本讲导读 本讲（Toda 第 3 讲）从无套利出发推导资产定价。§1 无套利资产定价：两期经济、资产张成 \(\langle W\rangle\)、无套利定义；资产定价基本定理（Thm 1，Harrison-Kreps 1979）——无套利 \(\iff\) 存在正状态价格 \(p\)；进而得风险中性测度 \(\{\nu_s\}\)、随机贴现因子 (SDF) \(m=p/\pi\)、\(\mathbb E[mR_j]=1\)、\(R_f=1/\mathbb E[m]\)、协方差定价公式 \(\mathbb E[R_j]-R_f=-R_f\mathrm{Cov}[m,R_j]\)。§2 线性因子模型：\(m=a-b'f\)（CAPM 为单因子 \(f=R_m\)）、beta、风险溢价 \(\gamma_k\)、零-beta 率 \(\gamma_0\)、\(\mu_j=\gamma_0+\sum_k\gamma_k\beta_{k,j}\)；Fama-MacBeth 两步回归、GMM、伪因子 (spurious factor) 问题。§3 二项期权定价（Cox-Ross-Rubinstein 1979）：3.1 欧式（复制、风险中性 \(p=\frac{R-D}{U-D}\)、买卖权平价 \(C-P=S_0-KR^{-T}\)）、3.2 美式（Prop 2 Merton：\(R\ge1\) 时美式看涨=欧式看涨）、3.3 连续时间极限与 Black-Scholes 公式。含习题与附录（凸集 / 超平面 / 分离超平面定理）。

设定：资产张成与无套利 / Setup: asset span and no-arbitrage 两期经济 \(t=0,1\)。\(t=1\) 时经济状态 \(s=1,\dots,S\)。\(J\) 个资产 \(j=1,\dots,J\)。资产 \(j\) 在 \(t=0\) 以价格 \(q_j\) 交易、在状态 \(s\) 支付 \(A_{sj}\)（可 \(A_{sj}<0\)）。价格向量 \(q=(q_1,\dots,q_J)\)、支付矩阵 \(A=(A_{sj})\)。定义 \((1+S)\times J\) 净支付矩阵（状态 0 即时间 0）A two-period economy \(t=0,1\). At \(t=1\) the state is \(s=1,\dots,S\). There are \(J\) assets \(j=1,\dots,J\). Asset \(j\) trades at price \(q_j\) at \(t=0\) and pays \(A_{sj}\) in state \(s\) (possibly \(A_{sj}<0\)). Price vector \(q=(q_1,\dots,q_J)\), payoff matrix \(A=(A_{sj})\). Define the \((1+S)\times J\) net-payment matrix (state 0 is time 0)

$$W=W(q,A)=\begin{bmatrix}-q'\\A\end{bmatrix}.$$

组合 \(\theta\in\mathbb R^J\)（\(\theta_j\) 为持有股数，$<0$ 即卖空）。其净支付向量 \(W\theta=\begin{bmatrix}-q'\theta\\A\theta\end{bmatrix}\in\mathbb R^{1+S}\)（在 \(t=0\) 付 \(q'\theta\)，在状态 \(s\) 收 \((A\theta)_s\)）。资产张成 \(\langle W\rangle=\{W\theta\mid\theta\in\mathbb R^J\}\subset\mathbb R^{1+S}\)。称 \(\langle W\rangle\) 无套利，若 \(\langle W\rangle\cap\mathbb R_+^{1+S}=\{0\}\)，即不存在某组合在每个状态支付非负、且至少一个状态严格为正。A portfolio \(\theta\in\mathbb R^J\) (\(\theta_j\) shares held, $<0$ is shortselling). Its net-payment vector \(W\theta=\begin{bmatrix}-q'\theta\\A\theta\end{bmatrix}\in\mathbb R^{1+S}\) (pay \(q'\theta\) at \(t=0\), receive \((A\theta)_s\) in state \(s\)). The asset span \(\langle W\rangle=\{W\theta\mid\theta\in\mathbb R^J\}\subset\mathbb R^{1+S}\). We say \(\langle W\rangle\) exhibits no-arbitrage if \(\langle W\rangle\cap\mathbb R_+^{1+S}=\{0\}\), i.e. no portfolio pays a non-negative amount in every state and a positive amount in at least one.

定理 1（资产定价基本定理，Harrison-Kreps 1979）/ Theorem 1 (FTAP) \(\langle W\rangle\) 无套利当且仅当存在 \(p\in\mathbb R_{++}^S\) 使 \([1,p']W=0\)。此时资产价格为 \(q_j=\sum_{s=1}^S p_s A_{sj}\)。\(p_s>0\) 称为状态 \(s\) 的状态价格。\(\langle W\rangle\) exhibits no-arbitrage if and only if there exists \(p\in\mathbb R_{++}^S\) such that \([1,p']W=0\). In this case the asset prices are \(q_j=\sum_{s=1}^S p_s A_{sj}\). \(p_s>0\) is called the state price in state \(s\).

