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Closed-Form Geometric Asian Option on the Logarithm of the Average Price

Article Quant Q&A · Author: Danny
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Summary

The document derives a closed-form price for an Asian option whose payoff is the positive part of the logarithm of the ratio between the geometric average stock price and the strike. Under the stated Black–Scholes geometric Brownian motion assumption, the log of the continuous geometric average is normally distributed. The derivation integrates the stock’s log price over time and rewrites the Brownian contribution as a stochastic integral with a deterministic weight, yielding the distribution’s mean and variance.

The payoff is then evaluated as the expectation of the positive part of a normal random variable and discounted at the risk-free rate. The resulting expression uses the standard normal cumulative distribution function and density. The document provides an analytical derivation for this particular continuous averaging convention and log payoff. Its result relies on the stated GBM setup and does not address discrete observation dates, dividends or alternative payoff definitions.

Key ideas

  • Under geometric Brownian motion, the continuously averaged log stock price is normally distributed.
  • The Brownian integral can be rewritten with a deterministic time weight to obtain its variance.
  • The payoff is the positive part of a normally distributed log ratio between the average price and strike.
  • Its expected payoff has a closed form involving the standard normal cumulative distribution and density.
  • The derivation applies to the stated continuous geometric averaging and model assumptions.

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OptionsDerivatives pricingStatistics
Full text
# Asian Option with Geometric Averaging


# Asian Option with Geometric Averaging












Can someone point me to any notes on how to derive the closed form formula for Asian geometric average option with payoff $\text{max}\left(\text{log}\left(\frac{A_T}{K}\right), 0\right)$ where $A_T$ is given by

$$ A_T=\text{exp}\left(\frac{1}{T}\int_0^T \text{log}(S_u) du \right) $$

We can assume that the stock follows a GBM as in the black scholes model.

## Answer by Gordon (score 6, accepted)

https://quant.stackexchange.com/a/18755

Note that \begin{align*} \int_0^T\ln S_u du &= \int_0^T\Big[\big(r-\frac{1}{2}\sigma^2\big)u + \sigma W_u \Big] du\\ &=\frac{1}{2}\big(r-\frac{1}{2}\sigma^2\big)T^2 + \sigma\int_0^T\int_0^u dW_s \,du\\ &=\frac{1}{2}\big(r-\frac{1}{2}\sigma^2\big)T^2 + \sigma\int_0^T\int_s^T du \,dW_s\\ &=\frac{1}{2}\big(r-\frac{1}{2}\sigma^2\big)T^2 + \sigma\int_0^T(T -s) \,dW_s,\\ \end{align*} which is normally distributed. Then \begin{align*} \ln \frac{A_T}{K} &= \frac{1}{T}\int_0^T\ln S_u\, du -\ln K\\ &\sim N\left(\frac{1}{2}\big(r-\frac{1}{2}\sigma^2\big)T-\ln K, \ \Big(\frac{\sigma T}{\sqrt{3}}\Big)^2 \right)\\ &=\mu + \Sigma\, \xi, \end{align*} where $\mu = \frac{1}{2}\big(r-\frac{1}{2}\sigma^2\big)T -\ln K$, $\Sigma = \frac{\sigma T}{\sqrt{3}}$, and $\xi$ is standard normal random variable. Consequently, the option payoff has value \begin{align*} e^{-rT} E\left(\max\Big(\ln \frac{A_T}{K}, \, 0 \Big) \right) &= e^{-rT} E\left(\max\big(\mu + \Sigma\, \xi, \, 0 \big) \right)\\ &=\frac{e^{-rT}}{\sqrt{2\pi}}\int_{-\frac{\mu}{\Sigma}}^{\infty} (\mu+\Sigma \, x) e^{-\frac{1}{2}x^2}dx\\ &=e^{-rT}\bigg[\mu \Phi\Big(\frac{\mu}{\Sigma}\Big)+\frac{\Sigma}{\sqrt{2\pi}} \, e^{-\frac{\mu^2}{2\Sigma^2}}\bigg], \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.