How to do stochastic integration $dS = a S^b dt + c S dW$, where $a$, $b$ and $c$ are constant, $b > 0$, and $W$ is the Wiener process.

I know how to do integration for $dS = aS dt + cS dW$, or $dS / S = a dt + c dW$:

According to Ito's lemma, for $dx = \mu dt + \sigma dW$, and $f(t,x)$, $$df(t,x) = (\frac{\partial f}{\partial t} + \mu \frac{\partial f}{\partial x} + \frac{1}{2} \sigma^2 \frac{\partial^2f}{\partial x^2} ) dt + \sigma \frac{\partial f}{\partial x} dW $$

Submitting in $S\rightarrow x$, $aS \rightarrow \mu$, $cS \rightarrow \sigma$, and $ln(S)\rightarrow f(t,x)$, will get:

$$d[ln(S)] = (a - \frac{1}{2} c^2) dt + c dW$$

, then we'll have $$S(t) = exp[(a - \frac{1}{2} c^2) t + cW_t]$$

But if $S^b$ is introduced, how could I solve it?


1 Answer 1


There are several possibilites to solve this SDE:

Solution 1 You already solved the case $b=1$, so without loss of generality $b \neq 1$.

First, consider the (deterministic) ODE

$$ds = a \cdot s^b \, dt$$

The solution $s(t)$ satisfies

$$\frac{s(t)^{-b+1}}{1-b} - a \cdot t = \text{const}$$

Therefore, we try the following Ansatz:

$$Y_t := \frac{S_t^{1-b}}{1-b} - a \cdot t \tag{1}$$

The idea behind this choice is simply the following: In the deterministic case, $Y_t$ is constant. But since we add a random perturbation, we should allow $Y$ to depend on $\omega$. By applying Itô's formula,

$$dY_t = c \cdot (1-b) \cdot (Y_t+a \cdot t) \, dW_t - \frac{b}{2} \cdot c^2 \cdot (Y_t+at) \, dt$$

Now let $Z_t := Y_t + a \cdot t$,

$$dZ_t = dY_t + a \, dt = c \cdot (1-b) \cdot Z_t \, dW_t + \left( - \frac{b}{2} \cdot c^2 \cdot Z_t+a \right) \, dt$$

Consequently, $Z_t$ solves a linear SDE. Sinc it is known, how to solve this SDE, we find an expression for $Z_t$ and -obviously- for $Y_t$. From (1), we get a candidate for the solution $S_t$.

(Hopefully, my calculations are correct. Don't hesitate to ask!)

Solution 2 For SDE's of the form $$dS_t = b(S_t) \, dt + \sigma(S_t) \, dW_t \tag{2}$$ there exist equivalent conditions for a transformation into a linear SDE, for example the following:

The SDE (2) can be transformed into a linear SDE $$dZ_t = (\alpha+ \beta \cdot Z_t) \, dt + (\gamma+\delta \cdot Z_t) \, dW_t$$ if and only if $$\frac{d}{dx} \left( \frac{\frac{d}{dx}(\kappa'(x) \cdot \sigma(x))}{\kappa'(x)} \right) = 0 \tag{3}$$ where $\kappa(x) := \frac{b(x)}{\sigma(x)}- \frac{1}{2} \sigma'(x)$. The transformation $Z_t = f(X_t)$ is given by $$f(x) := \begin{cases} e^{\delta \cdot d(x)} & \delta \neq 0 \\ \gamma \cdot d(x) & \delta = 0 \end{cases}$$ where $$d(x) := \int_0^x \frac{1}{\sigma(y)} \, dy \qquad \qquad \delta = - \frac{\frac{d}{dx}(\kappa'(x) \cdot \sigma(x))}{\kappa'(x)}$$

(cf. René L. Schilling/Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes, p.278.)

You can easily check that for the given SDE, condition (3) is satisfied and therefore this theorem tells you how to transform the SDE into a linear SDE.

  • $\begingroup$ thanks a lot for your detailed answer. sorry for the late reply. appreciate your time and elaboration! $\endgroup$
    – athos
    Oct 1, 2013 at 6:08
  • 2
    $\begingroup$ @athos You are welcome. By the way, you can upvote & accept answers which satisfy you. $\endgroup$
    – saz
    Oct 1, 2013 at 16:11
  • $\begingroup$ when you write $\sigma^\prime (x)$ do you mean the derivative? The same for $\kappa^\prime (x)$? $\endgroup$
    – user48672
    Aug 23, 2017 at 10:33
  • $\begingroup$ @user48672 Yes, exactly. $\endgroup$
    – saz
    Aug 23, 2017 at 12:35

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