Can one define the initial value of the exponential Ornstein-Uhlenbeck process $r$, defined by
$$r(t) = e^{y(t)}\quad\text{with}\quad dy(t) = k(θ −y(t)) \mathrm dt+\sigma \mathrm dW(t),$$
such that the process is strictly stationary? I would guess
$$ r(0)\sim\log(\mu,\sigma) $$
with
$$\mu=\exp\left(\theta + \frac{\sigma^2}{4k}\right)$$
and
$$\sigma=\exp\left(2\theta + \frac{\sigma^2}{2k}\right) \left[ \exp\left(\frac{\sigma^2}{2k}\right)−1\right]$$
would make it so. Is this correct?
Best Answer
For simplicity of notation suppose that $\theta=0$; then the solution of the SDE
$$dY_t = - k Y_t \, dt + \sigma dW_t$$
is given by
$$Y_t = e^{-kt} Y_0 +\sigma\int_0^t e^{-k(t-r)} \, dW_r.$$
If $r(0) \sim \log(\mu,\varrho^2)$, then we can write $r(0)= e^{Y_0}$ where $Y_0 \sim N(\mu,\varrho^2)$ is independent from $(W_t)_{t \geq 0}$. In particular, we see that $(Y_t)_{t \geq 0}$ is a Gaussian process with mean
$$\mathbb{E}Y_t = e^{-k t} \mu$$
and variance
$$\mathbb{V}Y_t = e^{-2kt} \varrho^2+ \frac{\sigma^2}{2k} \left(1-e^{-2kt} \right).$$
Since the exponential moments of the normal distribution are well-known, we get
$$\begin{align*}\mathbb{E}r_t &= \mathbb{E}e^{Y_t} = \exp \left( \mathbb{E}Y_t+ \frac{1}{2} \mathbb{V}Y_t \right) = \exp \left( e^{-kt} \mu + e^{-2kt} \varrho^2+ \frac{\sigma^2}{2k} \left(1-e^{-2kt} \right) \right). \end{align*}$$
This shows that the expectation of the process $(r_t)_{t \geq 0}$ depends on the time $t$. Consequently, it is not a stationary process.