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.

  • 1
    $\begingroup$ 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 $\mathcal{N}(0, \frac{1 - exp(-2\lambda t)}{2 \lambda})$ $\endgroup$
    – byouness
    Jun 11, 2018 at 12:32

1 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.


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