# Itô's formula and SDE

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 $(f^{-1})'=\sigma(x),(f^{-1})''=\sigma'(x)$

$dx=\frac{1}{2}(f^{-1})''(f^{-1})'dt+(f^{-1})'dW_t$

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

Help me!

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what problem are you trying to solve? –  mookid Jun 18 at 16:37
Oops! I edited. –  user157209 Jun 18 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).

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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. –  user157209 Jun 22 at 12:37
$g^{-1}$ 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 $d\langle X \rangle_t = \sigma^2(X_s) \, ds$. –  saz Jun 22 at 12:54
Is the answer of Solution 1 $X_t=g^{-1}(W_t)$? –  user157209 Jun 22 at 13:12
$(g^{-1}(x))'$ is $\sigma(g^{-1}(x))$? –  user157209 Jun 22 at 13:22
@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 $(1)$. –  saz Jun 22 at 14:28