# On the confinement to $[0;1]$ of the solution of $dX_{t}=(1-X_{t})X_{t}dB_{t}$

One considers the stochastic differential equation $$dX_{t}=(1-X_{t})X_{t}dB_{t},$$ with $B$ Brownian motion, and one assumes that $0\leq X_{0}\leq1$.

One wants to show that $\textbf{P}(X_{t}\in[0;1])=1$, and that, if $X_{s}=0$ then $X_{t}=0$ for every $t\geq s$.

Does anybody know how to prove this? Some help would be appreciated

## 1 Answer

Assume that $X_0$ is in $(0,1)$, let $T=\inf\{t\gt0\mid X_t=0\ \text{or}\ X_t=1\}$, thus $X_t$ is in $(0,1)$ for every $t\lt T$. Consider, for every $t\lt T$, $$Y_t=\log\left(\frac{X_t}{1-X_t}\right).$$ Then $T$ is also $T=\inf\{t\gt0\mid Y_t=\pm\infty\}$ and Itô's formula shows that, for every $t\lt T$, $$Y_t=Y_0+W_t+\frac12\int_0^tZ_s\mathrm ds,\qquad Z_t=\frac{\mathrm e^{Y_t}-1}{\mathrm e^{Y_t}+1}.$$ One sees that $|Z_t|\lt1$ for every $t\lt T$, in particular, $$|Y_t-Y_0-W_t|\leqslant\frac12t,$$ for every $t\lt T$. This proves that $|Y_t|$ is finite for every $t$ and that $T=+\infty$ almost surely, that is, $X_t$ is in $(0,1)$ almost surely for every $t$.

The same technique applies to the strong solution of $$\mathrm dX_t=C(X_t)\mathrm dW_t,$$ starting from $X_0$ in $(a,b)$, for every function $C:[a,b]\to\mathbb R$ positive on $(a,b)$, zero at $a$ and at $b$, and with bounded derivative on $[a,b]$. Then $(X_t)$ stays in $(a,b)$ forever, almost surely.

To show this, one considers the function $D$ defined on $(a,b)$, for some $c$ in $(a,b)$, by $$D(x)=\int_c^x\frac{\mathrm dz}{C(z)}.$$ Then $$D(X_t)=D(X_0)+W_t-\frac12\int_0^tC'(X_s)\mathrm ds,$$ thus, $$|D(X_t)-D(X_0)-W_t|\leqslant\frac12t\,\|C'\|_\infty,$$ for every $t$, and, since $\lim\limits_{x\to b^-}D(x)=+\infty$ and $\lim\limits_{x\to a^+}D(x)=-\infty$, the same reasoning applies.

• ok, i understand – John Jun 29 '14 at 18:34