返回题库

Black–Scholes 公式与衍生问题

Black–Scholes Formula

专题
Finance / 金融
难度
L4

题目详情

围绕 Black–Scholes 公式回答：

A. 公式依赖哪些关键假设？

B. 如何用风险中性测度推导无分红欧式看涨价格？

C. 如何通过解 BSM PDE 推导同一公式？

D. r=0r=0、无分红、现价 S0=1S_0=1。第一次触及 H>1H>1 时可行权并获得 1。该合约现在值多少？

E. 无分红股票服从 GBM。到期 TT 支付 1/ST1/S_T，其在 t=0t=0 的价值是多少？

For a European call/put on a stock with continuous dividend yield yy: >c>=>Se−yτ N(d1)>−>Ke−rτ N(d2),>>p>=>Ke−rτ N(−d2)>−>Se−yτ N(−d1),>> c > = > S e^{-y\tau}\,N(d_1) > - > K e^{-r\tau}\,N(d_2), > \quad > p > = > K e^{-r\tau}\,N(-d_2) > - > S e^{-y\tau}\,N(-d_1), > where >d1>=>ln⁡ ⁣(Se−yτK)+(r+σ22)τστ>=>ln⁡ ⁣(SK)+(r−y+σ22)τστ,>>d2=d1−στ,>> d_1 > = > \frac{\ln\!\bigl(\tfrac{S e^{-y\tau}}{K}\bigr)+(r + \tfrac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}} > = > \frac{\ln\!\bigl(\tfrac{S}{K}\bigr)+(r - y + \tfrac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}, > \quad > d_2 = d_1 - \sigma\sqrt{\tau}, > and N(⋅)N(\cdot) is the standard normal CDF.

A. “What assumptions underlie the Black-Scholes formula?”

B. “How to derive the BS formula for a European call on a nondividend-paying stock, using the risk-neutral measure?”

C. “How do you derive the same formula by solving the Black-Scholes PDE for a nondividend-paying stock?”

D. “Zero interest rate, nondividend stock at $1. The first time the stock hits level H>1H>1, you can exercise and receive 1. What is it worth now?”

E. “A nondividend-paying stock follows GBM. At maturity TT you receive 1/ST.1/S_T. What is this worth at time 0?”

解析

A.（无分红基础版）典型假设：

  • 无套利、可连续交易、可卖空、无交易成本、资产可无限分割；
  • rr 常数；
  • 标的价格服从 GBM，波动率 σ\sigma 常数。

B. 风险中性下 dS=rSdt+σSdWdS=rSdt+\sigma SdW，因此

c=e−rτ E[(ST−K)+],c=e^{-r\tau}\,\mathbb{E}[(S_T-K)^+],

利用 ln⁡ST\ln S_T 正态分布积分得到

c=SN(d1)−Ke−rτN(d2).c=SN(d_1)-Ke^{-r\tau}N(d_2).

C. 解 PDE 并用终端条件 V(S,T)=(S−K)+V(S,T)=(S-K)^+，变量代换到热方程可得同一公式。

D. 该合约价值为 1/H1/H（可用对冲/无套利论证）。

E. 一种常见结果写法为

Value(0)=e(σ2−r)TS0\mathrm{Value}(0)=\frac{e^{(\sigma^2-r)T}}{S_0}

（在具体建模与贴现约定下形式可能等价变换）。

Original Explanation

AnswerA:(Basic version c=S N(d1)−Ke−rτN(d2)c=S\,N(d_1)-K e^{-r\tau}N(d_2))

  1. No dividends.
  2. Constant rr.
  3. Stock follows geometric Brownian motion with drift μ\mu, volatility σ\sigma.
  4. No transaction costs; short sales allowed; full reinvestment.
  5. All securities are perfectly divisible.
  6. No arbitrage.

AnswerB: c=S N(d1)  −  K e−rτ N(d2),d1=ln⁡(S/K)+(r+σ22) τστ,d2=d1−στ.c = S\,N(d_1) \;-\; K\,e^{-r\tau}\,N(d_2), \quad d_1 = \frac{\ln(S/K) + (r + \tfrac{\sigma^2}{2})\,\tau}{\sigma\sqrt{\tau}}, \quad d_2 = d_1 - \sigma\sqrt{\tau}.

  • In risk-neutral measure, dS=rS dt+σS dW(t)dS=rS\,dt+\sigma S\,dW(t).
  • Payoff (ST−K)+(S_T-K)^+. Hence c=e−rτE[(ST−K)+]c=e^{-r\tau}\mathbb{E}[(S_T-K)^+].
  • Using ln⁡ST\ln S_T is normal with mean ln⁡S+(r−σ22)τ\ln S+(r-\tfrac{\sigma^2}{2})\tau and variance σ2τ\sigma^2\tau, we do the integral.

AnswerC: Solve ∂V∂t+r S ∂V∂S+12 σ2 S2 ∂2V∂S2=r V,\frac{\partial V}{\partial t} + r\,S\,\frac{\partial V}{\partial S} + \frac{1}{2}\,\sigma^2\,S^2\, \frac{\partial^2 V}{\partial S^2} = r\,V, with terminal condition V(S,T)=(S−K)+.V(S,T)=(S-K)^+. Using a change of variables to the heat equation, apply boundary conditions => the B-S formula.

AnswerD:
It is worth 1/H1/H. You can argue by constructing an arbitrage if its price differs from 1/H1/H.

AnswerE:
By Ito’s lemma on V=1/S,V=1/S, or a no-arbitrage argument, one can see

  • Vt=1/StV_t=1/S_t follows a certain SDE with drift (−r+σ2)(-r + \sigma^2), etc.
  • The discounted expectation under risk neutrality leads to a formula like
    Value=e(σ2−r)TS0 ,\mathrm{Value} = \frac{e^{(\sigma^2 - r)T}}{S_0}\,,
    if r≠0r\neq 0. E.g. if r=0r=0, we get eσ2T/S0.e^{\sigma^2 T}/S_0.