Expected Arithmetic Average Under Risk-Neutral Geometric Brownian Motion
Summary
The document addresses the expected value of an asset's time average when the price follows geometric Brownian motion, in the context of arithmetic Asian option pricing. The response derives the risk-neutral expected asset price at each future time as its forward level, then integrates that expectation over the averaging interval. For a constant interest rate, this gives the initial price multiplied by a factor involving the exponential of the rate over the remaining maturity.
The result is greater than the initial price when the rate is positive, and equals the initial price when the rate is zero. That resolves the apparent conflict in the question's zero-drift simulation: a zero physical drift is not necessarily the same as zero risk-neutral rate. The derivation concerns the expectation of the arithmetic average, not the distribution of that average or the option price itself. The question reports a simulation mismatch, but the answer asserts that a carefully specified simulation should agree and does not provide simulation details.
Key ideas
- Under risk-neutral geometric Brownian motion, the expected future asset price grows at the risk-free rate.
- The expected arithmetic average is obtained by integrating expected prices across the averaging interval.
- With a positive rate, the expected average exceeds the initial price; with a zero rate, they are equal.
- Zero physical drift does not by itself establish that the risk-neutral rate is zero.
- The expectation formula does not describe the full distribution needed to price an Asian option.
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Full text
# Understanding the expected value of the average
# Understanding the expected value of the average
I've been looking into Asian Options pricing. Part of the process is about looking for the expected value of the average of a time series undergoing e.g. geometric brownian motion.
I came across this paper: Pricing and hedging of arithmetic Asian options via the Edgeworth series expansion approach.
In the appendix, they claim to calculate the expected value of the average over a time period, see screenshot below:
I find this quite problematic, in that this value seems like it would necessarily be before the initial value, even if the drift of the underlying asset is assumed to be 0. It seems un-intuitive to say that the average must have a greater expected value than the initial value.
Furthermore, this value doesn't match what I get for the expected value when simulating as much as 10,000,000 price paths, but even more concerning perhaps is that the result from random simulations is still above the initial price.
I tested with the following parameters. Time to maturity: 0.25 years. Initial price: 1000. Underlying annualized volatility: 0.7. Underlying drift: 0
Would anyone happen to have an idea of what is going on here? Thanks.
## Answer by Kermittfrog (score 1, accepted)
https://quant.stackexchange.com/a/69121
The math checks out, and a carefully conducted simulation study will follow the result:
For $S_t$ following a geometric Brownian motion with constant parameters $\mu,\sigma$,we know that $$E(S_T|S_t)=S_te^{(\mu+0.5\sigma^2)(T-t)}$$
In this case, the risk neutral drift $\mu_{\mathbb{Q}}=r-\frac{1}{2}\sigma^2$, hence the risk-neutral expectation of $S_t$ at any future time (simply) becomes its forward level:
$$ Y(T)\equiv E_\mathbb{Q}(S_T|S_t)=S_te^{(\mu_{\mathbb{Q}}+\frac{1}{2}\sigma^2)(T-t)}=S_te^{r(T-t)} $$
Now integrating $Y(s)$ with respect to $s$ clearly yields the stated result,
$$ \begin{align} \frac{1}{T-t}\int_{s=t}^T Y(s)ds&=\frac{1}{T-t}\int_{s=t}^T E_\mathbb{Q}(S_s|S_t)ds\\ &=\frac{S_t}{T-t}e^{-rt}\int_{s=t}^T e^{rs}ds\\ &=\frac{S_t}{T-t}e^{-rt}\left[\frac{e^{rs}}{r}\right]_t^T\\ &=\frac{S_t}{T-t}e^{-rt}\left(\frac{e^{rT}}{r}-\frac{e^{rt}}{r}\right)\\ &=\frac{S_t}{T-t}\left(\frac{e^{r(T-t)}-1}{r}\right)\\ &=S_t\frac{e^{r(T-t)}-1}{r(T-t)} \end{align} $$
Do note that, for $x>0$, we have $\frac{e^x-1}{x}>1$. Hence, the expectation is clearly greater than $S_t$. Also, if $r=0$, the expectation becomes $S_t$ as expected.
HTH?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.