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.

  • 1
    $\begingroup$ That $\sigma$ is a constant parameter and not a function is given by the exercise and, in this context, I find the mention of $\sigma(x)=1$ misleading. It might refer to the more general situation where $$dX=\sigma(X)dW+b(X)dt$$ which implies that one should plug in $$d\langle X\rangle=\sigma^2(X)dt$$ in Itô's formula. In the present case though, one uses $f(x)=e^{\sigma x}$ hence there is no other choice than $$X=W$$ thus, of course, one should plug in $$dX=dW\qquad d\langle X\rangle=dt$$ in Itô's formula for $df(X)$. What else... $\endgroup$ – Did Dec 10 '16 at 8:40

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

$$dX_t = \sigma X_t \, dW_t + \frac{\sigma^2}{2} X_t \, dt, \qquad X_0 =1.$$

  • $\begingroup$ Not sure this is what the OP is asking about, see my comment on main. $\endgroup$ – Did Dec 10 '16 at 8:41
  • $\begingroup$ @Did Well, probably we will never know. Actually, I was thinking along similar lines, but from my point of view this involves too much guessing of what the author (of the solution) might have meant. So in the end I decided to give a (hopefully) clear answer to the original question. $\endgroup$ – saz Dec 10 '16 at 10:22

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