distribution of \int B(t)

[Bridgeport]

Let $B(t)$ be standard Brownian motion. I need to fund distribution of $\int_{0}^{t}B(s)ds$

Ogromnoe spasibo!

[Хорхе]

$N(0,t^2/2)$ if you mean marginal distribution.

[Bridgeport]

How did you compute it? Anywhere I should look for the explanations?

[Henrylee]

So, I got $N(0,t^3/3)$.
Really, let $S_n$ - Riemann's integral sum

$$ S_n=\frac{t}{n}\sum\limits_{k=1}^nB(s_k), $$

where $i_0=0,~s_1=t/n,\dots,s_k=tk/n,\dots,s_n=t$.
Then

$$ DS_n=\frac{t^2}{n^2}\sum\limits_{i,j}r_{ij}, $$

where $r_{ij}=cov(B(s_i),B(s_j))$.
As easy to see

$$ (r_{ij})=\frac{t}{n}\left[\begin{array}{ccccc} 1&1&1&\dots&1\\ 1&2&2&\dots&2\\ 1&2&3&\dots&3\\ \dots\\ 1&2&3&\dots&n \end{array}\right] $$

And $S_n$ is gaussian,

$$ DS_n=t^3\left(\frac13+\frac1{2n}+\frac1{6n^2}\right) $$

At last it is need to prove the convergence $S_n$ in $L^2$.

[Bridgeport]

Ogromnoe Spasibo!

And I thought that Riemann integral is obsolete.

One more question: Would $\int_{0}^{t}B(s)ds$ be a martingale? My guess is yes and I kind of can prove it using uniform integrability of Riemann approximation (since bounded in $L^2$ ) and a.s. convergence.

[Henrylee]

I would like to see your ideas about the proof.
As for me, I donk think it is martingale, and that's why.
Let $\sigma$-algebra ${\cal F}_t=\sigma\left\{B(s),s\in(0,t]\right\}$. As easy to suppose

$$ X_t=\int\limits_0^t B(s)\,ds\quad {\cal F}_t-\mbox{\rm measurable} $$

Let $0<p<t$, and for all $n$ let $k(n)\in\mathbb{N}$ such that

$$ s_{k(n)}\leqslant p<s_{k(n)+1} $$

Then (as I think)

$$ \mathbb{M}\left(S_n(t)|{\cal F}_p\right)=\frac{t}{n} \sum\limits_{k=1}^{k(n)}B(s_k)+\frac{t}{n}(n-k(n))B(p)\to X_p+(t-p)B(p) $$

It's not a proof, it's just something like idea.

[Bridgeport]

You are totally right! It will not be a martingale! I agree with your computations.

(Got too excited interchanging conditional expectation and limit and forgot about everything else)

[Хорхе]

Yes, I messed somewhat. Actually, one can use integration by parts:

$$ \int_{0}^{t}B(s)ds = tB(t) - \int_0^t s dB(s) $$

and then

$$ E\left[tB(t) - \int_0^t s dB(s)\right]^2 = t^3 + \int_0^t s^2 ds - 2t\int_0^t s ds = \frac{t^3}{3}. $$

(we use that

$$E\left[\int_0^t f(s)dB(s)\int_0^t g(s)dB(s)\right] = \int_0^tE[f(s)g(s)]ds$$

for adapted -- and non-random is adapted -- processes $f$, $g$).