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If $X_t=\int_0^t \sigma (s)dB_s$ is it possible to have information on the law of $X_t$ ? For example, if $h$ is very small, then $$X_{t+h}-X_t\approx \sigma (t)(B_{t+s}-B_t)\sim\mathcal N(0,t\sigma (t)^2).$$ Can we do better ?

I often heard that Itô integral $\int_0^t \sigma (t)dB_t$ can be seen as a local brownian motion with volatility $\sigma (t)$, but I'm not sure what it mean, any idea ?

user657324
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1 Answers1

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Suppose $\sigma$ is a deterministic function. The mean will always be zero by the mean value property of ito integrals. The variance is $\int_0^t \sigma(s)^2 ds$ by the Ito isometry. It is a gaussian random variable.

See: to show it is gaussian The ito integral is gaussian

Also, you can apply both arguments in the random process $\sigma(s,B_s)$

The result is mean zero and variance $\mathbb{E}\int_0^t \sigma(s, B_s)^2 ds$

But the distribution isn't necessarily gaussian: Not necessarily gaussian: Is the Ito integral of a predictable process Gaussian distributed?

fGDu94
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  • That's indeed well remarked. Congrat. But how do you know that it will be Gaussian ? And in the case where $\sigma (t)$ is not deterministic ? – user657324 May 10 '19 at 14:41
  • To show gaussian: https://math.stackexchange.com/questions/997061/the-ito-integral-is-gaussian – fGDu94 May 10 '19 at 14:46
  • Think of it as a continuous limit of discrete sums of gaussians – fGDu94 May 10 '19 at 14:46
  • For the integral of a random process we take the expected value of the ito isometry to get the variance – fGDu94 May 10 '19 at 14:49
  • But if $\sigma (t)$ is also random, does the ito integral will be Gaussian as well ? – user657324 May 10 '19 at 15:00
  • Not necessarily gaussian: https://math.stackexchange.com/questions/2480786/is-the-ito-integral-of-a-predictable-process-gaussian-distributed – fGDu94 May 10 '19 at 15:12