Showing a Brownian Motion Integral Is Gaussian

# Show that Riemann integral over BM is gaussian process


# 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.

## Answer by Daneel Olivaw (score 3)

https://quant.stackexchange.com/a/52930

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 $\mathcal{F}_t$.

Background:

- Lévy’s Characterization of Brownian Motion

- Time-changed Brownian Motion