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 $\lim_{t\to\infty}E[Y_t^\gamma]$, $\gamma \neq 1$?

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

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

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.

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 $\lim_{t\to\infty}E[Y_t^\gamma]$, $\gamma \neq 1$?

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.