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I want to find the autocovariance function of $y_{t} = \exp(x_{t})$ which is a nonlinear time series. I am assuming $x_{t}$ is stationary with mean $\mu$ and covariance $\gamma(h)$. When I try to find the autocovariance function I use the moment generating function approach, I get that $E[\exp(x_{t}) \, \exp(x_{t+h})] - E[exp(x_{t})] \, E[exp(x_{t+h})] =0$. I am not sure if this is correct.

I know $E[\exp(x_{t}) \, \exp(x_{t+h})] = \exp(2\mu + \gamma(0))$ since both $x_{t}$ and $x_{t+h}$ are stationary, meaning they have the same $\mu$ and $\gamma(0)$. The individual products also evaluate to the above expression. For the derivation let $z = x_{t} + x_{t+h}$. Since we are assuming that the series is stationary and normal z is a normal random variable with mean $\mu_{t} + \mu_{t+h}$ = $2\mu_{x}$ and variance $2\gamma(0).$ I want to find $\gamma_{y}(h) = E[\exp(z)] - E[x_{t}]E[x_{t+h}]$. The first term is $\exp(\mu_{x} + \gamma(0))$ and the second term is also the same.

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  • $\begingroup$ To begin with, you could start elaborating the last paragraph - which looks wrong. $\endgroup$
    – leonbloy
    Commented Jan 16, 2013 at 18:53
  • $\begingroup$ Let $z= x_{t} + x_{t+h}$. $M_{z}(1) = E[exp(z)]=M_{x}(1)*M_{x+h}(1) = exp(\mu_{x} + 1/2\sigma_{x}^2))*exp(\mu_{x+h} + 1/2\sigma_{x+h}^2)) = exp(2\mu+\gamma(0)$ since the mean and variance are the same for $x_{t} and x_{t+h}$. $\endgroup$
    – lord12
    Commented Jan 16, 2013 at 19:36
  • $\begingroup$ I dont' get it. What is $M$? Why are you assuming that you can factor the expectation (second =)? I neither understand the next =. YOu should write your derivation in detail in the body of your question. $\endgroup$
    – leonbloy
    Commented Jan 16, 2013 at 19:53
  • $\begingroup$ M is the moment generating function $\endgroup$
    – lord12
    Commented Jan 16, 2013 at 19:54
  • $\begingroup$ any comments or insights. I realize xt and xt+h may not be independent. $\endgroup$
    – lord12
    Commented Jan 16, 2013 at 21:15

1 Answer 1

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Let $y_t = \exp(x_t)$

Assuming $x_t$ is gausssian (I guess you forgot to tell us), we have that, in general for a lognormal $Y$, $E[Y]=E[e^X] = \exp({\mu+\sigma^2/2})$.

Further, $E[y_t y_{t+h}]= E[\exp(x_t + x_{t+h})]=E[\exp(z)]$ where $z=x_t+x_{t+h}$ is gaussian and stationary, with mean $\mu_z=2 \mu_x$ and variance $\sigma_z^2=2 \gamma(0)+2 \gamma(h)$

(In general, $\sigma^2_{A+B}=\sigma^2_A+\sigma^2_B+2 \sigma_{AB}$. So, $Var(z)=Var(x_t)+Var(x_{t+h})+ 2Cov(x_t x_{t+h})=2 \sigma_x^2 + 2 \gamma_h$)

Hence, $$E[y_t y_{t+h}] = \exp({\mu_z+\sigma_z^2/2})= \exp({2\mu_x+\gamma(0)+ \gamma(h)})$$

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  • $\begingroup$ Why does $\sigma_{z}^2$= $2\gamma(0) + 2\gamma(h)$ as opposed to $\gamma(0) + \gamma(h)$? $\endgroup$
    – lord12
    Commented Jan 17, 2013 at 3:51
  • $\begingroup$ Also when you take $E[y_{t}y_{t+h}]-E[y_{t}]E[y_{t+h}]$ doesn't that = 0. $\endgroup$
    – lord12
    Commented Jan 17, 2013 at 4:06
  • $\begingroup$ @lord12 No. $E(A B) \ne E(A) E(B)$ unless $A$ and $B$ are uncorrelated $\endgroup$
    – leonbloy
    Commented Jan 17, 2013 at 11:28

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