Let $(t,\omega )\mapsto f(t,\omega )$ a "nice" function. Consider the process $$X_t:=\int_0^t f(s,\cdot )dW_s,$$ what does $(X_t)$ represent exactly ? My teacher say that it's the random position at time $t$ where $\mathbb E[X_t]=0$ and $\mathbb E[X_t]=\int_0^t \mathbb E[f(s,\cdot )^2]ds.$
My Question : Can we rather say that it's the random position of a particule with random speed $V_t$, where $\mathbb E[V_t]=0$ and $\mathbb E[V_t^2]=\mathbb E[f(s,\cdot )^2]$ ? And if $f(t,\cdot )=f(t)$ (i.e. it's deterministic), then $V_t\sim \mathcal N(0,f(s)^2)$ ? Or this doesn't make sense ? And since $$W_t=\int_0^t dB_s,$$ then Brownian motion would have speed $V_t\sim \mathcal N(0,1)$ for all $t$.
Does such interpretation works ?
One can think of the original Brownian motion $W$ as a process whose infinitesimal increment $dW_t$ has mean $0$ and variance $dt$. Moreover, these increments are independent for different $t$. The analogous statement about $X$ would be that an infinitesimal increment $dX_t$ has mean $0$ and variance $f(t,\cdot)^2dt$. In most cases the increments of $X$ won't be independent.