Stochastic-Calculus

How to show that this process is ’normally distributed'?

  • July 5, 2016

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

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

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 $$ \int_0^t e^{-a(t-s)}dW_s\sim N\left(0\quad,\quad\frac{1}{2a}(1-e^{-2at})\right) $$

引用自:https://quant.stackexchange.com/questions/27924