# What is the autocovariance of $y_t = \exp{(x_t)}$ where $x_t$ is a stationary normal process.

Let $$x_t$$ be a stationary normal process with mean $$\mu_x$$ and autocovariance function $$\gamma(h)$$. Define the nonlinear time series $$y_t=\exp{(x_t)}$$.

(a) Express the mean function $$E(y_t)$$ in terms of $$\mu_x$$ and $$\gamma(0)$$. The moment generating function of a normal random variable $$x$$ with mean $$\mu$$ and variance $$\sigma^2$$ is

$$M_x(\lambda)=E[\exp{(\lambda x)}]=\exp{(\mu \lambda + > \frac{1}{2}\sigma^2 \lambda^2})$$

(b) Determine the autocovariance function of $$y_t$$. The sum of the two normal randomvariables $$x_{t+h}+x_t$$ is still a normal random variable.

EDIT: I figured out the first part, but still struggling with part b.

First we not the MGF for a normal rv $$x$$ with mean $$\mu_x$$ and variance $$\sigma_x^2$$ is $$E[\exp(xt)] = \exp\left[\mu_x t + \frac{\sigma_x^2 t^2}{2}\right]$$

so that $$E[y_t] = E[\exp(x_t)]$$ is the same as the normal MGF evaluated at $$t=1$$, which implies that the mean function of $$y_t$$ is

$$E[y_t] = E[\exp(x_t)]= \exp\left[\mu_x + \frac{\gamma_x(0)}{2}\right]$$

EDIT 5: I think I finally see where the dependence of $$x_t$$ plays in. How is this?

Note that since $$x_t$$ and $$x_{t+h}$$ are identically distributed normal but not independent then the mean and variance of $$x_t+x_{t+h}$$ are

$$\mu_{x_t+x_{t+h}} = 2\mu_x$$

$$\sigma^2_{x_t+x_{t+h}} = \gamma_x(0) + \gamma_x(0) + 2\rho_x(h)\gamma_x(0) = 2\gamma_x(0)(1+\rho_x(h))$$

Thus the MGF for $$x_t+x_{t+h}$$ is

$$M_{x_t+x_{t+h}}(\lambda) = E[\exp(\lambda(x_t+x_{t+h}))] = \exp\left[2\mu_x \lambda + \gamma_x(0)(1+\rho_x(h))\lambda^2]\right]$$

By the same argument above using the MGF evaluated at $$\lambda = 1$$ we have

$$E[y_ty_{t+h}] = E[\exp(x_t+x_{t+h})] = M_{x_t+x_{t+h}}(1) = \exp\left[2\mu_x + \gamma_x(0)(1+\rho_x(h))]\right]$$ And we can calculate the autocovariance:

$$\gamma_y(h) = E[y_ty_{t+h}] - \mu_y^2 = \exp[2\mu_x + \gamma_x(0) + \gamma_x(0)\rho_x(h)] - \exp[2\mu_x +\gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\exp[\gamma_x(0)\rho_x(h)] - \exp[2\mu_x + \gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\left(\exp[\gamma_x(0)\rho_x(h)] - 1\right)$$

• $E(y_ty_{t+h})$ is a function of $h$, but not of $t$, since it stationary. Mar 11, 2021 at 22:06
• I came to the conclusion that the autocovariance is 0. Is that right? Mar 11, 2021 at 22:37
• I realized that my actual mistake was not squaring the coefficient for the variance in the MGF. If I do this then I believe I get the correct answer, I created a 4th edit with the solution. I also added a complete solution below which might be easier to understand without all the fumbling around above. Mar 16, 2021 at 15:57
• $\gamma_y(h)$ must be a function of $\gamma_x(h)$.It does not show up in your solution. Mar 17, 2021 at 17:27
• If you look at my answer you will see the dependence on $\gamma_x(h)$. " series is in no way based on a lag," is irrelevant. You are getting function of two points in the series, so it depends on separation. Mar 17, 2021 at 17:40

Wrong! $$E(exp(x_t+x_{t+h}))$$ is not $$E(x_t)^2$$. You need to take into account the autocorrelation., since they are not independent.

