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 ?
Best Answer
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?