# How to show that this process is “normally distributed”?

Say we have following SDE (Vasicek): $$dr(t) =(b-ar_t) dt + \sigma dW_t$$

I am able to reach an integral form of this SDE : $$r(t) = r(0) e^{-at} + \frac{b}{a}[1 - e^{-at}] + \sigma e^{-at}\int_0^t e^{as}dW_s$$

From here, I would like to conclude that $r(t)$ is Gaussian but I don't know how to proceed.

I somehow understand that

$$E[r(t)] = r(0) e^{-at} + \frac{b}{a}[1 - e^{-at}]$$

and that

$$Var[r(t)]= \sigma e^{-at}\int_0^t e^{as} dW_s$$

• You should look for the definition of Ito integrals and study their properties. This will show you that the last term on the RHS is Gaussian with zero mean and variance given by Ito isometry. Noting that the first 2 terms of the RHS are on the other hand deterministic concludes the demonstration. – Quantuple Jul 4 '16 at 10:35

First, note $$\mathbb{E^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=0$$ and $$\mathbb{Var^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=\mathbb{E^Q}\left[\int_{0}^{t} e^{-2a(t-s)}ds\right]=\frac{1}{2a}(1-e^{-2at})$$ therefore $$\mathbb{E^Q}[r_t]=r_0 e^{-at} + \frac{b}{a}(1 - e^{-at})$$ $$\mathbb{Var^Q}(r_t)=\frac{\sigma^2}{2a}(1-e^{-2at})$$ second
The Itô integral can be defined in a manner similar to the Riemann–Stieltjes integral, that is as a limit in probability of Riemann sums; such a limit does not necessarily exist pathwise. Suppose that $W_t$ is a Wiener process and that $X_t$ is a right-continuous (cadlag), adapted and locally bounded process if $I=\{t_0,t_1,\cdots,t_n\}$ is a sequence of partitions of $[0,t]$ with mesh going to zero, then the Itô integral of $X_t$ with respect to $W_t$ up to time t is a random variable $$\int_{0}^{t}X_sdW_s=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{X({{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})$$ Set $X_s=e^{as}$, $X_s$ is a deterministic function thus $$\int_0^t e^{-a(t-s)}dW_s\sim N\left(0\quad,\quad\frac{1}{2a}(1-e^{-2at})\right)$$
• For an ito integral, you may need to replace $X(t_i)$ by $X(t_{i-1})$. – Gordon Jul 5 '16 at 14:07