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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?

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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: $$ \Bbb E\left[M_t\cdot\int_0^t B_s\,ds\right]=\int_0^t s^2/2\,ds = t^3/6. $$

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  • $\begingroup$ 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. $\endgroup$ – jdods Apr 20 '16 at 22:42
  • $\begingroup$ I can't find a reference for this version of the Ito isometry. Could you point me towards one. $\endgroup$ – jdods Apr 21 '16 at 11:20
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    $\begingroup$ 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 $\int_0^s 1\,dB_u$. $\endgroup$ – John Dawkins Apr 21 '16 at 13:06
  • $\begingroup$ 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. $\endgroup$ – jdods Apr 21 '16 at 19:13
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    $\begingroup$ Yes, for $W$ a Brownian motion. $\endgroup$ – John Dawkins Apr 21 '16 at 20:12

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