Finding the Variance of a Brownian Exponential Path Average
Summary
The document asks for the variance of the time average of the exponential of standard Brownian motion. The questioner applies Itô’s formula to the exponential process and tries to derive the variance from an endpoint term, a stochastic integral, and their covariance. The answer points out that this application omitted the one-half drift term in Itô’s formula, making that derivation incorrect as written.
The answer then offers a direct moment calculation. It exchanges expectation and integration to obtain the mean, and expresses the second moment as a double integral over pairs of times. Because two Brownian values are jointly normal, their sum has variance determined by both times and their minimum; applying the normal exponential moment gives the integrand. The remaining integral can be evaluated by splitting the square into the regions where one time precedes the other. The response outlines the method but does not carry out the final integration or state a closed-form variance.
Key ideas
- Itô’s formula for the exponential of Brownian motion includes a one-half drift term.
- The mean of the path integral can be found by exchanging expectation and time integration.
- The second moment becomes a double integral involving the joint normal distribution of Brownian values.
- The double integral can be split into regions according to which time is earlier.
- The answer gives a calculation route but does not provide the final variance expression.
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# Ito Calculus Solution for Path Average of Geometric Brownian Motion
# Ito Calculus Solution for Path Average of Geometric Brownian Motion
I am struggling with the following question and cannot find a solution online.
> Let ${X_t = \frac{1}{t} \int_0^t e^{W_u} du}$, where ${W_u}$ is a standard Brownian > motion. What is the Variance of ${X_t}$?
It is clear to me that this would be the variance of a path of a geometric Brownian motion with ${\sigma = 1}$ and ${\mu = \frac{1}{2}}$. Because of this, I am expecting there to be some sort of Ito Calculus solution that doesn't involve converting things to series and taking limits, but I may be on the wrong track with that.
So far, I have done the following using Ito Calculus:
${F_t = e^{W_t}}$
${\rightarrow dF_t = e^{W_t}dt + e^{W_t}dW_t}$
${\rightarrow e^{W_t} - e^{W_0} = \int_0^t e^{W_u} du} + \int_0^t e^{W_u}dW_u$
${\rightarrow Var(\int_0^t e^{W_u} du) = Var(e^{W_t}) + Var(\int_0^t e^{W_u}dW_u) + 2 Cov(e^{W_t}, \int_0^t e^{W_u}dW_u)}$
The first component is obvious, using MGF of normal distribution. The second component is obvious using Ito's Isommetry. But I am struggling to determine ${Cov(e^{W_t}, \int_0^t e^{W_u}dW_u)}$.
Any hints in this regard or confirmation whether I'm on the right track with solving this one are much appreciated.
## Answer by Andrea (score 2, accepted)
https://quant.stackexchange.com/a/81223
You application of the Ito formula to $F_t$ missed the $\frac{1}{2}$ term for $\frac{1}{2} e^W_t dt$.
But, I would chose an alternative way to do it.
Focusing on $\int_0^T e^{W_t} dt$ you need
$E \left [ \int_0^T e^{W_t} dt \right ] = \int_0^T E[e^{W_t}] dt = \int_0^T e^{\frac{1}{2}t} dt $
for the mean: remember $E$ and the integral commute.
The 2nd moment requires a minor ingenuity
$E \left [ \left ( \int_0^T e^{W_t} dt \right ) ^2 \right ] = E \left [ \left ( \int_0^T e^{W_t} dt \right ) \left ( \int_0^T e^{W_s} ds \right ) \right ] $
so you can write it as a double integral
$\int_0^T \int_0^T E \left [ e^{W_t} e^{W_s} \right ] ds \, dt$
which is
$\int_0^T \int_0^T E \left [ e^{W_t + W_s} \right ] ds \, dt$
Now $W_t + W_s$ is a normal with 0 mean and variance = $t + s + 2 \min(s, t)$ so
$\int_0^T \int_0^T e^{\frac{1}{2} (t + s + 2 \min(s, t))} ds \, dt$
which can be solved splitting the square in 2 regions where $s < t$ and $s > t$.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.