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Deriving the Covariance of Geometric Brownian Motion Prices

Article Quant Q&A · Author: RRG
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Summary

The document works through the covariance between prices at two times under geometric Brownian motion. It writes the process as an exponential of drift and Brownian motion, uses the lognormal expectation to obtain each price’s mean, and calculates the joint expectation by separating independent Brownian increments. Subtracting the product of the means gives the covariance in exponential form for an earlier and a later observation.

The derivation is conditional on the stated GBM model and uses a unit initial price without loss of generality; a different initial value scales the covariance accordingly. The response leaves the expression unsimplified and does not discuss estimation from market data, model fit, or implications for trading. It is therefore a mathematical result for a specified stochastic process, rather than evidence that observed asset prices follow GBM.

Key ideas

  • Under GBM, each price is lognormally distributed.
  • Price covariance is obtained from the joint expectation minus the product of the individual means.
  • For ordered times, independent Brownian increments simplify the joint expectation.
  • The derivation assumes the GBM model and a unit initial price.
  • The final covariance expression is left in an unsimplified exponential form.

Tags

StatisticsVolatilityRisk management
Full text
# Auto-correlation of GBM


# Auto-correlation of GBM












The GBM is defined by $ dS(t) = \mu S(t)dt + \sigma S(t) dW_t, $ with analytical solution $ S(t^\prime) = S(t) exp\left[\left(\mu-\frac{\sigma^2}{2}\right)\left(t^\prime-t\right)+\sigma\left(W(t^\prime)-W(t)\right)\right]. $ What is the analytical expression for the covariance, $ Cov[S(t),S(t^\prime)] $ ?

## Answer by math (score 4, accepted)

https://quant.stackexchange.com/a/7717

W.l.o.g we use the initial condition $S(0)=1$ and define $\gamma:=\mu-\frac{\sigma^2}{2}$. Hence we have the dynamics

$$S_t=e^{\gamma t +\sigma W_t}$$

By definition $Cov(X,Y)=E((X-E(X))(Y-E(Y))=E(XY)-E(X)E(Y)$, where the last equality follows from the linearity of the expectation. Note $\gamma t+\sigma W_t$ is normal distributed with mean $\gamma t$ and variance $\sigma^2t$. Hence $S_t$ is lognormal with mean $\phi_t:=E[S_t]=e^{\gamma t+\frac{\sigma^2t}{2}}$. For $t>r$ this yields,

$$Cov(S_t,S_r) =E((S_t-\phi_t)(S_r-\phi_r))=E(S_rS_t)-\phi_r\phi_t$$

The last step is to calculate $E(S_tS_r)$.

$$E(S_rS_t)=e^{\gamma(t+r)}E(e^{\sigma(W_t+W_r)})=e^{\gamma(t+r)}E(e^{\sigma(W_t-W_r)}e^{2\sigma(W_r)})=e^{\gamma(t+r)}E(e^{\sigma(W_t-W_r})E(e^{2\sigma W_r})$$

where we have used that $W_t-W_r$ and $W_r$ are independent. Again, using the formula for the mean of a lognormal distribution we get $$E(e^{\sigma(W_t-W_r)})=e^{\frac{\sigma^2(t-r)}{2}}$$ and $$E(e^{2\sigma W_r})=e^{2\sigma^2 r}$$

Hence

$$Cov(S_t,S_r)=e^{\gamma(t+r)}e^{\frac{\sigma^2(t-r)}{2}}e^{2\sigma^2 r}-e^{\gamma t+\frac{\sigma^2t}{2}}e^{\gamma r+\frac{\sigma^2r}{2}}$$

I leave it to you to check if this can be further simplified.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.