定理 1 证明（用分离超平面定理）/ Proof of Theorem 1 (via the separating hyperplane theorem) （充分性） 若这样的 \(p\) 存在，对 \(0\neq w=(w_0,\dots,w_S)\in\mathbb R_+^{1+S}\) 有 \([1,p']w=w_0+\sum_{s=1}^S p_s w_s>0\)，故 \(w\notin\langle W\rangle\)，即 \(\langle W\rangle\cap\mathbb R_+^{1+S}=\{0\}\)。（必要性） 设无套利，则 \(\langle W\rangle\cap\Delta=\emptyset\)（\(\Delta=\{w\in\mathbb R_+^{1+S}\mid\sum_{s=0}^S w_s=1\}\) 单纯形）。\(\langle W\rangle,\Delta\) 凸非空、\(\Delta\) 紧，由（强）分离超平面定理存在 \(0\neq\lambda\in\mathbb R^{1+S}\) 使 \(\langle\lambda,w\rangle<\langle\lambda,d\rangle\) 对 \(w\in\langle W\rangle,d\in\Delta\)。若存在 \(\theta\) 使 \(\langle\lambda,W\theta\rangle\neq0\)，令 \(w=W\alpha\theta\)、\(\alpha\to\pm\infty\) 矛盾，故 \(\lambda'W=0\)，不等式变为 \(0<\langle\lambda,d\rangle\)。取 \(d=e_s\) 得 \(\lambda_s>0\)。令 \(p_s=\lambda_s/\lambda_0\) 则 \(p\gg0\)、\([1,p']W=0\)，逐分量即 \(q_j=\sum_s p_s A_{sj}\)。\(\blacksquare\)(Sufficiency) If such \(p\) exists, for \(0\neq w=(w_0,\dots,w_S)\in\mathbb R_+^{1+S}\), \([1,p']w=w_0+\sum_{s=1}^S p_s w_s>0\), so \(w\notin\langle W\rangle\), i.e. \(\langle W\rangle\cap\mathbb R_+^{1+S}=\{0\}\). (Necessity) Suppose no-arbitrage; then \(\langle W\rangle\cap\Delta=\emptyset\) (\(\Delta=\{w\in\mathbb R_+^{1+S}\mid\sum_{s=0}^S w_s=1\}\) the unit simplex). \(\langle W\rangle,\Delta\) are convex nonempty and \(\Delta\) compact, so by the (strong) separating hyperplane theorem there is \(0\neq\lambda\in\mathbb R^{1+S}\) with \(\langle\lambda,w\rangle<\langle\lambda,d\rangle\) for \(w\in\langle W\rangle,d\in\Delta\). If some \(\theta\) had \(\langle\lambda,W\theta\rangle\neq0\), taking \(w=W\alpha\theta\), \(\alpha\to\pm\infty\) contradicts it, so \(\lambda'W=0\) and the inequality becomes \(0<\langle\lambda,d\rangle\). Taking \(d=e_s\) gives \(\lambda_s>0\). Letting \(p_s=\lambda_s/\lambda_0\), \(p\gg0\), \([1,p']W=0\), and component-wise \(q_j=\sum_s p_s A_{sj}\). \(\blacksquare\)

风险中性测度与 SDF / Risk-neutral measure and the SDF 无风险资产在每个状态支付 1，其价格 \(\dfrac1{1+r}=\sum_{s=1}^S p_s>0\)。令 \(\nu_s=p_s/\sum_s p_s>0\)（\(\sum_s\nu_s=1\)），则The risk-free asset pays 1 in every state, so its price is \(\dfrac1{1+r}=\sum_{s=1}^S p_s>0\). Letting \(\nu_s=p_s/\sum_s p_s>0\) (\(\sum_s\nu_s=1\)),

$$q_j=\frac1{1+r}\sum_{s=1}^S\nu_s A_{sj}=\frac1{1+r}\tilde{\mathbb E}[A_{sj}],$$

