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This was an exam question at Cambridge University.

Let $S_t = S_0\exp(\sigma W_t + (r-\dfrac{1}{2}\sigma^2)$ 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

$S_0\exp(-\dfrac{1}{2}(r+\sigma^2/6)T)N(-d_1+\sigma\sqrt{T/3}) - Ke^{-rT}F(-d_1)$

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

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

share|improve this question
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. – Richard 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
up vote 1 down vote accepted
  • first - a nice and short note for the calculation can be found here.

  • 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.

Another reference Theory of Continuously-sampled Asian Option pricing.

share|improve this answer
The 2nd ref does not open on my iPad. Will try again on a computer later. Thank you – Lost1 Dec 30 '13 at 11:17
Google the title and you will get a download link. I can not post the direct link. – Richard Dec 30 '13 at 11:54

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