Using Jensen’s Inequality to Compare Asian and European Call Values
Summary
The document asks for the value of a continuously sampled geometric-average Asian call and why it should cost less than a European call. One response points to the connection between the geometric Asian’s distribution and a European option with reduced effective volatility, while noting an apparent error in the question’s proposed formula. A later response offers a comparison that avoids deriving the explicit Asian pricing formula.
The argument applies Jensen’s inequality to the convex payoff function of the log price, bounding the payoff on the average log price by the average of payoffs at individual times. Taking expectations turns that bound into an average of European call values across expiries; the response then invokes the claim that call value increases with expiry. It says the reasoning extends to an arithmetic-average Asian call. The comparison relies on the stated setup and assumptions behind the expectation and expiry comparison; the post does not work through those assumptions or resolve the proposed formula in detail.
Key ideas
- The geometric-average call payoff can be expressed as a convex function of the average log price.
- Jensen’s inequality bounds that payoff by an average of payoffs evaluated over the sampling period.
- The expected bound is related to an average of European call values at intermediate expiries.
- The response uses increasing European call value with expiry to compare the average with the terminal-expiry call.
- The same outline is said to apply to an arithmetic-average Asian call.
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Full text
# How to prove price of Asian option under geometric averaging is cheaper than a European call?
# 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.
## Answer by Richi Wa (score 3, accepted)
https://quant.stackexchange.com/a/9823
- 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.
## Answer by user name (score 4)
https://quant.stackexchange.com/a/81288
An easier way to show it is less than Europen call option (without calculating the explicit formula) is to make use of Jensen's inequality.
Define $f(x) = (\exp(x) - k)^+$, this function is convex. So $f(\int_0^T\frac{1}{T}\log(S_u)du) \leq \int_0^T\frac{1}{T}f(\log(S_u))du$. Note we need the $\frac{1}{T}$ to make sure the new measure integrates to 1, and satisfies the assumption of Jensen's inequality.
So the price of Asian option is less than $e^{-rT}E[\int_0^T\frac{1}{T}f(log(S_u))du]$. Assuming exchangeability of expectation and integration, it is just $\int_0^T\frac{1}{T}EC(S_0,K,r,u,\sigma)du$. Since the price of European call is an increasing function of expiry time, this expression is less than $EC(S_0,K,r,T,\sigma)$.
This approach can also be applied to prove the arithmetic mean version of Asian call also worth less than the corresponding European call.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.