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.

  • $\begingroup$ I changed the small $t$ to capital $T$ in the definition of $A_T$. $\endgroup$
    – Gordon
    Jul 8, 2015 at 13:11

1 Answer 1


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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