first question here :)
I am reading an article about the Heston model and stochastic volatility, in which they state the following: "because the process for $V_t$ is independent of the brownian motion $W_t$, the distribution of $\int_u^t \sqrt{V_s} \, dW_s$ is normal with mean $0$ and variance $\int_u^t V_s \, ds$". This statement is very similar to the one I know, which is the case of a deterministic function $f(t)$ instead of the variance process $V_t$. However I have never heard before a version of this where we consider a stochastic process independent of a brownian motion, hence I don't quite understand why this is true.
Does anyone have a (sketch of) proof of the first statement ? Or any reference where I could find it ? Thanks !
A stochastic integral $\int_0^tU_s\,dW_s$ is in general not Gaussian when the integrand is not deterministic, even if $U$ and the Brownian motion $W$ are independent. Take for example $$ U_s=1_{\{s\le \tau\}} $$ where $\tau$ is independent of $W$ and exponentially distributed with parameter $\lambda\,.$ Then $$ \int_0^tU_s\,dW_s=W_{\tau\wedge t}\,. $$ Conditional on $\tau\,,$ $W_{\tau\wedge t}$ has variance $\tau\wedge t\,.$ Therefore its characteristic function is \begin{align} \mathbb E\left[e^{i\,x\,W_{\tau\wedge t}}\right]&= \mathbb E\left[e^{-\frac12x^2(\tau\wedge t)}\right] =\lambda\int_0^te^{-\frac12x^2u}e^{-\lambda u}\,du+\lambda\int_t^\infty e^{-\frac12x^2t}e^{-\lambda u}\,du\\ &=\lambda \frac{1-e^{-(\frac{x^2}2+\lambda) t}}{\frac{x^2}{2}+\lambda}+e^{-(\frac{x^2}2+\lambda) t}\,. \end{align} If $W_{\tau\wedge t}$ were Gaussian this should however be of the form $e^{-\frac12x^2\sigma^2}\,.$