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I have an AR(1) process that looks like this:

$$ \ln(g_t) = (1 - \rho_g)(\ln(\mu_g) - c) + \rho_g\ln(g_{t-1}) + \epsilon^g_t $$

where $|\rho_g| < 1$, $\epsilon^g_t \sim N(0, \sigma^2_g)$, and $c = \cfrac{1}{2} \left( \cfrac{\sigma^2_g}{1 - \rho^2_g} \right)$

and I want to find $E(g_t)$ and $Var(g_t)$.


I thought about starting out with a simple substitution $y = \ln(g_t)$, to make it a standard AR(1) process that I can find the variance of:

$$ y_t = B + \rho_g y_{t-1} + \epsilon^g_t $$

where $B = (1 - \rho_g)(\ln(\mu_g) - c)$. I can find the expected value:

\begin{align} E(y_t) &= E(B + \rho_g y_{t-1} + \epsilon^g_t) \\ &= B + \rho_g E(y_{t-1}) \\ \mu_y &= B + \rho_g \mu_y \\ \mu_y &= \cfrac{B}{1 - \rho_g} \\ &= \ln(\mu_g) - c \end{align}

and variance of this (using stationarity):

\begin{align} Var(y_t) &= Var(B + \rho_g y_{t-1} + \epsilon^g_t) \\ &= Var(\rho_g y_{t-1} + \epsilon^g_t) \\ \sigma^2_y &= \rho^2_g \sigma^2_y + \sigma^2_g \\ &= \cfrac{\sigma^2_g}{1 - \rho^2_g} \end{align}

but I can't see how this transformation helps me find $\sigma^2_g$. Apart from that I don't really know where to begin.

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    $\begingroup$ Hint: If you can show that $\log(g_t)$ has a normal distribution, then all you need to do is look up the properties of lognormal distributions. $\endgroup$ – whuber May 23 '14 at 20:58
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    $\begingroup$ @whuber That was super helpful, and I think I got it. Does my edit look correct? Basically it works out (I think) that $\log(g_t)$ is a linear combination of i.i.d normal random variables, so it has a normal distribution too. $\endgroup$ – Michael A May 23 '14 at 21:18
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    $\begingroup$ A simpler way to demonstrate normality is by induction. Its hypothesis is that $\log(g_{t-1})$ is Normal and you need to show that implies $\log(g_t)$ is Normal--but obviously it is, since it's a linear combination of two independent Normals. BTW, a good way to post your work--and perhaps eventually your answer--is to put it in a separate answer rather than as an edit. That lets people vote separately on the question and the answer (giving you valuable feedback) and it helps improve the rate of our officially answered questions (something we would like to increase!). $\endgroup$ – whuber May 23 '14 at 21:44
  • $\begingroup$ @whuber It's been a while since I've taken much math... but induction in the sense of 1) prove that $\log(g_1)$ is normal, then 2) prove that if we assume that $\log(g_t)$ is normal, then $\log(g_{t+1})$ is also normal, right? That's pretty much what you said, I think, but I wanted to check. $\endgroup$ – Michael A May 23 '14 at 21:46
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    $\begingroup$ Yes, that's right. (To get the induction started you also have to show $\log(g_0)$ is Normal, but that's trivial.) The appeal of this approach--besides its inherent rigor--is that it obviates having to expand $\log(g_t)$ into a potentially complicated, arbitrarily-long linear combination. $\endgroup$ – whuber May 23 '14 at 21:49
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whuber mentioned in the comments that I just need to show that $\log(g_t)$ has a normal distribution.

Since this is a standard AR(1) process we know that $\epsilon^g_t \sim N(0, \sigma^2_g)$ is i.i.d. I used a simple example for this, without the messy constant term:

$$ \log(g_t) = \rho_g \log(g_{t-1}) + \epsilon^g_t $$

That means that (working forward from $g_0$):

\begin{align} \log(g_1) &= \rho_g \log(g_0) + \epsilon^g_1 \end{align}

and then

\begin{align} \log(g_2) &= \rho_g \log(g_1) + \epsilon^g_2 \\ &= \rho_g \left(\rho_g \log(g_0) + \epsilon^g_1 \right) + \epsilon^g_2 \\ &= \rho_g^2 \log(g_0) + \rho_g \epsilon^g_1 + \epsilon^g_2 \end{align}

etc. so

\begin{align} \log(g_t) &= \rho_g^t \log(g_0) + \rho_g^{t-1} \epsilon^g_1 + \rho_g^{t-2} \epsilon^g_2 + \dots + \epsilon^g_t\\ \end{align}

so $\log(g_t)$ is just a linear combination of independent, normally distributed random variables (the $\epsilon_t$), so it has a normal distribution.

Once you know that $\log(g_t)$ has a normal distribution, you know that $g_t$ has a log-normal distribution, so you can look up the properties of the distribution and get the mean and variance that way.

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