Skip to content
All library documents

Lognormal Stock Returns: CDF, Density, and Serial Correlation

Article Quant Q&A · Author: kroneckersdelta
Read the original

Summary

The document derives the distribution of a one-period stock price ratio under a geometric Brownian motion model. Taking the ratio cancels the initial price and leaves an exponential of the drift-adjusted return and the Brownian increment. Because the increment over a unit interval is standard normal, the ratio is lognormally distributed, and its cumulative distribution function follows by applying the logarithm and the standard normal CDF. Differentiating that CDF gives the probability density, with the usual change-of-variable factor proportional to the inverse price ratio.

The answer also states that adjacent one-period ratios have zero correlation under the model, since they depend on independent Brownian increments. These conclusions assume constant model parameters and nonoverlapping unit intervals; they describe the specified mathematical model rather than empirical stock returns, which may exhibit dependence or other departures from lognormality.

Key ideas

  • A one-period price ratio under geometric Brownian motion is lognormally distributed.
  • The normal CDF applied to the standardized log ratio gives its cumulative distribution.
  • Differentiating the CDF yields the density, including an inverse-ratio change-of-variable factor.
  • Adjacent ratios are uncorrelated in this model because their Brownian increments are independent.

Tags

StatisticsEquities
Full text
# CDF&density of stock price modeled by standard brownian motion


# CDF&density of stock price modeled by standard brownian motion












Assume that the price of the stock follows the model $S(t) = S(0) exp ( mt − ((σ^2)/2 ) t + σW(t) )$ , (1) where W(t) is a standard Brownian motion; σ > 0, S(0) > 0, m are some constants.

Derive the CDF and PDF for $S(t)/S(t-1)$.

CDF:

$S(t)/S(t-1) = \frac{S(0)exp(mt-\frac{\sigma^2}{2}t+\sigma W(t))}{S(0)exp(m(t-1) -\frac{\sigma^2}{2}(t-1)+\sigma W(t-1))}$ using standard algebra and rewriting I get

$S(t)/S(t-1)=exp(m-\frac{\sigma^2}{2}+\sigma(W(t)-W(t-1))$. Using that $W(t)-W(t-1)$ is $N(0,1)$ I get that $S(t)/S(t-1) = exp(m-\frac{\sigma^2}{2}+\sigma Z)$, where Z is $N(0,1)$

CDF is thus $F(x)=P(exp(m-\frac{\sigma^2}{2}+\sigma Z)\le x)$ = $F(x)=P(Z\le \frac{ln(x)-m+\frac{\sigma^2}{2}}{\sigma})$ which is $\Phi(\frac{ln(x)-m+\frac{\sigma^2}{2}}{\sigma})$.

Is this correct?

If the CDF is correct, can I use it to derive the PDF? Or how would I go about calculating the PDF?

is it possible to calculate the correlation between $S(t)/S(t − 1)$ and $S(t − 1)/S(t − 2)$ from this? Or how can it be done?

## Answer by ZRH (score 1, accepted)

https://quant.stackexchange.com/a/44632

I agree with your derivation.

$\mathrm{pdf}=\frac{d(\mathrm{CDF})}{dx}=\frac{d\Phi(\frac{ln(x)-m+\sigma^2/2}{\sigma})}{dx}=\frac{1}{x\sigma}\phi(\frac{ln(x)-m+\sigma^2/2}{\sigma})$.

As for your question on the correlation between $S(t)/S(t-1)$ and $S(t-1)/S(t-2)$, there is none because the Z~(0,1) are not serially correlated.

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.