Showing the expectation of a lognormal AR(1) process Suppose I have a lognormal AR(1) process:
$$\log(y_{t+1}) = (1-\theta)c + \theta \log (y_t) + \varepsilon_{t+1},$$ $$\varepsilon \sim N(0,\sigma^2)$$
To show $\operatorname{E}(y_{t+1})$, is it enough to say that because it's a lognormal AR(1) process, then it follows a lognormal distribution and hence use the formula $\operatorname{E}[Y]=e^{\mu+\frac{1}{2} \sigma^2}$. If that is not enough, what would be the correct way to take the expectation?
 A: Hi: I'm not sure if the expectation is conditional or unconditional but both are pretty straightforward. ( after I unconfused my confusion ).
if it's conditional, then 
$E(log(y_{t+1})| log(y_{t})) = (1-\theta) \times c + \theta \times log(y_t) = \mu^{*}$
$Var(log(y_{t+1})| log(y_{t})) = \sigma^2$. Denote this variance as $\sigma^2*$
So, then you use your expression for the mean of the transformed variable, $y_{t+1}$, and you end up with $E(y_{t+1}|y_{t}) = \exp(\mu^{*} + \frac{\sigma^2*}{2})$
If it's unconditional, then, as you showed,
$E(log(y_{t})) = c = \mu^{**} ~~\forall t$.
Also, it's easy to show that $Var(log(y_{t})) = \frac{\sigma^2}{1-\theta^2} = \sigma^2**$.
So, again using your expression for the expectation of the transformed variable, $y_{t+1}$, you end up with $E(y_{t+1}) = \exp(\mu^{**} + \frac{\sigma^2**}{2})$
My apologies for earlier confusion but I'm pretty sure this is correct and it's mostly just your original answer !!!!!
A: Your basic reasoning is correct, but your expression for the variance of the process is wrong.  If we let $Z_t \equiv \log Y_t$ then the process $\{ Z_t | t \in \mathbb{Z} \}$ is a standard Gaussian $\text{AR}(1)$ model with mean $c$ and error variance $\sigma^2$.  Thus, assuming stationarity, we have the stationary marginal distributions:
$$Z_t \sim \text{N}\Bigg( c, \frac{\sigma^2}{1-\theta^2} \Bigg)
\quad \quad \quad 
Y_t \sim \text{LN}\Bigg( c, \frac{\sigma^2}{1-\theta^2} \Bigg).$$
Thus, you have the marginal moments:
$$\begin{aligned}
\mathbb{E}(Y_t) 
&= \exp \Bigg( c + \frac{\sigma^2}{2(1-\theta^2)} \Bigg) \\[12pt]
\mathbb{V}(Y_t) 
&= \Bigg[ \exp \Bigg( \frac{\sigma^2}{1-\theta^2} \Bigg) - 1 \Bigg] \exp \Bigg( 2c + \frac{\sigma^2}{1-\theta^2} \Bigg).
\end{aligned}$$
