Covariance of Log Geometric Brownian Motion Values

Summary

The document derives the covariance between the logarithms of a geometric Brownian motion at two times. Applying Itô’s lemma gives the log process as its initial log value, a deterministic drift proportional to time, and a Brownian component scaled by volatility. For times ordered as s less than t, the deterministic terms do not contribute to covariance because they cancel when subtracting the product of expectations.

The remaining covariance comes from the shared Brownian path: the value at the earlier time is contained in the later value, so their Brownian covariance is the earlier time. Multiplication by the squared volatility gives the result, expressed generally as volatility squared times the minimum of the two times. This corrects the proposed derivation, which incorrectly retained a drift-related product term. The result assumes the standard GBM setup and deterministic initial value and parameters; it describes log values, not the covariance of the price levels themselves.

Key ideas

• The logarithm of a geometric Brownian motion has a linear drift and a scaled Brownian component.
• The covariance of two log values comes from their shared Brownian increments.
• For times s and t, the covariance is volatility squared times the smaller time.
• Deterministic drift terms cancel from covariance after centering.
• The result concerns logarithms of prices rather than price levels.

Tags

# Covariance of logarithms of geometric Brownian motion


# Covariance of logarithms of geometric Brownian motion












Suppose I have a Geometric Brownian Motion process, $$dX_t=\mu X_t dt + \sigma X_t dW_t$$

I'd like to find the covariance of $\log(X_t)$ and $\log(X_s)$ where $s<t$. We can write $\log(X_t)$ in differential form as $$d\log(X_t)=\sigma dW_t+\left(\mu-\frac{\sigma^2}{2}\right)dt$$

That's $$cov(\log(X_t),\log(X_s))=E[\log(X_t)\log(X_s)] - E[\log(X_t)]E[\log(X_s)]$$ $$=\sigma^2 s - ts\left(\mu-\frac{\sigma^2}{2}\right)^2$$

Is there anything wrong with my derivation? As my intuition tells my there shouldn't be any term associated with $\sigma^4$. Any help is appreciated!

## Answer by Jónás Balázs (score 3, accepted)

https://quant.stackexchange.com/a/50726

Let $Y = \log X$, then:

$$\begin{align} Y &= Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t \\ EY_t &=Y_0 + (\mu-\frac{\sigma^2}{2})t \\ EY_tEY_s &= Y_0^2 + Y_0 (\mu-\frac{\sigma^2}{2}) (t+s) + (\mu-\frac{\sigma^2}{2})^2 t s \\ E(Y_tY_s) &= E\left((Y_0 + (\mu-\frac{\sigma^2}{2})t + \sigma W_t) (Y_0 + (\mu-\frac{\sigma^2}{2})s + \sigma W_s)\right) \\ &= Y_0^2 + Y_0 (\mu-\frac{\sigma^2}{2}) (t+s) + (\mu-\frac{\sigma^2}{2})^2 t s + \dots + \sigma ^2 \min(t,s) \end{align}$$

What remains: $$C\text{ov}(Y_t, Y_s) = \sigma^2 \min(t,s)$$

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