[Math] How to solve a linear stochastic differential equation

brownian motionstochastic-differential-equationsstochastic-integrals

I don't know how to find a solution of this stochastic differential equation:

$dX_{t}=(1+\delta \mu X_{t})dt+\delta X_{t}dB_{t}$

Where $B_{t}$ is a standard Brownian motion and $\mu$ and $\delta$ are real numbers.

Context

I've to demonstrate that $X_{t}=\int_{0}^{t}\exp{[\delta (B_{t}-\mu t -B_{s}+\mu s)-\frac{1}{2}(t-s)]ds}$

I know Ito's formula $dY=\dfrac {\partial g}{\partial t} dt + \dfrac {\partial g}{\partial x} dX_t + \dfrac{1}{2} \dfrac {\partial^2 g}{\partial x^2} dt$ for $Y=g(t,X)$.

Best Answer

Hints:

  1. First, solve the homogeneous SDE $$dX_t = \delta \mu X_t \, dt + \delta X_t \, dB_t. \tag{1}$$ To this end, apply Itô's formula to $f(x) := \log x$ and $(X_t)_{t \geq 0}$.
  2. In order to solve the SDE $$dX_t = (1+\delta \mu X_t) \, dt + \delta X_t \, dB_t,$$ we use a variation of constants, i.e. we make the Ansatz $$X_t = Z_t X_t^0$$ where $X_t^0$ is a solution of the SDE $(1)$ with initial value $X_0^0=1$. To find the stochastic differential of $(Z_t)_{t \geq 0}$, apply Itô's formula.

Literature: The monograph Brownian Motion - An Introduction to Stochastic Processes by René Schilling & Lothar Partzsch treats (linear) SDEs.

Solution

  1. By Itô's formula, $$\begin{align*} \log X_t-\log X_0 &= \int_0^t \frac{1}{X_s} \, dX_s - \frac{1}{2} \int_0^t \frac{1}{X_s^2} d\langle X \rangle_s \\ &= \int_0^t \delta \, dB_s + \int_0^t \left(\delta \mu - \frac{1}{2} \delta^2 \right) \, ds \\ &= \delta B_t + \left( \delta \mu - \frac{1}{2} \delta^2 \right) \cdot t. \end{align*}$$ Consequently, $$X_t = X_0 \exp \left( \delta B_t + \left( \delta \mu - \frac{1}{2} \delta^2 \right) t \right).$$

  2. Set $Y_t := 1/X_t^0$, i.e. $$Y_t = \exp \left(-\delta B_t - \left( \delta \mu - \frac{1}{2} \delta^2 \right)t \right). \tag{2}$$ Applying Itô's formula yields, $$Y_t - Y_0 =- \delta \int_0^t Y_s \, dB_s + \frac{\delta^2}{2} \int_0^t Y_s \, ds + \left(\frac{\delta^2}{2}-\delta \mu \right) \int_0^t Y_s \, ds,$$ i.e. $$dY_t = -\delta Y_t \, dB_t+(\delta^2- \delta \mu) Y_t \, dt.$$ From $$X_t \cdot Y_t = \int_0^t Y_s \, dX_s + \int_0^t X_s \, dY_s + \langle X,Y \rangle_t$$ it follows that $$\begin{align*} Z_t &= X_t Y_t \\ &= \int_0^t Y_s ((1+\delta \mu X_s) \, ds + \delta X_s) \,dB_s) + \int_0^t X_s \, (-\delta Y_s dB_s + (\delta^2- \delta \mu) Y_s \, ds) \\ &\quad -\delta^2 \int_0^t Y_s X_s \, ds \\ &= \int_0^t Y_s \, ds. \end{align*}$$ Note that $(Y_t)_{t \geq 0}$ is known explicitly, cf. $(2)$. Consequently, we get $$\begin{align*} X_t &= \frac{Z_t}{Y_t} \\ &= \exp\left( \delta B_t + \left( \delta \mu- \frac{\delta^2}{2} \right)t\right) \cdot \left[X_0 + \int_0^t \exp \left(-\delta B_s - \left( \delta \mu - \frac{\delta^2}{2} \right) s \right) \, ds \right]. \end{align*}$$

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