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}}.

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