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

Question

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.

Answer 1

Note that the Ito integral of a deterministic integrand $f: \mathbb{R}_+ \rightarrow \mathbb{R}$ is normally distributed

\begin{equation} \int_0^t f(u) \mathrm{d}W_u \sim \mathcal{N} \left( 0, \int_0^t f^2(u) \mathrm{d}u \right). \end{equation}

In your case, we have $f(t) = e^{-\lambda t}$ and thus

\begin{equation} \int_0^t f^2(u) \mathrm{d}u = \frac{1}{2 \lambda} \left( 1 - e^{-2 \lambda t} \right). \end{equation}

When it comes to simulation, the first approach simulates the SDE while the second uses the known distribution of its solution. The first approach converges as the number of time steps becomes small while in the second, you can directly simulate the time point of interest without intermediate sample points.