Solving a Simple SDE

Question

I am wondering, does, $F(X) = e^X$ a solution to the problem of

$dX = \mu dt + \sigma dW$?

My understanding of solving SDE is simply start with a guess solution, then expand using Ito's formula, if we can get back the SDE that is being asked, then we are done. But what exactly is the example below demonstrate?

Answer 1

It seems that you don't have a clear understanding of what is going on. The example is simply an application of the Ito's Formula.

Ito's Formula. Suppose that $X(\cdot)$ has a stochastic differential $$ dX = Fdt+GdW \tag{1} $$ for $F \in L^1(0,T)$, $G \in L^2(0,T)$. Assume $u:\mathbb{R} \times[0,T] \to \mathbb{R}$ is continuous and that $\frac{\partial u}{\partial t}$,$\frac{\partial u}{\partial x}$,$\frac{\partial^2 u}{\partial x^2}$ exist and continuous. $$ Z \overset{define}{=} u(X(t),t) $$ Then $Z$ has the stochastic differential $$\begin{align} dZ &= \frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}dX + \frac12 \frac{\partial^2 u}{\partial x^2}G^2dt \\ &= \left(\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}F+\frac12 \frac{\partial^2 u}{\partial x^2}G^2\right)dt + \frac{\partial u}{\partial x}GdW \tag{2} \end{align} $$

In your example, $F = \mu$, $G=\sigma$, and $Z =u(X(t),t)= e^X$. The partial derivatives are computed as $$ \frac{\partial u}{\partial t} = 0, \quad \frac{\partial u}{\partial x}=e^X=Z, \quad \frac{\partial^2 u}{\partial x^2}=e^X=Z $$ Substitute into the equation (2), one gives $$ \begin{equation} dZ= \left(0+\mu Z+\frac12 \sigma^2Z\right)dt + \sigma ZdW = \left(\mu +\frac12 \sigma^2\right)Zdt + \sigma ZdW \tag{3} \end{equation} $$ If $Z$ is a solution, we should have $$ \mu = \left(\mu +\frac12 \sigma^2\right)Z\quad \text{and}\quad \sigma = \sigma Z $$