I was given the following problem:
Let $W$ denote a standard one-dimensional Brownian motion.
Let $S_t = e^{\sigma W_t}$ for $\sigma>0$. Use Ito's lemma to write a stochastic differential equation (SDE) for $dS_t$ in terms of $S_t$.
The solution key starts off as:
$f(x,t) = e^{\sigma x}$. Then $f_x = \sigma f$, and $f_{xx} = \sigma^2 f$. Then $\sigma(x) = 1$…
Here, I'm not able to work out how they got $\sigma(x)=1$. If I have a SDE $dX_t=bdt +\sigma dW_t$ then I know the $\sigma(x)$ function is just the coefficient of $dW_t$ but in this case where I have a solution to a SDE which I have to find I don't understand how to find the $\sigma(x)$ function.
Any help would be much appreciated.
Best Answer
For $f(x) := e^{\sigma x}$ we have
$$f'(x) = \sigma f(x) \qquad \text{and} \qquad f''(x) = \sigma^2 f(x).$$
Applying Itô's formula, we find
$$\begin{align*} e^{\sigma W_t}-1 = f(W_t)-f(0) &= \int_0^t \sigma f(W_s) \, dW_s + \frac{1}{2} \int_0^t \sigma^2 f(W_s) \, ds \\ &= \sigma \int_0^t S_s \, dW_s + \frac{\sigma^2}{2} \int_0^t S_s \, ds \end{align*}$$ where we used that, by definition, $S_s = f(W_s) = e^{\sigma W_s}$. Hence, $(S_t)_{t \geq 0}$ satisfies
$$S_t - 1 = \sigma \int_0^t S_s \, dW_s + \frac{\sigma^2}{2} \int_0^t S_s \, ds$$
which is equivalent to saying that $(S_t)_{t \geq 0}$ solves the SDE
$$dX_t = \sigma X_t \, dW_t + \frac{\sigma^2}{2} X_t \, dt, \qquad X_0 =1.$$