Check solution to the SDE $dX_t = - \mu X_t \, dt+ \sigma \, dW_t$

Question

I get stuck in this problem. I just can't get the hang of how we need to "guess" a function first and almost everything along the process of solving depends on it; It's not entirely logical to me when it comes to guessing it.

Let $dX_t=-\mu X_t \, dt+ \sigma \, dW_t$ for all $t$ where $W_t$ denotes a standard Brownian motion and $\mu,\sigma, X_0$ are positive constants. Prove that

$$X_t=e^{-\mu t}X_0+\sigma \int_0^t e^{-\mu (t-u)} \, dW_u$$

for every $t$.

My attempt

Let $f(t,x)=\ln x$ then $Y_t:=\ln W_t$. Using Ito's equation, we have $\frac{\partial f}{\partial t}=0$,$\frac{\partial f}{\partial x}=\frac{1}{x}$ and $\frac{\partial ^2 f}{\partial x^2}=-\frac{1}{x^2}$.

But using this simply doesn't get me exactly where I want. Something slightly close though. My choice of $f$ is due to seeing $e$ in the final expression I want.

At times like this I am very stuck and wonder if I am not getting what I want because of my choice of $f$ or because I still haven't manipulated my expression enough(given my choice of $f$ was correct), or if I have made some minor miscalculation. I just don't know which one.

Can someone tell me how to prove this at all please??

Thank you

Answer 1

Since you already know how the solution looks like, there is no need to guess functions first. Just use Itô's formula to check that the given process is indeed a solution to the SDE.

Obviously, we have

$$X_t = e^{-\mu t} Y_t \tag{1}$$

for

$$Y_t := X_0 + \sigma \int_0^t e^{u \mu} \, dW_u. \tag{2}$$

Note that $(Y_t)_{t \geq 0}$ is an Itô process. Applying Itô's formula (for Itô processes) with

$$f(t,y) := e^{-\mu t} y$$

we get

$$\begin{align*} X_t- X_0 &= f(t,Y_t)-f(0,Y_0) \\ &= \int_0^t \partial_y f(s,Y_s) \, dY_s + \int_0^t \partial_t f(s,Y_s) \, ds + \frac{1}{2} \int_0^t \partial_{y}^2 f(s,Y_s) \, d\langle Y \rangle_s \\ &\stackrel{\star}{=} \underbrace{\int_0^t e^{-\mu s} \, dY_s}_{\stackrel{(1)}{=} \sigma \int_0^t \, dW_s} - \mu \int_0^t \underbrace{e^{-\mu s} Y_s}_{\stackrel{(2)}{=} X_s} \, ds \\ &= \sigma \int_0^t \, dW_s -\mu \int_0^t X_s \, ds \end{align*}$$

where we have used in $(\star)$ that

$$\partial_y f(t,y) = e^{-\mu t} \quad \partial_y^2 f(t,y) = 0 \quad \partial_t f(t,y) = - \mu e^{-\mu t} y.$$