Geometric Asian option (${\left(\exp (T^{-1}{\int }_0^T\log S_{t}dt)-K\right)}^{+}$)

An Asian option whose averaging uses the geometric mean. In the constant-coefficient Black-Scholes model, the logarithm of the continuous geometric average is Gaussian with mean $\log S_0+(r-{\sigma }^2/2)T/2$ and variance ${\sigma }^{2}T/3$, because $\mathrm{Var}({\int }_0^T{W}_{t}dt)=T^3/3$. Its forward mean is $S_0{e}^{rT/2-{\sigma }^{2}T/12}$. The discounted lognormal call formula gives its price. It is strictly cheaper than the terminal European call when $\sigma ,K,T>0$ and $r\ge -{\sigma }^2/6$, since both its forward mean and log-variance are no larger, with strictly smaller variance. No unconditional comparison holds for arbitrary negative interest rates: for $r<-{\sigma }^2/6$, sufficiently small positive strikes reverse the inequality.

For $I_t={\int }_0^t\log S_{u}du$ and $G_T=\exp (T^{-1}{I}_T)$ in the Black-Scholes model, the conditional law of $\log G_T$ under the risk-neutral measure has a normal distribution with mean $m_t$ and variance $q_t$. The future Brownian integral equals ${\int }_t^T(T-u)d{W}_u$, so the Itô isometry gives the variance. For strike $K>0$ and $t<T$, the call price is $e^{-\rho (T-t)}[e^{m_t+q_t/2}\Phi (d_1)-K\Phi (d_2)]$, where $d_2=(m_t-\log K)/\sqrt{q_t}$ and $d_1=d_2+\sqrt{q_t}$. This follows by completing the square in the truncated normal exponential moment; the known past integral must be included in the mean.

For sampling times $t_i$ and $\bar{t}=n^{-1}{\sum }_i{t}_i$, the geometric mean $A=({\prod }_i{S}_{t_i}{)}^{1/n}$ has $\log A$ normally distributed under the stock-numeraire measure in the Black-Scholes model. Its mean is $\log S_0+(\rho +{\sigma }^2/2)\bar{t}$, and its variance is $v={\sigma }^2{n}^{-2}{\sum }_{i,j}\min (t_i,t_j)$. For descending times the sum is ${\sum }_i(2i-1)t_i$. The drift shift comes from exponential tilting, and must be included when the delivered payment contains a factor $S_T$.

A delivery $S_T\max (r,c/A)$ has value $S_0[r\Phi (d)+c{e}^{-m+v/2}\Phi (-d+\sqrt{v})]$, where $m,v$ are the mean and variance of $\log A$ under the stock-numeraire measure in the Black-Scholes model, and $d=(m-\log (c/r))/\sqrt{v}$. Split the normal integral at $\log (c/r)$ and complete the square in $e^{-y}$ times its normal probability density.