6
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.