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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.

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1 Answer 1

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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.

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