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衍生品定价

Price of a derivative

专题
Finance / 金融
难度
L4

题目详情

原始来源未公开完整题干;当前保留的解析内容是用换计价资产方法推导衍生品定价与 Black-Scholes 看涨期权公式。

英文原题

Price a derivative in the Black-Scholes framework using a change of numeraire. In particular, explain how taking the stock as numeraire leads to the probability terms in the vanilla call price.

解析

一般定价:选计价资产(numeraire)NtN_t,则

V0=N0EQN[PayoffTNT].\boxed{V_0=N_0\,\mathbb{E}^{\mathbb{Q}^N}\left[\frac{\text{Payoff}_T}{N_T}\right]}.

取股票为计价资产 Nt=StN_t=S_t(stock numeraire),可以把含 STS_T 的 payoff 简化,并得到如 N(d1)N(d_1)N(d2)N(d_2) 的概率解释:

QS(ST>K)=N(d1),Q(ST>K)=N(d2).\mathbb{Q}^S(S_T>K)=N(d_1),\qquad \mathbb{Q}(S_T>K)=N(d_2).

因此 vanilla call 的价格可写为

C0=S0N(d1)KerTN(d2).\boxed{C_0=S_0N(d_1)-Ke^{-rT}N(d_2)}.

英文解析

General pricing by a numeraire: choose a numeraire asset NtN_t. Then

V0=N0EQN[PayoffTNT].\boxed{V_0=N_0\,\mathbb{E}^{\mathbb{Q}^N}\left[\frac{\text{Payoff}_T}{N_T}\right]}.

Taking the stock as numeraire, Nt=StN_t=S_t, can simplify payoffs involving STS_T and gives the probability interpretations of N(d1)N(d_1) and N(d2)N(d_2):

QS(ST>K)=N(d1),qquadQ(ST>K)=N(d2).\mathbb{Q}^S(S_T>K)=N(d_1),\\qquad \mathbb{Q}(S_T>K)=N(d_2).

Therefore the vanilla call price is

C0=S0N(d1)KerTN(d2).\boxed{C_0=S_0N(d_1)-Ke^{-rT}N(d_2)}.