2

Consider the Itō integral $X_t := \int_0^t s \,dB_s$.

Here is my attempt. Let $f(t,x) = tx$. By Itō's formula, $$ d f(t, B_t) = B_t dt + t dB_t $$ so $$ t B_t = \int_0^t B_s\, ds + \int_0^t s \,dB_s. $$ But how is $\int_0^t B_s\,ds$ calculated?

Are $\int_0^t f(s) \,dB_s$ and $\int_0^t f(s, B_s)\,ds$ not solvable for any $f$, and generally left as they are when computing Itō integrals and solving Itō SDEs?

Thanks and regards!

kahen
  • 15,760
Tim
  • 47,382

2 Answers2

4

I don't think there is an explicit formula for $\int_0^t B_s ds$ but as it is a limit of normal random variables it is still a normal random variable and thus you just need to calculate its variance and mean. It is not difficult to see that $E\left( \int_0^t B_s ds \right)$ is zero. The variance is a little more tricky we have

\begin{align}E\left( \left( \int_0^t B_s ds \right)^2 \right) &= E\left( \int_0^t \int_0^t B_u B_s \, du \, ds \right) \\ \, &= \int_0^t \int_0^t E(B_u B_s) \, du \, ds \\ \, &= \int_0^t \int_0^t \min(u, s) \, du \, ds = \frac{t^3}{3} \end{align} The Fubini steps can be all well justified.

Bunder
  • 2,423
  • Thanks! When solving ito integrals and Ito SDEs, are we satisfied and stop when the results contain only $\int f(B_s) ds$ and/or $\int f(s) dB_s$, without continuing to solve $\int f(B_s) ds$ and/or $\int f(s) dB_s$? – Tim Mar 30 '13 at 22:31
  • I think so. I mean, you could use Ito's formula to find alternative presentations of these integrals, but if you already used it once you are most likely going back to the beginning. – Bunder Mar 30 '13 at 22:56
0

The Fubini trick is nice. But consider finding distribution of $\int_0^t B_s ds$ using integration by parts and Ito isometry: $$X_t=\int_0^t B_s ds=tB_t-\int_0^tsdB_s=\int_0^t(t-s)dB_s$$ So as $t-s$ is deterministic, this thing is a mean 0 Gaussian process with independent increments with covariance function $$Cov(X_s,X_t)=\int_0^{\min(s,t)} (t-u)(s-u)du$$

Daniel Li
  • 3,200