Asian options III: the Geometric Asian

I’ve introduced the Asian Option before, it is similar to a vanilla option but its payoff is based on the average price over a period of time, rather than solely the price at expiry. We saw that it was exotic – it is not possible to give it a unique price using only the information available from the market.

Today I’m going to introduce another type of option – the geometric asian option. This option is similar, but its payoff is now based on the geometric average, rather than the arithmetic average, of the spot over the averaging dates

    \[C(T) = \Bigl( \bigl( \prod_{i=1}^{N} S_i \bigr)^{1 \over N} - K \Bigr)^+\]

This option is exotic, just like the regular (arithmetic-average) asian option. At first sight it seems even more complicated; and what’s more, it is almost never encountered in practice, so why bother with it? Well, it’s a very useful as an exercise because in many models where the arithmetic asian’s price has no closed form, the geometric asian happens to have one! Further, since an arithmetic average is ALWAYS higher than a geometric average for a set of numbers, the price of the geometric asian will give us a strict lower bound on the price of the arithmetic asian.

Considering the Black-Scholes model in its simplest form for the moment (although the pricing formula can be extended to more general models), let’s consider what the spot will look like at each of the averaging times, and as we did in the earlier post, considering a simple geometric asian averaging over only two times t_1 and t_2 so that the payoff is C(T) = \Bigl( \sqrt{ S_1 \cdot S_2 } - K \Bigr)^+

At t_0S(t_0) = S_0. At t_1,

    \[S(t_1) = S_1 = S_0 \exp{ \Bigl\{ (r - {1\over 2} \sigma^2 )t_1 + \sigma \sqrt{t_1} x_1 \Bigr\} }\]

where x_1 \sim {\mathbb N}(0,1). At t_2,

    \[S(t_2) = S_2 = S_1 \exp{ \Bigl\{ (r - {1\over 2} \sigma^2 )(t_2 - t_1) + \sigma \sqrt{t_2-t_1} x_2 \Bigr\} }\]

where x_2 \sim {\mathbb N}(0,1) also, and importantly x_2 is uncorrelated with x_1 due to the independent increments of the Brownian motion.

Now the reason we couldn’t make a closed form solution for the arithmetic asian was that S_1 + S_2 is the sum of two lognormal distributions, which itself is NOT lognormally distributed. However, as discussed in my post on distributions, the product S_1 \cdot S_2 of two lognormal distributions IS lognormal, so valuing an asian that depends on the product of these two is similar to pricing a vanilla, with a slight change to the parameters that we need

    \[\begin{matrix} \sqrt{ S_1 \cdot S_2 } & = & S_0 \exp \Bigl\{ {1\over 2}(r - {1\over 2} \sigma^2 )(t_1 + t_2) + \sigma ( \sqrt{t_1} x_1 + {1\over 2} \sqrt{t_2 - t_1} x_2 ) \Bigr\} \\ & = & S_0 \exp \Bigl\{ {1\over 2}(r - {1\over 2} \sigma^2 )(t_1 + t_2) + {1 \over 2} \sigma \sqrt{ 3 t_1 + t_2 } x_3 \Bigr\} \end{matrix}\]

where x_3 is another normal variable (we’ve used the result about the sum of two normally distributed variables here).

If we re-write this as

    \[\begin{matrix} \sqrt{ S_1 \cdot S_2 } & = & S_0 \exp \Big\{ {1 \over 2} r ( t_1 + t_2 ) \Bigr\} \exp \Big\{ -{1 \over 2} \sigma^2 (t_1 + t_2) + {1 \over 2} \sigma \sqrt{ 3 t_1 + t_2 } x_3 \Big\} \\ & = & \sqrt{ F_1 \cdot F_2 } \exp \Big\{ {1 \over 2} \sigma^2 ( \tilde{t} - \tilde{\tilde{t}} ) \Big\} \exp \Big\{ -{1 \over 2} \sigma^2 \tilde{t} + \sigma \tilde{t} x_3 \Big\} \end{matrix}\]

where F_1 and F_2 are the forwards at the respective times and \tilde{t} and \tilde{\tilde{t}} are defined below. This is just the same as the vanilla pricing problem solved here. So, we can use a vanilla pricer to price a geometric asian with two averaging dates, but we need to enter transformed parameters

\begin{matrix} \tilde{t} & = & {1 \over 2} \sqrt{ 3t_1 + t_3 } \\ \tilde{\tilde{t}} & = & {1 \over 2} (t_1 + t_2) \\ F & \to & \sqrt{ F_1 \cdot F_2 } \cdot e^{{1 \over 2} \sigma^2 (\tilde{t} - \tilde{\tilde{t}}) } \\ \sigma^2 t & \to & \sigma^2 \tilde{\tilde{t}} \end{matrix}

In fact this result is quite general, we can price a geometric asian with any number of averaging dates, using the general transformations below (have a go at demonstrating this following the logic above)

 \begin{matrix} \tilde{t} & = & {1 \over n} \sqrt{  \sum_{i , j=0}^n {\rm min}(t_i, t_j) } \\ \tilde{\tilde{t}} & = & {1 \over n} \sum_{i =1}^n t_i \\ F & \to & \Big( \prod_{i=1}^n F_i \Big)^{1 \over n} \cdot e^{{1 \over 2} \sigma^2 (\tilde{t} - \tilde{\tilde{t}}) } \\ \sigma^2 t & \to & \sigma^2 \tilde{\tilde{t}} \end{matrix}