Ito Integral as time changed Brownian motion

Question

Consider a probability space $(\Omega, \mathcal F, \mathbb P)$ with a Brownian motion $B_t$. Let $\mathcal F_t$ be the natural filtration generated by $B_t$. Let $Y_t$ be a $B_t$ measurable process such that $$X_t := \int_0^t Y_s dB_s$$ is well-defined, finite etc.

Define $\theta_t := \int_0^t Y_s^2 ds$.

We know that there exists a Brownian motion $W_s$ such that $X_t = W_{\theta_t}$.

The question I have is the following: Is $W_{\theta_t}$ an $\mathcal F_t$ Brownian motion?

I know that the theorems say that some enlargement of probability space maybe required. So I am a little confused.

Answer 1

not necessarily. See here for counterexamples: https://mathoverflow.net/questions/84216/a-non-degenerate-martingale/84289#84289

eg.

Another counter-example: Let $f:R\to R$ be a non-constant $C^2$ funtion with $f, f', f''$ all bounded, and such that (i) $f$ vanishes in a non-empty open interval $I$, (ii) $f'>0$ outside the closure of $I$, and (iii) $E[f(W_1)]=0$. Consider the random variable $Y_1:= f(W_1)$. Clearly $P[Y_1=0]= P[W_1\in I]>0$. The process $$ \sigma_t(\omega):=h(t,W_t(\omega)), $$ where $$ h(t,x) :=\cases{\int_R {1\over\sqrt{2\pi(1-t)}} \exp[-(y-x)^2/2(1-t)] f'(y) dy,& 0\le t<1,\cr 0,&t\ge 1,\cr} $$ is predictable and bounded, and $\sigma_t(\omega)>0$ fo all $t\in[0,1)$ and all $\omega\in \Omega$. Moreover, by Ito's formula, $$ Y_1=\int_0^1 \sigma_t dW_t, $$ almost surely. In fact, $$ E[f(W_1)|{\mathcal F}_s]=\int_0^s \sigma_t\,dW_t,\qquad\forall s>0, $$ almost surely.