Formula for $$A=E(exp(x_t+x_{t+h}))$$: To save writing, simplify notation. $$w=x_t-\mu$$, $$y=x_{t+h}-\mu$$, $$\rho=\gamma(h)/\gamma(0)$$ = auto-correlation, $$\alpha=\sqrt{1-\rho^2}$$.

$$A=\frac{1}{2\pi \sigma^2 \alpha}\int_{-\infty}^\infty\int_{-\infty}^\infty e^Kdwdy$$ where $$K=2\mu+\frac{1}{2\sigma^2 \alpha^2}(2\sigma^2 \alpha^2(w+y)-w^2+2\rho wy-y^2)$$

The variables can be decoupled by $$b=w+y$$ and $$c=w-y$$ so that:

$$K=2\mu+\frac{1}{2\sigma^2 \alpha^2}(2\sigma^2 \alpha^2 b-\frac{b^2+c^2}{2}+\rho\frac{b^2-c^2}{2})$$

and $$dwdy=\frac{1}{2}dbdc$$

I'll let you do the calculation using the product of two normal distributions.

• Can you be a little more specific. I don't see it. Mar 12, 2021 at 13:22
• Are you trying to say that $E(y_t^2)$ is not $E(\exp(x_t+x_{t+h}))$? Because I did not suggest $E(x_t^2) = E(\exp(x_t+x_{t+h}))$ in my answer. Actually I didn't suggest either of those statements in my answer. Mar 12, 2021 at 15:04
• Sorry for the typos - it was late! I corrected my answer. You have $E(exp(x_t+x_{t+h}))=E(exp(x_t))E(exp(x_{t+h})=E(exp(x_t))^2$ Breaking up into a product requires independence. Mar 13, 2021 at 3:02
• I expanded my answer to include detailed development. Suggest you check it - arithmetic is not my strong suit. Mar 13, 2021 at 5:39
• Ok, I will look at this a little deeper, but to be clear, I did not use the equation above, I simply substituted the new Normal rv created by adding $x_t + x_{t+h}$, which is still a normal rv, into the normal MGF. I did not assume independence anywhere. Mar 16, 2021 at 14:33

I figured it out so I'll answer my own question.

First we note the MGF for a normal rv $$x$$ with mean $$\mu_x$$ and variance $$\sigma_x^2$$ is $$E[\exp(x\lambda)] = \exp\left[\mu_x \lambda + \frac{\sigma_x^2 \lambda^2}{2}\right]$$

so that $$E[y_t] = E[\exp(x_t)]$$ is the same as the normal MGF evaluated at $$\lambda=1$$, which implies that the mean function of $$y_t$$ is

$$E[y_t] = E[\exp(x_t)]= \exp\left[\mu_x + \frac{\gamma_x(0)}{2}\right]$$

Note that since $$x_t$$ and $$x_{t+h}$$ are identically distributed normal but not independent then the mean and variance of $$x_t+x_{t+h}$$ are

$$\mu_{x_t+x_{t+h}} = 2\mu_x$$

$$\sigma^2_{x_t+x_{t+h}} = \gamma_x(0) + \gamma_x(0) + 2\rho_x(h)\gamma_x(0) = 2\gamma_x(0)(1+\rho_x(h))$$

Thus the MGF for $$x_t+x_{t+h}$$ is

$$M_{x_t+x_{t+h}}(\lambda) = E[\exp(\lambda(x_t+x_{t+h}))] = \exp\left[2\mu_x \lambda + \gamma_x(0)(1+\rho_x(h))\lambda^2]\right]$$

By the same argument above using the MGF evaluated at $$\lambda = 1$$ we have

$$E[y_ty_{t+h}] = E[\exp(x_t+x_{t+h})] = M_{x_t+x_{t+h}}(1) = \exp\left[2\mu_x + \gamma_x(0)(1+\rho_x(h))]\right]$$ And we can calculate the autocovariance:

$$\gamma_y(h) = E[y_ty_{t+h}] - \mu_y^2 = \exp[2\mu_x + \gamma_x(0) + \gamma_x(0)\rho_x(h)] - \exp[2\mu_x +\gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\exp[\gamma_x(0)\rho_x(h)] - \exp[2\mu_x + \gamma_x(0)]$$

$$=\exp[2\mu_x + \gamma_x(0)]\left(\exp[\gamma_x(0)\rho_x(h)] - 1\right)$$

• Wrong! see my full answer. Mar 12, 2021 at 4:23
• I think it is corrected now, but still not totally sure. Mar 12, 2021 at 13:58
• My answer looks similar to yours, but there differences in detail. Arithmetic has always been a problem for me. Mar 18, 2021 at 21:28