# How does one prove $\int_{0}^{a}B(t)dt\sim \mathcal N(0,\frac{a^3}{3})$? [closed]

Let $B(t)$ is Brownian Motion. I want to prove the integral $\int_{0}^{a}B(t)dt$ has normal distribution , $\mathcal N(0,\frac{a^3}{3})$.

• What have you tried and where are you stuck? What properties of Brownian motion are you using? This answer might be helpful. – cardinal Nov 17 '13 at 17:59
• This question was simultaneously crossposted to math.SE, where it has an accepted answer. For future reference, please do not crosspost simultaneously. This meta post contains more details regarding the philosophy behind this. – cardinal Nov 18 '13 at 0:44

All you need to do is to consider a partition like $\bigtriangleup_a=\dfrac{a}{n}$ for $n>0$ and $a_k=k\bigtriangleup_a$ for $k=0,...,n-1$. Here $B(t)$ is a continuous function so you can approximate it by a Riemann integral as $Y_a=\bigtriangleup_a\sum_{k=0}^{n-1}B(a_k)$. The normality distribution of $Y_a$ comes from the fact that you have a linear summation of normally distributed random variables. The $E(Y_a)$ is also easy to find because $E(B(a_k))=0$. To find the variance you need to find $E(Y_a^2)$. Hint: try to write it in term of a double integral. If you have problems in finding $E(Y_a^2)$, then have a look at here. But try to get it yourself after reading some lines since this is a very standard question.