Power transformation of OU process

Question

Say I have $X$ that follows an Ornstein-Uhlenbeck process:

$$ dX_t = \phi (\mu - x_t) d_t + \sigma d W_t $$

Let $Y_t = \exp(X_t)$. Is there anything that helps me compute $\lim_{t\to\infty}E[Y_t^\gamma]$, $\gamma \neq 1$?

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Here is my approach:

We know that the stationary solution for $X_t$ is Gaussian with mean $\mu$ and variance $\sigma^2/(2\phi)$. Hence, the expected value of $\lim_{t\to\infty} Y_t$ is that of the log-normal distribution,

$$\lim_{t\to\infty} E[Y_t] = \exp\left(\mu + \frac{\sigma^2}{2\phi}\right)$$

Now,

$$Y_t^\gamma = \exp(\gamma X_t)$$

I would intuitively guess that the process $\gamma X_t$ follows

$$ d\gamma X_t = \phi \gamma (\mu - x_t) d_t + \gamma \sigma dW_t$$

and therefore compute

$$\lim_{t\to\infty} E[Y_t^\gamma] = \exp\left(\mu + \frac{\gamma^2\sigma^2}{2\gamma\phi}\right)$$

But clearly, I'm out of my reach here.

Answer 1

It can be shown that (by using Itô's lemma on $X_t e^{\phi t}$): $$X_t = X_0e^{-\phi t} + \mu (1 - e^{-\phi t}) + \int_0^t \sigma e^{\phi (t-s)} dW_s $$

Which leads to: $$\gamma X_t = \gamma X_0e^{-\phi t} + \gamma \mu (1 - e^{-\phi t}) + \int_0^t \gamma \sigma e^{\phi (t-s)} dW_s $$

This process has:

• mean $E[\gamma X_t] = \gamma X_0e^{-\phi t} + \gamma \mu (1 - e^{-\phi t})$
• variance $Var[\gamma X_t] = \int_0^t \gamma^2 \sigma^2 e^{2 \phi(t-s)} ds = \frac{\gamma ^2 \sigma ^2}{2 \phi} (1 - e^{-2\phi t}) $

At $t \rightarrow \infty$, the process $\gamma X_\infty$ is gaussian with:

• mean $E[\gamma X_\infty] = \gamma \mu$
• variance $Var [\gamma X_\infty] = \frac{\gamma^2 \sigma^2}{2\phi}$

So its exponent is lognormal with mean: $$E[Y_\infty ^\gamma] = E[e^{\gamma X_\infty}] = e^{E[\gamma X_\infty] + \frac{1}{2} Var[\gamma X_\infty]} = e^{\gamma \mu + \frac{\gamma^2\sigma^2}{4\phi}}$$

You were almost there but were a bit too quick.

PS: one can also apply Itô's lemma to get the expression of $Y_t^\gamma$ then get its mean and take the limit. I chose to do $\gamma X_t$ and take the exponent at the end to follow your approach.