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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?

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  • $\begingroup$ No. What that model means is $y_{t}$ conditionally lognormal with mean $(1-\theta c) \times y_{t-1}$ and variance $\sigma^2$. But your question is referring to the unconditional mean. The easiest way to answer the question is to calculate the unconditional expectation of a "regular" AR(1) where $y_t $ is on the LHS and RHS. ( so same model but don't use logs ). Then, once you have that one, the case for log(y_t) will be straightforward. The unconditional calculation for the mean of the AR(1) ( without the logs ) is probably all over the net. If you can't find, let me know. $\endgroup$
    – mlofton
    May 11, 2019 at 5:18
  • $\begingroup$ Hi @mlofton I'm unsure if I understand this right: So you mean I should take the expectation of the same model of AR(1) without the logs. Does this mean exponentiating the whole expression and then taking the expectation or just dropping the logs and then taking the expectation. If it's just dropping the logs then I understand the expectation would be $E(y) = \frac{(1-\theta)c}{1-\theta} = c$. However, I am unsure how to transform this to the $log(y_t)$ case as you mentioned. $\endgroup$
    – Julian5S
    May 11, 2019 at 5:54
  • $\begingroup$ 5S. I didn't explain it well or correctly so let me try to work it out and send an answer. you are correct that the unconditional mean is $c$ but I don't think that's what the question is looking for. Hold on a bit and I'll try to work it out. And forget what I said earlier. That's not gonna help as you showed. $\endgroup$
    – mlofton
    May 11, 2019 at 18:36
  • $\begingroup$ if you knew $\sigma$ you could do something along those lines, but if you need to estimate it then it gets tricky. see my answer to a similar question $\endgroup$
    – Aksakal
    Dec 12, 2019 at 3:21

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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 !!!!!

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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:

$$\begin{aligned} \mathbb{E}(Y_t) &= \exp \Bigg( c + \frac{\sigma^2}{2(1-\theta^2)} \Bigg) \\[12pt] \mathbb{V}(Y_t) &= \Bigg[ \exp \Bigg( \frac{\sigma^2}{1-\theta^2} \Bigg) - 1 \Bigg] \exp \Bigg( 2c + \frac{\sigma^2}{1-\theta^2} \Bigg). \end{aligned}$$

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