Integrating a Zero-Volatility Stock Price Model to Derive Continuous Compounding
Summary
The document shows how to solve the stock-price differential equation when volatility is set to zero. In that case, the proportional price change is the constant drift rate multiplied by the change in time. Separating variables and integrating from the initial time to a later time gives the change in the natural logarithm of price; exponentiating yields the familiar exponential growth formula for the stock price.
The accepted explanation also frames the equation as a separable differential equation. A second answer emphasizes that, with no diffusion term, the ordinary chain rule applies to the logarithm of price, producing the same result without a stochastic correction. This is a derivation of deterministic continuous compounding, not a solution to the full stochastic model with nonzero volatility. The document offers the algebraic steps but no empirical evidence about whether a constant-drift model describes actual stock returns.
Key ideas
- With volatility set to zero, the proportional stock-price change equals drift times elapsed time.
- Separating variables turns the equation into an integral of the reciprocal price.
- Integrating the reciprocal-price differential gives a change in log price.
- Exponentiating the integrated equation produces exponential growth at the drift rate.
- The derivation is deterministic and does not address the stochastic correction when volatility is nonzero.
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# Missing step in stock price movement equations
# Missing step in stock price movement equations
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?
## Answer by vonjd (score 7, accepted)
https://quant.stackexchange.com/a/520
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): http://plus.maths.org/content/have-we-caught-your-interest
## Answer by Jase (score 3)
https://quant.stackexchange.com/a/4735
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 $(1)$:
$$\displaystyle \ \ \int_0^T d\ln{S(t)} = \mu T$$
$$\therefore \ln{S(T)} = \ln{S(0)} + \mu T$$
$$\therefore e^{\ln{S(T)}} = e^{\ln{S(0)}}e^{\mu T}$$
$$\therefore \boxed{S(T) = S(0)e^{\mu T}}$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.