1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.