# Integral of a Brownian motion

We are going to show that ${M_{t} = \int_{0}^{t} B_{s} ds}$ is a normal distribution ${\mathcal N(0, \frac 1 3 t^{3})}$ for each fixed ${t\ge 0}$, where ${B}$ is a Brownian motion.

First, we are going to find out the distribution of ${M_{1}}$. We can write ${M_{1}}$ using the definition of Riemann integral,

$\displaystyle M_{1} = \lim_{n\rightarrow \infty} \frac 1 n \sum_{i=1}^{n} B_{i/n}.$

If we write the increment on a segment by ${\Delta B_{i/n} = B_{i/n} - B_{(i-1)/n}}$, then

$\displaystyle M_{1} = \lim_{n\rightarrow \infty} \frac 1 n \sum_{i=1}^{n} \sum_{j=1}^{i} \Delta B_{j/n} = \lim_{n\rightarrow \infty} \sum_{i = 1}^{n} \frac{n+1- i}{n} \Delta B_{i/n}.$

Note that ${\{\Delta B_{i/n}: i= 1, \ldots n\}}$ are iid random variables of ${\mathcal N(0, 1/n)}$. Together with the fact

$\displaystyle \lim_{n\rightarrow \infty}\sum_{i=1}^{n} \frac{(n+1 - i)^{2}}{n^{3}} = 1/3,$

we have ${M_{1} \sim \mathcal N(0, 1/3)}$.

Next, we simply use change of variables to conclude the result as follows,

$\displaystyle M_{t} = \int_{0}^{1} B_{ts} d ts = t^{3/2} \int_{0}^{1} \bar B_{s}ds \sim t^{3/2} M_{1},$

where ${\bar B_{s} = t^{-1/2} B_{ts}}$.