How to show $Y(t)=\ln(\frac{X(t)}{1-X(t)})$ has a constant diffusion coefficient

Question

A Process $X(t)$ on $(0,1)$ has a stochastic differential with coefficient $\sigma(x)=x(1-x)$. Assuming $0<X(t)<1$, show that the process defined by $Y(t)=\ln(\frac{X(t)}{1-X(t)})$ has a constant diffusion coefficient.

Answer 1

By assumption, $(X_t)_{t \geq 0}$ satisfies the SDE

$$X_t-X_0 = \int_0^t X_s \cdot (1-X_s) \, dB_s \qquad X(0) \in (0,1).$$

Applying Itô's formula to the function $f(x) := \ln(x)-\ln(1-x)$ yields

$$\begin{align*} Y_t-Y_0 &= \ln(X_t) - \ln (1-X-X_t) - \ln (X_0)-\ln(1-X_0) \\ &= \int_0^t \left( \frac{1}{X_s} + \frac{1}{1-X_s} \right) \, dX_s + \frac{1}{2} \int_0^t \left( - \frac{1}{X_s^2} + \frac{1}{(1-X_s)^2}\right) \, d\langle X \rangle_s \tag{1} \end{align*}$$

where

$$dX_s = X_s \cdot (1-X_s) \, dB_s \qquad d \langle X \rangle_s = X_s^2 (1-X_s)^2 \, ds \tag{2}$$

Combining $(1)$ and $(2)$, we find

$$Y_t-Y_0 = \int_0^t 1 \, dB_s + \frac{1}{2} \int_0^t (2X_s-1) \, ds = B_t + \frac{1}{2} \int_0^t (2X_s-1) \, ds$$