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 $dB_t$. $\endgroup$ – Richard Dec 13 '12 at 9:24

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 $(1)$:

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

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

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

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


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