Covariance of Exponentials of Brownian Motion at Two Times

Summary

The document derives the covariance between exponentials of Brownian motion evaluated at two nonnegative times. It assumes the earlier time is no later than the later time, writes covariance as a joint expectation minus the product of expectations, and decomposes the later Brownian value into an earlier value plus an independent increment. The normal moment-generating formula then gives the result for that ordering.

The accepted response confirms the derivation and rewrites it in a form that is symmetric in the two time arguments by using their minimum. This provides a compact general expression and resolves the apparent asymmetry caused by initially assuming one time order. The calculation relies on standard Brownian motion, whose increments are independent and whose variance at time t is t; it does not cover drifted or otherwise modified processes. The result is a probability calculation, not a trading strategy or market-data analysis.

Key ideas

• The covariance can be computed by subtracting the product of marginal expectations from the joint expectation.
• Brownian independent increments allow the later value to be split into an earlier value and an independent increment.
• For normal variables, exponential expectations follow from the normal moment-generating formula.
• The general expression depends on the smaller of the two time points and is symmetric in them.

Tags

# Calculate $Cov(e^ {B_t} ,e^{B_s})$


# Calculate $Cov(e^ {B_t} ,e^{B_s})$












Let $(B_t)_{t \geq0} $ be a Brownian Motion. Calculate $Cov(e^ {B_t} ,e^{B_s})$ I would verify the following solution which the result looks a bit weird.

My solution: let $0 \leq s \leq t$. $$Cov(e^ {B_t} ,e^{B_s})=E[e^{B_t + B_s}]-E[e^{B_t}]E[e^{B_s}] \\=E[e^{B_t -B_s} e^{2B_s}]-e^{t/2}e^{s/2} \\ =E[e^{B_t -B_s}]\ E[e^{2B_s}]-e^{t/2}e^{s/2} (\because independence) \\= e^{(t-s)/2}e^{2s} - e^{t/2}e^{s/2} \\=e^{(t+2s)/2} (e^{s/2}-e^{-s/2})\\ =2e^{t/2+s} \sinh(s/2) $$

Here I have extensively used the fact $ E[e^{X}]=e^{E[X]+Var(X)/2}$ whenever $X$ is normal.

## Answer by hypernova (score 2, accepted)

https://quant.stackexchange.com/a/39580

Your result seems correct.

If you were worried about the symmetry of $s$ and $t$, you may simply put, provided that $0\le s\le t$, \begin{align} \text{Cov}\left(e^{B_t},e^{B_s}\right)&=e^{\left(t-s\right)/2}e^{2s}-e^{t/2}e^{s/2}&&\text{(your result)}\\ &=e^{t/2}e^{3s/2}-e^{t/2}e^{s/2}\\ &=e^{t/2}e^{s/2}\left(e^s-1\right). \end{align} Thus for general $t\ge 0$ and $s\ge 0$, $$ \text{Cov}\left(e^{B_t},e^{B_s}\right)=e^{t/2}e^{s/2}\left(e^{t\wedge s}-1\right), $$ where $t\wedge s=\min\left\{t,s\right\}$.

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