Solve this SDE: $dX_t=\frac{1}{2}\sigma(X_t)\sigma'(X_t)dt+\sigma(X_t)dW_t$ with $X_0=x_0$

My try is let $f(x)=\int_{x_0}^{x}\frac{dy}{\sigma(y)}$ and $(f^{-1})'=\sigma(x),(f^{-1})''=\sigma'(x)$


However, I don't understand how to use Ito's formula. So I cannot solve this.

Help me!

  • $\begingroup$ what problem are you trying to solve? $\endgroup$ – mookid Jun 18 '14 at 16:37
  • $\begingroup$ Oops! I edited. $\endgroup$ – user157209 Jun 18 '14 at 16:48

First of all, note that we need further assumptions on $\sigma$, e.g. that $\sigma$ is stricly positive or negative, bounded and at least twice differentiable. Let us assume without loss of generality that $\sigma$ is strictly positive.

Solution 1 We set

$$g(x) := \int_{x_0}^x \frac{dy}{\sigma(y)}.$$

Since $g$ is strictly positive, we know that $g$ has an inverse, $f := g^{-1}$. It follows from differential calculus that

$$\begin{align*} f'(x) &= \frac{1}{g'(f(x))} =\sigma(f(x)) \tag{1} \\ f''(x) &= \sigma'(f(x)) f'(x) = \sigma'(f(x)) \cdot \sigma(f(x)) \tag{2} \end{align*}$$

By Itô's formula,

$$f(B_t+c)-f(c)= \int_0^t f'(B_s+c) \, dB_s + \frac{1}{2} \int_0^t f''(B_s+c) \, ds$$

for any $c \in \mathbb{R}$. Plugging $(1)$ and $(2)$ into this equation shows that $X_t := f(B_t+c)$ solves the given SDE. As $X(0) \stackrel{!}{=} x_0$, we get $c=g(x_0)$.

Solution 2 Solution 1 works fine if we already know how the solution looks like - otherwise, we are stuck. So, here is an alternative approach: If we want to solve an SDE of the form

$$dX_t = \alpha(X_t) \, dB_t + \beta(X_t) \, dt,$$

i.e. an SDE where the coefficients do not depend explicitely on the time $t$, then the substitution

$$g(x) := \int^x \frac{1}{\alpha(y)} \,dy$$

is worth a try. Here, we have $\alpha(y) = \sigma(y)$ and $\beta(y) = \sigma(y) \cdot \sigma'(y)$. Hence, we define

$$g(x) := \int_0^x \frac{1}{\sigma(y)} \, dy.$$

Then, obviously, $$g'(x) = \frac{1}{\sigma(x)} \qquad \qquad g''(x) = -\frac{\sigma'(x)}{\sigma(x)^2}.$$

Applying Itô's formula yields

$$\begin{align*} g(X_t)-g(X_0) &= \int_0^t g'(X_s) \, dX_s + \frac{1}{2} \int_0^t g''(X_s) \, d\langle X \rangle_s \\ &= \int_0^t dB_s + \frac{1}{2} \int_0^t \sigma'(X_s) \, ds - \frac{1}{2} \int_0^t \sigma'(X_s) \, ds \\ &= B_t \end{align*}$$

Here we used that the quadratic variation $\langle X \rangle_t$ is given by $$\langle X \rangle_t = \int_0^t \sigma(X_s)^2 \, ds.$$

Consequently, we find that $X_t = g^{-1}(B_t+g(X_0))$. Mind that we have to ensure that all expressions are well-defined; in general, we will only obtain a solution for $t<\tau$ where $\tau$ is some suitable stopping time.

Remark The result is a special case of a theorem by Sussmann and Doss, see e.g. Karatzas/Shreve or Schilling/Partzsch (2nd edition).

  • $\begingroup$ In solution 2, $$g(x) := \int^x \frac{1}{\sigma(y)} \,dy$$, but what is $g^{-1}$? Moreover, I can't understand why $\int_0^t g′(Xs)dXs+\frac{1}{2}\int_0^tg′′(Xs)d<X>s=\int_0^tBs+\frac{1}{2}\int_0^tσ′(Xs)ds−\frac{1}{2}\int_0^tσ′(Xs)ds$ can transform. $\endgroup$ – user157209 Jun 22 '14 at 12:37
  • $\begingroup$ $g^{-1}$ is simply the inverse function of $g$. Concerning your second question: This follows from the fact that $$dX_t = \frac{1}{2} \sigma'(X_t) \sigma(X_t) \, dt + \sigma(X_t) \, dW_t$$ and $d\langle X \rangle_t = \sigma^2(X_s) \, ds$. $\endgroup$ – saz Jun 22 '14 at 12:54
  • $\begingroup$ Is the answer of Solution 1 $X_t=g^{-1}(W_t)$? $\endgroup$ – user157209 Jun 22 '14 at 13:12
  • $\begingroup$ $(g^{-1}(x))'$ is $\sigma(g^{-1}(x))$? $\endgroup$ – user157209 Jun 22 '14 at 13:22
  • $\begingroup$ @user157209 Sorry, there was a typo in the first solution. Obviously, the solution of the SDE has to be the same for solution I and II. And concerning your second question: Yes, see equation $(1)$. $\endgroup$ – saz Jun 22 '14 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.