# exercise autocovariance function

I don't know how to obtain the autocovariance function of the following process, having a multiplication makes it difficult for me.

$$X_t = Z_t + \theta Z_tZ_{t-1}$$

with $$Z_i \sim N(0, \sigma^2)$$ (white noise)

$$\gamma_X(k) = \sigma^2(1+\theta\sigma^2), \hspace{1cm} k = 0$$

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– Sycorax
Nov 27, 2022 at 16:41
• That answer cannot possibly be correct, because for $\theta \lt -1/\sigma^2$ and $\sigma \ne 0$ it is negative. I believe you need $\theta^2$ in the formula in place of $\theta.$ As far as the calculation goes, do you know how to compute variances of products of independent variables?
– whuber
Nov 27, 2022 at 18:16

Given that $$\mathbb{E}[X_t]=0$$, we have
$$\gamma_X(k)=\mathbb{E}[(Z_t+\theta Z_tZ_{t-1})(Z_{t+k}+\theta Z_{t+k}Z_{t+k-1})]$$ If $$h=0$$, then $$\gamma_X(k)=\mathbb{E}[Z_tZ_t + \theta Z_tZ_tZ_{t-1} + \theta Z_tZ_{t-1}Z_t + \theta^2Z_tZ_{t-1}Z_tZ_{t-1}] \\=\mathbb{E}[Z_tZ_t] + \mathbb{E}[\theta Z_tZ_tZ_{t-1}] + \mathbb{E}[\theta Z_tZ_{t-1}Z_t] + \mathbb{E}[\theta^2Z_tZ_{t-1}Z_tZ_{t-1}]$$ The trick is that in effect
$$\mathbb{E}[Z_sZ_tZ_u]=\mathbb{E}[Z_s]\mathbb{E}[Z_tZ_u]$$ and $$\mathbb{E}[Z_sZ_tZ_uZ_v]=\mathbb{E}[Z_sZ_t]\mathbb{E}[Z_uZ_v]$$ So $$\gamma_X(k)=\sigma^2+\theta^2\sigma^4=\sigma^2(1+\theta^2\sigma^2)$$