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)

pd: the answer is:

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

  • 2
    $\begingroup$ Please add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$
    – Sycorax
    Nov 27, 2022 at 16:41
  • $\begingroup$ 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? $\endgroup$
    – whuber
    Nov 27, 2022 at 18:16

1 Answer 1


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 $$\gamma_X(k)=\sigma^2+\theta^2\sigma^4=\sigma^2(1+\theta^2\sigma^2)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.