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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:

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

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

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

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  • $\begingroup$ $E(y_ty_{t+h})$ is a function of $h$, but not of $t$, since it stationary. $\endgroup$ Mar 11, 2021 at 22:06
  • $\begingroup$ I came to the conclusion that the autocovariance is 0. Is that right? $\endgroup$
    – Hasselhoff
    Mar 11, 2021 at 22:37
  • $\begingroup$ 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. $\endgroup$
    – Hasselhoff
    Mar 16, 2021 at 15:57
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    $\begingroup$ $\gamma_y(h)$ must be a function of $\gamma_x(h)$.It does not show up in your solution. $\endgroup$ Mar 17, 2021 at 17:27
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    $\begingroup$ 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. $\endgroup$ Mar 17, 2021 at 17:40

2 Answers 2

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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.

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  • $\begingroup$ Can you be a little more specific. I don't see it. $\endgroup$
    – Hasselhoff
    Mar 12, 2021 at 13:22
  • $\begingroup$ 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. $\endgroup$
    – Hasselhoff
    Mar 12, 2021 at 15:04
  • $\begingroup$ 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. $\endgroup$ Mar 13, 2021 at 3:02
  • $\begingroup$ I expanded my answer to include detailed development. Suggest you check it - arithmetic is not my strong suit. $\endgroup$ Mar 13, 2021 at 5:39
  • $\begingroup$ 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. $\endgroup$
    – Hasselhoff
    Mar 16, 2021 at 14:33
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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:

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

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

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

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

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