# Prove the given stochastic integral are equally distributed

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.

• Should it be $dX_t^i = \frac \gamma 2dW^i_t + \frac \alpha 2 X^i_t dt$? – Gordon Jun 19 '18 at 17:10