European call option v.s. Asian call option

Let a stock follow a Geometric Brownian motion

$\textnormal{d}S_t = S_t(r \textnormal{d} t + \sigma \textnormal{d} W_t)$

with constant $r,\sigma > 0$ . Let

$A_T = \frac{1}{T}\int_0^T S_u \textnormal{d} u$

European Asian call option has the payoff $(A_T-K)^+$ and European vanilla call option has the payoff $(S_T-K)^+$ . Then European vanilla option has higher value.

Proof.

By (1)Jensen’s inequality, (2)Fubini, and (3)the fact that the longer the maturity is the higher the vanilla European option’s value is, we have

$\begin{aligned} &~ \mathbb{E}[e^{-rT}(A_T-K)^+] \\= &~ \mathbb{E}[e^{-rT}(\frac{1}{T}\int_0^T S_u \textnormal{d} u -K)^+] \\\leq & ~ \mathbb{E}[e^{-rT}\frac{1}{T}\int_0^T (S_u-K)^+ \textnormal{d} u] \\= &~ \frac{1}{T}\int_0^T \mathbb{E}[e^{-rT} (S_u-K)^+ ] \textnormal{d} u \\\leq &~ \frac{1}{T}\int_0^T \mathbb{E}[e^{-ru} (S_u-K)^+ ] \textnormal{d} u \\\leq &~ \frac{1}{T}\int_0^T \mathbb{E}[e^{-rT} (S_T-K)^+ ] \textnormal{d} u \\ = &~ \mathbb{E}[e^{-rT} (S_T-K)^+ ] \\ \end{aligned}$

Notice that we didn’t use the dynamics of Geometric Brownian motion. Above deduction is true as long as (3) is true.