# Show that Riemann integral over BM is gaussian process

I am looking at the process

$$X_t = \int_0^tB_udu$$

I know that this is a gaussian process with variance $$t^3/3$$. However, I would like to manually show the first statement directly.

For this, I would like to calculate the Laplace transform and show that it is the gaussian MGF. But the integral in the process makes that a tad difficult. My thought process that leads me to a slightly wrong result is:

We know that $$B_t \sim \sqrt{t}B_1$$. Thus: $$E\left[e^{\alpha X_t}\right] = E\left[e^{\alpha\int_0^tB_udu}\right] = E\left[e^{\alpha B_1\int_0^t\sqrt{u}du}\right],$$ which I can now integrate w.r.t. $$B_1$$. But that lands me at a MGF that is slightly off, so I suspect I can't do this transformation.

What would be the mathematically right way to calculate the MGF of $$X$$ here?

Edit: Just to be clear, of course you can do this with several ways, however I'm curious how to treat this specific situation - if possible at all.

We assume we work on a probability space $$(\Omega,\mathcal{F},\mathbb{P})$$ equipped with the filtration $$\{\mathcal{F}_t\}_t$$. By Itô's Lemma: $$B_t\text{d}t=\text{d}\left(tB_t\right)-t\text{d}B_t$$ Hence: $$X_t=tB_t-\int_0^tu\text{d}B_u$$ Let us define the function $$\theta(t) :=\sqrt[3]{3t}$$ and the filtration $$\mathcal{F}^\theta_t:=\mathcal{F}_{\theta(t)}$$. Introduce the following process: $$Y_t=\int_0^{\theta(t)}s\text{d}B_s$$ $$Y_t$$ is a local martingale with respect to the Brownian Motion with $$Y_0=0$$, hence its quadratic variation is: \begin{align} [Y,Y]_t=\int_0^{\theta(t)}s^2\text{d}s=\left[\frac{s^3}{3}\right]_0^{\theta(t)}=t \end{align} It follows from Levy's characterization theorem that $$Y_t$$ is a Brownian Motion on the filtration $$\mathcal{F}^\theta_t$$ and thus is Gaussian.
Moreover, the time-changed Brownian Motion $$\theta(t)B_{\theta(t)}$$ also remains Gaussian with respect to the filtration $$\mathcal{F}^\theta_t$$. Indeed the function $$\theta:\mathbb{R}^+\rightarrow \mathbb{R}^+$$ is a mere deterministic bijection from $$\mathbb{R}^+$$ into itself with $$\theta(0)=0$$ hence $$\theta(t)B_{\theta(t)}$$ can be represented as $$sB_s$$ where $$s\in\mathbb{R}^+$$.
As a difference of Gaussian variables, $$X_t$$ is Gaussian with respect to the filtration $$\mathcal{F}^\theta_t$$ with same distribution as $$\eta(t)+\xi(t)Z$$, where $$Z$$ is a standard normal variable and $$\eta(t), \xi(t)$$ some deterministic functions of $$t$$. Changing back to the filtration $$\mathcal{F}_t$$, it is easy to see we are merely applying some deterministic transformation to the functions $$\eta(t)$$ and $$\xi(t)$$ thus $$X_t$$ remains Gaussian under the original filtration $$\mathcal{F}_t$$.
• Good answer! However, since $\int f dB$ ist a gaussian process if f in L2, we would have the result immediately from application of the ito formula (difference of GP is GP). Currently revisiting a lot of that material for interview purposes and think I should revisit the time change, that sometimes yields really elegant solutions. However, it's not exactly what I'm looking for as I wanted to explicitly derive the MGF in this case (which involves an integral that one can't derive explicitly). Mar 30, 2020 at 16:05