Assuming a naive stochastic process for modelling movements in stock prices we have:

$dS = \mu S dt + \sigma S \sqrt{dt}$

where S = Stock Price, t = time, mu is a drift constant and sigma is a stochastic process.

I'm currently reading Hull and they consider a simple example where volatility is zero, so the change in the stock price is a simple compounding interest formula with a rate of mu.

$\frac{dS}{S} = \mu dt$

The book states that by, "Integrating between time zero and time T, we get"

$S_{T} = S_{0} e^{\mu T}$

i.e. the standard continuously compounding interest formula. I understand all the formulae but not the steps taken to get from the second to the third. This may be a simple request as my calculus is a bit rusty but can anyone fill in the blanks?

  • $\begingroup$ Is this really $ \cdots + \sigma S \sqrt{dt}$? Maybe this doesn't matter, but I would assume that it is $\cdots + \sigma S dB_t$. The $\sqrt{dt}$ could come from a discrete simulation of the path of $S_t$ where $\sqrt{t}$ is the volatility of $dB_t$. $\endgroup$
    – Richi Wa
    Commented Dec 13, 2012 at 9:24

2 Answers 2


This is the separable differential equation for simple continuous compounding!

See this very accessible article for a step-by-step derivation (esp. under continuous compounding):


Do his first step first; integrate both sides:

$$\displaystyle \ \ \int_0^T \frac{dS(t)}{S(t)} = \mu T - 0 \,\,\,\,\,\,\,\,\,\,\,(1)$$

With zero diffusion, we know that $\langle S_.\rangle_t = 0$. Therefore, by applying Ito's lemma (or actually normal calculus):

$$d\ln{S(t)} = \frac{1}{S(t)}dS(t)\,\,\,\,\,\,\,\,\,\,\,(2)$$

Sub this into $(1)$:

$$\displaystyle \ \ \int_0^T d\ln{S(t)} = \mu T$$

$$\therefore \ln{S(T)} = \ln{S(0)} + \mu T$$

$$\therefore e^{\ln{S(T)}} = e^{\ln{S(0)}}e^{\mu T}$$

$$\therefore \boxed{S(T) = S(0)e^{\mu T}}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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