# Showing bounds of Stochastic Process

Suppose that we have the SDE:

$$dZ_t = 2Z_t(1-Z_t)dt + 4Z_t(1-Z_t)dB_t$$

With $$Z_0 = \frac{1}{3}$$. How can I show that $$0 \leq Z_t \leq 1$$.

I have tried solving the 'alalogous' differential equation, replacing $$B$$ with some function $$f$$ giving a solution of the form (assuming $$f(0)=0$$):

$$Z(t) = \frac{\exp[2t+4f(t)]}{1+\exp[2t+4f(t)]}$$

Which is recognized as the logistic function which is known to be within the desired bound, however I'm not sure how I can translate this for the Brownian case (or if this at all relevant to the Brownian case), or if there is a different solution entirely. Any help would be appreciated.

## 1 Answer

Consider the equation

$$\mathrm{d}Y_t = \biggl( 2 + 8 \frac{e^{Y_t} - 1}{e^{Y_t} + 1} \biggr) \, \mathrm{d}t + 4 \, \mathrm{d}B_t. \tag{1}\label{e:1}$$

The coefficients $$\mu(y, t) = 2+\operatorname{tanh}(y/2)$$ and $$\sigma(y, t) = 4$$ are bounded and Lipschitz, hence this equation has a unique strong solution for all time $$t \geq 0$$ and for each sensible initial condition. Similarly, the equation

$$\mathrm{d}Z_t = 2Z_t(1-Z_t) \, \mathrm{d}t + 4Z_t(1-Z_t) \, \mathrm{d}B_t \tag{2}\label{e:2}$$

has bounded and Lipschitz coefficients as functions of $$(t, Z_t)$$, hence admits a unique strong solution. Now, these two equations are related via the relation

$$Z_t = \frac{e^{Y_t}}{e^{Y_t} + 1}. \tag{3}\label{e:3}$$

Indeed, assuming $$Y_t$$ solves $$\eqref{e:1}$$,

\begin{align*} \mathrm{d}Z_t &= \frac{e^{Y_t}}{(e^{Y_t} + 1)^2} \, \mathrm{d}Y_t - \frac{e^{Y_t}(e^{Y_t} - 1)}{2(e^{Y_t}+1)^3} \, (\mathrm{d}Y_t)^2 \\ &= Z_t(1 - Z_t) \, \mathrm{d}Y_t - \frac{Z_t(1-Z_t)}{2} \frac{e^{Y_t} - 1}{e^{Y_t} + 1} \, (\mathrm{d}Y_t)^2 \\ &= Z_t(1-Z_t) \biggl(2 + 8 \frac{e^{Y_t} - 1}{e^{Y_t} + 1} \biggr) \, \mathrm{d}t + 4 Z_t(1-Z_t) \, \mathrm{d}B_t - 8 Z_t(1-Z_t) \frac{e^{Y_t} - 1}{e^{Y_t} + 1} \, \mathrm{d}t \\ & = 2Z_t(1-Z_t) \, \mathrm{d}t + 4Z_t(1-Z_t) \, \mathrm{d}B_t, \end{align*}

which is precisely $$\eqref{e:2}$$. Therefore, by the uniqueness, the solution of $$\eqref{e:2}$$ with initial condition $$Z_0 = \frac{1}{3}$$ is of the form $$\eqref{e:3}$$ with $$Y_t$$ solving $$\eqref{e:1}$$ and $$Y_0 = \log\frac{1}{2}$$. Finally, it is clear from $$\eqref{e:3}$$ that $$0 < Z_t < 1$$ for all $$t \geq 0$$ almost surely.