Let $x_t$ be an AR(1) process, such that $x_t=\alpha_0+\alpha_1 x_{t-1}+\epsilon_t$, where $\alpha_0$ is a constant, $|\alpha_1|\lt 1$ and $\epsilon_t$ is a white noise process with mean $0$ and variance $\sigma^2$.
By backwards substitution we can see that $x_t=\sum^\infty_{i=0}(\alpha_0+\epsilon_{t-i})\alpha_1^i=\sum^\infty_{i=0}\alpha_0\alpha_1^i+\sum^\infty_{j=0}\epsilon_{t-j}\alpha_1^j$ and because $|\alpha_1|<1$ we have that $\Bbb{E}[x_t]=\frac{\alpha_0}{1+\alpha_1}$ and $\gamma(k)=\text{Cov}(x_t,x_{t-k})=\sum^\infty_{i=0}\alpha_1^{2i}·\text{Cov}(\epsilon_t,\epsilon_{t-k})=\frac{\sigma^2}{1-\alpha_1^2}$.
Unless I made a mistake, this seems quite straightforward. Now my real question is, how do we find the expectation and the autocovariance $\gamma(k)$ when instead we have the following process $y_t=\exp(\alpha_0+\alpha_1y_{t-1}+\epsilon_t)$? Clearly we already have $\Bbb{E}[\ln(y_t)]$ by the above, but using that how can we find an expression for $\Bbb{E}[y_t]$ and the autocovariance? Is there a general way to do this for a function $f(x)$ and a process $y_t=f(\alpha_0+\alpha_1y_{t-1}+\epsilon_t)$ for example?
Thanks in advance for the help!

  • $\begingroup$ I think maybe you mean that the transformed process is given by the transformation $y_t=e^{x_t}$. If this is what you mean then this is straightforward using stats.stackexchange.com/a/132879/77222 (if you assume Gaussian white noise). But the process you describe satisfying $y_t=\exp(\alpha_0+\alpha_1y_{t-1}+\epsilon_t)$ have much more complicated behaviour and may not even be stationary. Btw, $k$ is missing in your formula for $\gamma(k)$. $\endgroup$ May 27, 2021 at 8:24
  • $\begingroup$ @JarleTufto Thanks for your reply! I was simply wondering, out of curiosity, if there was a way to find expectation and auto covariance for processes like for example $\exp(\alpha_0+\alpha_1y_{t-1}+\epsilon_t)$ and $\cos(\alpha_0+\alpha_1y_{t-1}+\epsilon_t)$ and how they behave in general. Do you happen to know any resources where similar processes are discussed in any way at all? $\endgroup$
    – user619755
    May 27, 2021 at 8:34
  • 1
    $\begingroup$ No, if $y_t$ satisfies $y_t=\exp(\alpha_0+\alpha_1y_{t-1}+\epsilon_t)$, then neither $y_t$ or $\ln y_t$ would be a linear Gaussian process. What you could consider would be $y_t=\exp(\alpha_0+\alpha_1\ln y_{t-1}+\epsilon_t)$ which would be equivalent to the transformation I suggest ($\ln y_t$ would then be a linear Gaussian process and $y_t$ at any collection of time points would be jointly lognormally distributed. $\endgroup$ May 27, 2021 at 8:45

1 Answer 1


I think that this is a tricky question, since in general $E(f(X_t))\neq f(E(X_t))$. Depending on $f(.)$, the process $f(X_t)$ may be complete different from $X_t$. Because of that, I don't think that there is a general solution for this kind of problem. However, there are some cases where the solution is straightforward as mentioned by @JarleTufto.

Let me give an example with another model where this problem occurs and that is related to your AR(1)-model, a simple EGARCH(1,1) model: \begin{align} \epsilon_t&=\sigma_t u_t \\ \ln(\sigma_t^2)&=\gamma_0+\gamma_1\ln(\sigma_{t-1}^2)+g(u_{t-1}) \\ g(u_{t-1})&=\theta u_{t-1}+\lambda(\vert u_{t-1} \vert -E(\vert u_{t-1}\vert)) \quad, u_t \overset{i.i.d.}{\sim}(0,1) \end{align} It is straightforward to calculate the expected value of $\ln(\sigma_t^2)$: \begin{align} E(\ln(\sigma_t^2))=\frac{\gamma_0}{1-\gamma_1} \end{align} Notice that $\ln(\sigma_t^2)$ is basically an AR(1)-process with WN innovation $g(u_t)$. But when you want to calculate the expected value of $\sigma_t^2=\exp(\ln(\sigma_t^2))$ given by $E(\sigma_t^2)=E(\exp(\ln(\sigma_t^2)))=E(\epsilon_t^2)$, you get: \begin{equation} E(\epsilon_t^2)=\exp\left(\frac{\gamma_0}{1-\gamma_1}\right)\prod_{i=1}^{\infty}E\left(\exp\left(\gamma_1^{i-1}g(u_{t-i})\right)\right) \end{equation} Assuming that $u_t \overset{i.i.d.}{\sim}{\cal N}(0,1)$, you get: \begin{align*} E(\epsilon_t^2)=\exp\left(-\lambda\sqrt{\frac{2}{\pi}}\right)\left(\exp(0.5(\theta-\lambda)^2\Phi(\theta-\lambda)+\exp(0.5(\theta+\lambda)^2\Phi(\theta+\lambda)))\right) \end{align*} Compare the expressions for $E(\ln(\sigma_t^2))$ and $E(\epsilon_t^2)$. It is hard to imagine, that there is a general solution for this type of problem.


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