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.
Best Answer
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.