2
$\begingroup$

Let $W^i_t$ and $W_t$ be pairwise independent Brownian motions for $i \in \{1, \dots , d\}$.

Let $X_t^i$ be $d$ independent Ornstein–Uhlenbeck processes for $i \in \{1, \dots , d\}$, i.e. each $X_t^i$ fulfills the SDE:

$$dX_t = \frac \gamma 2dW^i_t + \frac \alpha 2 X^i_t dt$$

To show: the following stochastic integrals are equally distributed:

$$\sum\int^T_0X^i_tdW^i_t \overset{d}= \int_0^T\sqrt{\sum (X^t_i)^2} dW_t$$

My attempt: if the integrands were deterministic, the Wiener integrals would be normally distributed with mean zero. However, the integrands are random.

One possibly useful observation is that the solution $X$ of the above Ornstein–Uhlenbeck equation is normally distributed with

$$X \sim \mathcal N \left(X_0e^{\,\alpha t / 2}, \frac {\gamma ^2 (e^{\alpha t} - 1)}{4\alpha} \right)$$

I thought I could apply Ito's isometry at some point, though in this situation it might only help me to calculate the variance of the random variable and not it's distribution.

I also tried to apply Ito's lemma to the differential $d\left(W_t\cdot\sqrt{\sum (X_t^i) ^ 2}\right)$ without any success.

Any help appreciated.

$\endgroup$
1
  • $\begingroup$ Should it be $dX_t^i = \frac \gamma 2dW^i_t + \frac \alpha 2 X^i_t dt$? $\endgroup$
    – Gordon
    Jun 19, 2018 at 17:10

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.