\varepsilon_{t-1}] = \mathbb{E}[\varepsilon_t]\mathbb{E}[\varepsilon_{t-1}] = 0 $$ so $$ \mathbb{E}[(\varepsilon_t+\beta\varepsilon_{t-1})^2] = \delta^2 + 0 + \beta^2\delta^2 = (1+\beta^2)\delta^2 $$

Autocorrelation function of MA(1)

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