# How to prove price of Asian option under geometric averaging is cheaper than a European call?

This was an exam question at Cambridge University.

Let $$S_t = S_0 \exp \left(\sigma W_t + (r-\dfrac{1}{2}\sigma^2) \right)$$ and a bank account returns a continuously-compounded rate of interest $$r$$. Consider the derivative which pays

$$Y = (\exp(T^{-1}\int^T_0\log(S_u)\text{d}u) - K)^+$$ at time T.

What is the time-0 price for this derivative, and show it is less than the price of a European call.

The price of this, if I am not wrong, is

$$S_0\exp(-\dfrac{1}{2}(r+\sigma^2/6)T) N(d_2) - Ke^{-rT}F(-d_1)$$

where $$d_1 = \log(K/S_0-1/2(r-\sigma^2)T)/(\sigma\sqrt{T/3})$$ and $$d_2 = -d_1+\sigma\sqrt{T/3}$$.

I don't see how this is less than the European call.

• Hi, the pay-off that you describe here is that of an Asian option with geometric averaging. Maybe I find time to formulate an answer later. – Ric Dec 16 '13 at 7:59
• @Richard thank you – Lost1 Dec 16 '13 at 11:16
• @Richard nudge if you cba, a reference would also do. – Lost1 Dec 23 '13 at 13:38
• The volatility of this product is sigmasqrt(T/3), which is smaller than that of the European option sigmasqrt(T). Thus the price of it should be lower than that of the European option. – user6908 Jan 6 '14 at 12:19

• second: what do they mean by cheaper? The pay-off is different - so what can we compare. The only meaning is that if the stock has an implied volatility of $\sigma$ then the continuously sampled Asian option has an implied vol of $\sigma/3$ (check your $d_1$ there is something wrong in the numerator). So we can say that given the same moneyness the Asian option looks like a standard European option but with a third of its implied vol. Thus the Asian is cheaper.