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I am reading a textbook on a chapter about SDEs. The author mentions that the stochastic logistic differential equation can be solved by using a suitable integration factor.

$$dX_t=rX_t(K-X_t)dt+\alpha X_t dB_t$$ where $r,K,\alpha$ are all constants and $B_t$ is a standard Wiener process.

The integration factor is $Y_t=\exp\left\{\frac{\alpha^2t}2-2\alpha B_t\right\}$. Let $Z_t=X_tY_t$, and after applying Ito's product rule, we reach

$$dZ_t=rZ_t\left(K-\frac{Z_t}{Y_t}\right)dt$$

This looks like the logistic ODE but there is an annoying term $Y_t$. I do not know how to proceed further to solve it. The author comments that the solution reads

$$X_t=\frac{\exp\left\{(rK-\frac12\beta^2)t+\beta B_t\right\}}{x_0^{-1}+r\int_0^t\exp\left\{(rK-\frac12\beta^2)s+\beta B_s\right\}ds}$$ without mentioning what is $\beta$.

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  • $\begingroup$ I cannot. As far as I know, the book is still under internal review for publishing. $\endgroup$ Mar 9 at 9:04

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Define $$\begin{align} &Z_t =\frac{1}{X_t}\\ &M_t = \exp\left\{(rK-\frac12\alpha^2)t+\alpha B_t\right\} \end{align}$$ By using the Ito's lemma, we have: $$\begin{align} &\frac{dZ_t}{Z_t} =\left(\alpha^2-rK+r\frac{1}{Z_t}\right)dt-\alpha dB_t\\ &\frac{dM_t}{M_t}= rKdt+\beta dB_t\\ \end{align}$$ Then $$\begin{align} d(M_tZ_t)&=M_tdZ_t+Z_tdM_t+\frac{1}{2}\cdot2\cdot d\langle M,Z\rangle _t\\ &=M_tZ_t\left(\frac{dZ_t}{Z_t}+ \frac{dM_t}{M_t} +\frac{\langle M,Z\rangle _t}{M_tZ_t}\right)\\ &=M_tZ_t\left( \left(\left(\alpha^2-rK+r\frac{1}{Z_t}\right)+rK-\alpha^2 \right) dt +\left(-\alpha+\alpha\right)dB_t \right)\\ &=rM_tdt\\ \end{align}$$ $$\begin{align}&\implies M_tZ_t = M_0Z_0 +r\int_0^t M_sds = x_0^{-1} + r\int_0^t M_sds\\ &\implies X_t = \frac{1}{Z_t} =\color{red}{ \frac{M_t}{ x_0^{-1} + r\int_0^t M_sds}}\end{align}$$ Q.E.D

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