# $tB_t$ Integral representation, Brownian Motion

I never learned stochastic differential equations, and so am trying to do some self study. I've arrive at this question: $tB_t\sim N(0,t^3)$? $B_t$ is standard brownian motion. $B_t\sim N(0,t)$, so just applying standard probability, we get the $t^3$. Is that ok?

But I have derived the formula: $tB_t=\int_0^t sdB_s+\int_0^t B_sds$. I think that both integrals on the right are $N(0,\frac{1}{3}t^3)$ random variables. So I can't just add the variances on the left to get $\frac{4}{3}t^3$.

It seems that the integrals have a non-zero covariance, so this doesn't surprise me, but I can't seem to figure out what I am misunderstanding.

So my question(s): Is $$tB_t=\int_0^t sdB_s+\int_0^t B_sds$$ correct? If so, how am I to see that the left side as a $N(0,t^3)$? Since these are stochastic integrals, am I to think of the $B_s$ in each integral as a distinct random process?

Your stochastic integral formula is correct, as are your values for the variances of the two integrals on the right side. This implies that the covariance of those integrals is $t^3/6$. To see this directly, let $M_t$ denote the first stochastic integral on the right; this is a martingale. Therefore, $$\Bbb E\left[M_t\cdot\int_0^t B_s\,ds\right]=\int_0^t\Bbb E[M_tB_s]\,ds= \int_0^t \Bbb E[M_sB_s]\,ds.$$ But by the Ito Isometry formula, $$\Bbb E[M_sB_s]=\int_0^s u \,du= s^2/2,$$ Inserting this into the previous formula we get the covariance: $$\Bbb E\left[M_t\cdot\int_0^t B_s\,ds\right]=\int_0^t s^2/2\,ds = t^3/6.$$
• Thanks for the concise solution! Perfect. Would you be so kind as to make a comment about a conceptual way to think about the integral equation. I'm trying to understand it in a probability theoretic sense, in terms of the $B_t$'s being sample path realizations. Are all three the same path? I'm assuming that isn't the right way to think about it. The equation is in a distributional sense, that is both sides have the same distribution but aren't strictly equal for a given sample path. – jdods Apr 20 '16 at 22:42
• The wikipedia article en.wikipedia.org/wiki/It%C3%B4_isometry has what you need. We are looking at the $L^2$ inner product between the stochastic integral $\int_0^s u\,dB_u$ and the stochastic integral $\int_0^s 1\,dB_u$. – John Dawkins Apr 21 '16 at 13:06
• So then $E(\int XdW\cdot\int YdW)=E(\int XYdt)$ is a general result (for appropriate cases)? It was just a little unclear from the Wikipedia article. – jdods Apr 21 '16 at 19:13
• Yes, for $W$ a Brownian motion. – John Dawkins Apr 21 '16 at 20:12