# Two papers - two different solutions of the Ornstein-Uhlenbeck process

Bernal 2016 says that the solution of $$dr_{t}=\lambda*(\mu-r_{t})*dt+\sigma dW_{t} \qquad (eq.1)$$

equals $$r_{t}=r_0*exp(-\lambda t)+\mu(1-exp(-\lambda t))+\sigma \int_{0}^{t} exp(-\lambda t)dW_{t} \qquad (eq.2)\\$$

which leads to following Euler Maruyana Scheme: $$r_{t+\delta t}=r_t+\lambda (\mu -r_t)\delta t+\sigma \sqrt{\delta t} *\mathcal{N}(0,1) \qquad (eq.3)\\$$

On the other side this paper tells us that the solution of the SDE should be $$S_{i+1}=S_i*exp(-\lambda \delta)+\mu(1-exp(-\lambda t))+\sigma \sqrt{\frac{(1-exp(-2\lambda t)}{2\lambda}}*\mathcal{N}(0,1) \qquad (eq.4)\\$$

Bernal uses $(eq.3)$ for calibration whereas GE and Berg use $(eq.4)$.

Why the difference? Bernals method makes completely sense to me.

• There seems to be a type in your equation 4 ($S_0$ should replace $S_i$). Other than that, equation 2 is identical to equation 4. Because the distribution of $\int_0^t exp(-\lambda t) dW_t$ is $\mathcal{N}(0, \frac{1 - exp(-2\lambda t)}{2 \lambda})$ Jun 11 '18 at 12:32

Note that the Ito integral of a deterministic integrand $f: \mathbb{R}_+ \rightarrow \mathbb{R}$ is normally distributed
$$\int_0^t f(u) \mathrm{d}W_u \sim \mathcal{N} \left( 0, \int_0^t f^2(u) \mathrm{d}u \right).$$
In your case, we have $f(t) = e^{-\lambda t}$ and thus
$$\int_0^t f^2(u) \mathrm{d}u = \frac{1}{2 \lambda} \left( 1 - e^{-2 \lambda t} \right).$$