Moments of Integrated Geometric Brownian Motion for Asian Options
Summary
The document asks how to compute higher moments of the time integral of an exponential Brownian motion, a quantity relevant to arithmetic Asian options. It first gives the expected value, then presents a general expression for the m-th moment using divided differences.
The formula specifies nodes based on the interest rate and volatility, and uses a recursive definition of divided differences to evaluate the result. The answer points to published derivations and related discussion, but provides no numerical example or comparison with simulation. Applying the formula requires careful handling of repeated or closely spaced nodes, which the short answer does not address. It gives a method for calculating moments, but does not itself derive option prices or assess approximation accuracy.
Key ideas
- The time integral of geometric Brownian motion appears in arithmetic Asian option calculations.
- The first moment can be obtained by integrating the expected value of the process over time.
- Higher moments are expressed with divided differences at nodes determined by the rate and volatility.
- The brief answer cites derivations elsewhere and does not work through a numerical example.
Tags
Full text
# Moments of the integral of the exponential of Brownian motion/Normal random variable
# Moments of the integral of the exponential of Brownian motion/Normal random variable
I'm studying arithmetic Asian options and there is integral of the following form: $$X_T=\int_0^T e^{\sigma W_t+\left(r-\frac{\sigma^2}{2}\right)t}dt,$$
where $W_t$ is a Brownian motion/Wiener process.
Is it possible to calculate momnets of this integral, i.e. $E[X_T^k]$?
Clearly, $$E[X_T]=E\left[\int_0^T e^{\sigma W_t+\left(r-\frac{\sigma^2}{2}\right)t}dt\right]=\int_0^T E\left[e^{\sigma W_t+\left(r-\frac{\sigma^2}{2}\right)t}\right]dt=\int_0^T e^{rt}dt=\frac{e^{rT}-1}{r}$$
What about other moments?
Thank you in advance.
## Answer by Achrbot (score 1, accepted)
https://quant.stackexchange.com/a/78010
You can use the formulas of Baxter and Brummelhuis (2011). They provide moment formulas for the time integral of geometric brownian motion X_t, using divided differences.
The full derivation is in the paper, but the formula is $$ \mathbb{E}\left[X_T^m\right] = T^m m! \exp [Tb_0, Tb_1, \ldots, Tb_m] $$ where $b_k = kr +\frac{\sigma^2}{2}k(k-1)$, and the divided difference $f[a_0, \ldots, a_k]$ is defined recursively as $$ f[a_0, \ldots, a_k] = \frac{f[a_1, \ldots, a_{k}]-f[a_0, \ldots, a_{k-1}]}{a_k-a_0}, \quad f[a,b] = \frac{f(b)-f(a)}{b-a}. $$
You might also be interested in the paper by Levy (2018), which further discusses the divided difference approach.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.