Brownian motion with absorbing boundary Let $a\in (0,1)$, suppose to have $X_t$ the unique solution of the following SDE
$$dX_t=\sqrt {X_t(1-X_t)}dW_t$$
where $W_t$ is the standard Brownian motion.
It is known that the process $X_t$ is a Brownian motion which is absorbed in $0$ and $1$.
If I call $\tau =\inf\{t\geq 0: X_t=0 \text{ or } X_t=1\}$. How can I prove that $\tau$ is an absorbing time in the sense that once the process tauches  the boundary it stays there forever?
A second question: how can I deduce that the solution of the SDE above is the Brownian motion absorbed in $0$ and $1$, that is the process whis generator is $Lf=\frac{1}{2}x(1-x)f''$?
 A: *

*Consider the process $Y_{t} := X_{t \wedge \tau}$. As $\sigma(x):=x(1-x)$ satisfies $\sigma(1)=0=\sigma(0)$, you can easily verify that $(Y_t)_{t \geq 0}$ is a solution to the SDE $$dY_t = Y_t (1-Y_t) \, dB_t, \qquad Y_0 =a.$$ Since you know that the SDE has a unique solution, it follows that $X_t = Y_t$, i.e. $X_t$ is trapped once it has reached $0$ or $1$.

*It's a classical result that if $(X_t)_{t \geq 0}$ is a solution to an SDE $$dX_t = \sigma(X_t) \, dB_t$$ then its generator is given by $$Lf(x) = \frac{1}{2} \sigma(x)^2 f''(x), \qquad f \in C_b^2$$ (some assumptions on the coefficient $\sigma$ are required, e.g. boundedness); you can find results of this form in various books on this topic (Karatzas + Shreve, Ethier + Kurtz, Schilling + Partzsch,...). The basic idea is just to apply Itô's formula: $$\mathbb{E}f(X_t)-\mathbb{E}f(X_0) = \underbrace{\mathbb{E} \left( \int_0^t f'(X_s) \, dX_s \right)}_{0} + \frac{1}{2} \mathbb{E} \left( \int_0^t f''(X_s) \sigma^2(X_s) \, ds \right).$$ Now divide both sides by $t$ and let $t \to 0$: $$\lim_{t \to 0} \frac{\mathbb{E}f(X_t)-f(x)}{t} = \frac{1}{2} \sigma^2(x) f''(x).$$
