3

Let $X_t$ follows a Ornstein-Uhlenbeck process

$dX_t = -\theta X_t dt + dW_t$

for $\theta < 1$ and $X_0 = 0$.

I am interested in the stochastic integral

$Y_T = \int_0^T X_s dX_s$ and I wonder if what I do to solve it is the right approach. I am asking because I am aware of the many conditions under which solution to stochastic integral exist, but haven't studied those in detail.

I recall the solution to

$Z_T = \int_0^T W_s dW_s$ is $\frac{1}{2} (W_T^2 - [W, W]_T)$ = $\frac{1}{2} (W_T^2 - T)$, which follows a gamma distribution.

If I use the same trick for the case of OU process above, I get

$Y_T = \frac{1}{2}(X_T^2 - [X,X]_T) = \frac{1}{2}(X_T^2 - T)$, where we know $X_T \sim N(0, \frac{1}{2\theta}(1-e^{2\theta T}))$, so $X_T^2$ also follows a gamma distribution.

However, I feel somewhat unsettled with this solution. It is with the fact that $W_s$ and $dW_s$ are independent but $X_s$ and $dX_s$ are not independent, and somehow I think that would make the solution of $Y_T$ different from $Z_T$ (in some profound way.)

My question is, what is the solution of $Y_T$ and does the dependence of increments affect the stochastic integral?

Steve Yang
  • 43
  • Related: "integrated OU process" (instead of integrating the OU process wrt itself you integrate it in time) https://math.stackexchange.com/q/4431021/532409 – Quillo Apr 30 '22 at 18:21

1 Answers1

2

Note that \begin{align*} d\left(X_t^2\right) &= 2X_t dX_t + d\langle X, X\rangle_t \\ &= 2X_t dX_t + dt. \end{align*} Then \begin{align*} \int_0^T X_s\, dX_s = \frac{1}{2}\left(X_T^2 -T\right). \end{align*}

Gordon
  • 4,541