美式 vs 欧式期权：提前行权与套利

American vs. European Options

专题: Finance / 金融
难度: L4
来源: QuantQuestion

题目详情

A. 为什么无分红股票上的美式看涨在到期前从不最优提前行权？

B. 同一无分红标的：欧式看跌 $K=80$ 的价格为 8， $K=90$ 的价格为 9。是否存在套利？

A. “American options can be exercised earlier, so they are often more valuable than European with the same features. But for a non-dividend-paying stock, an American call and a European call should be worth the same, because early exercise is never optimal. Why never exercise an American call on a nondividend-paying stock before maturity?”

B. “A European put with strike=80 is priced at 8; a European put with strike=90 is priced at 9, same underlying no dividend. Is there an arbitrage?”

解析

A. 无分红时，美式看涨不提前行权：

提前行权只能拿到内在价值 $S-K$ ，但看涨在到期前通常具有正的时间价值。
更优做法是卖出期权而不是行权（保留时间价值）；严格证明可用组合支配或风险中性下凸性（Jensen）论证。

B. 有套利。看跌价格对 $K$ 应当是凸的且单调递增。给定价格 $(80,8)$ 与 $(90,9)$ 违反凸性（80 的太贵）。

一种套利：

做多 8 份 $K=90$ 的看跌，
做空 9 份 $K=80$ 的看跌。

初始成本为 0（ $8\cdot 9 - 9\cdot 8=0$ ），到期支付非负且在某些 $S_T$ 下严格为正。

Original Explanation

AnswerA:

By exercising the call early, you only get the intrinsic value. But a call’s time value is positive for a nondividend-paying stock. So you would sell the call in the market rather than exercise early.
Detailed portfolio argument: If $S_{t_0}>K$ at $t_0<T$ , compare the immediate exercise vs. continuing to hold. A combination of short stock plus a borrowed fraction of the strike strictly dominates early exercise.
Mathematical approach with risk-neutral valuation and Jensen’s inequality: The payoff $(S-K)^+$ is convex in $S$ . In the risk-neutral measure, the discounted expected payoff is always greater than the immediate-exercise value.

AnswerB: A put payoff $(K-S)^+$ is convex in $K$ . Because a put with $K=0$ is worthless, the slope at $K=0$ suggests the price function in $K$ is convex from 0. Then from convexity we get $\dfrac{8}{9}\,\text{Put}(K=90) \;>\;\text{Put}(K=80).$ Hence the 80-strike put is overpriced.
Construct an arbitrage: go long 8 of the 90-strike puts, short 9 of the 80-strike puts. Initial cost is zero, final payoff is always $\ge0$ and sometimes $>0$ .