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

  • $\begingroup$ What have you tried and where are you stuck? What properties of Brownian motion are you using? This answer might be helpful. $\endgroup$ – cardinal Nov 17 '13 at 17:59
  • $\begingroup$ 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. $\endgroup$ – 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.

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    $\begingroup$ Just as a side note: One needs to consider a little more than just the partition argument. Namely, the associated limit must be shown to also be a normal random variable. This is true, but doesn't simply follow "naively" from the fact that each element of the sequence is normal. (+1) $\endgroup$ – cardinal Nov 18 '13 at 0:22
  • $\begingroup$ My intention was not to provide a rigorous proof. But it is good that you pointed this out. $\endgroup$ – Stat Nov 18 '13 at 3:59
  • $\begingroup$ That's fine; I was not criticizing, but rather simply making an auxiliary remark, in part so that casual readers would not take the phrase "all you need to do..." too literally. :-) $\endgroup$ – cardinal Nov 18 '13 at 23:22

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