即资产价格是在风险中性概率 \(\{\nu_s\}\) 下的贴现期望支付。再令 \(\pi_s\) 为客观概率、\(m_s=p_s/\pi_s\)，则i.e. the asset price is the discounted expected payoff under the risk-neutral probability \(\{\nu_s\}\). Letting \(\pi_s\) be the objective probability and \(m_s=p_s/\pi_s\),

$$q_j=\sum_{s=1}^S p_s A_{sj}=\sum_{s=1}^S\pi_s m_s A_{sj}=\mathbb E[mA_j].$$

随机变量 \(m\) 即随机贴现因子 (SDF)。令 \(R_j=A_j/q_j\) 为毛收益，则对任意资产 \(\mathbb E[mR_j]=1\)。无风险利率 \(R_f\) 满足 \(\mathbb E[mR_f]=1\iff R_f=1/\mathbb E[m]\)。The random variable \(m\) is the stochastic discount factor (SDF). Letting \(R_j=A_j/q_j\) be the gross return, \(\mathbb E[mR_j]=1\) for any asset. The risk-free rate \(R_f\) satisfies \(\mathbb E[mR_f]=1\iff R_f=1/\mathbb E[m]\).

协方差定价公式 / The covariance pricing formula 由 \(\mathrm{Cov}[X,Y]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]\)，Using \(\mathrm{Cov}[X,Y]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]\),

$$0=\mathbb E[m(R_j-R_f)]=\mathbb E[m](\mathbb E[R_j]-R_f)-\mathrm{Cov}[m,R_j-R_f],$$

整理得which rearranges to

$$\mathbb E[R_j]-R_f=-\frac1{\mathbb E[m]}\mathrm{Cov}[m,R_j-R_f]=-R_f\,\mathrm{Cov}[m,R_j].$$

因子模型与 beta 定价 / Factor model and beta pricing 实证资产定价的核心问题是 SDF \(m\) 究竟是什么。从业者更关心线性因子定价模型，假设 SDF 形如 \(m_t=a-b'f_t\)（\(f_t\) 因子向量，\(a,b\) 常数）。最著名的例子是 CAPM：单因子 \(f_t=R_{mt}\)（市场收益）。若有 \(K\) 个因子、\(b=(b_1,\dots,b_K)'\)，协方差定价公式变为A central question in empirical asset pricing is what the SDF \(m\) is. Practitioners care more about linear factor pricing models, assuming the SDF takes the form \(m_t=a-b'f_t\) (\(f_t\) a vector of factors, \(a,b\) constants). The most famous example is the CAPM: a single factor \(f_t=R_{mt}\) (the market return). With \(K\) factors and \(b=(b_1,\dots,b_K)'\), the covariance pricing formula becomes

$$\mathbb E[R_j]-R_f=R_f\sum_{k=1}^K b_k\,\mathrm{Cov}[f_k,R_j].$$

定义资产 \(j\) 关于因子 \(k\) 的 beta \(\beta_{k,j}=\dfrac{\mathrm{Cov}[f_k,R_j]}{\mathrm{Var}[f_k]}\)，风险溢价 \(\gamma_k=R_f\mathrm{Var}[f_k]\)，零-beta 率 \(\gamma_0=R_f\)，则Define the beta of asset \(j\) w.r.t. factor \(k\), \(\beta_{k,j}=\dfrac{\mathrm{Cov}[f_k,R_j]}{\mathrm{Var}[f_k]}\), the risk premium \(\gamma_k=R_f\mathrm{Var}[f_k]\), and the zero-beta rate \(\gamma_0=R_f\); then

$$\mu_j=\mathbb E[R_j]=\gamma_0+\sum_{k=1}^K\gamma_k\beta_{k,j}.$$

Fama-MacBeth 两步回归与 GMM / Fama-MacBeth two-pass regression and GMM 经典估计法是 Fama and MacBeth (1973) 两步回归。第一步：把资产收益 \(R_{j,t}\) 对常数与因子回归，估计 \(\alpha,\beta\)（\(R_{j,t}=\alpha_j+\sum_{k=1}^K\beta_{k,j}f_{k,t}+\epsilon_{j,t}\)）；模型成立则 \(\alpha_j\) 跨资产相同。第二步：把 \(\hat\mu_j\)（\(R_{j,t}\) 的时序样本均值）对常数与估计的 beta 回归，估计零-beta 率 \(\gamma_0\) 与风险溢价 \(\gamma_k\)；第二步回归的 \(R^2\) 视为模型拟合优度。现代做法是用 GMM 利用矩条件 \(\mathbb E_t[m_{t+1}R_{j,t+1}]=1\)（或 \(\mathbb E_t[m_{t+1}(R_{j,t+1}-R_{f,t})]=0\)），并用 \(J\) 检验。伪因子问题：若 \(m_t=a-b'f_t\) 近似成立，但纳入与所有收益独立的伪因子 \(g_t\)，矩条件 \(\mathbb E_t[(a-b'f_t-cg_t)(R_{j,t}-R_{f,t})]=0\) 在 \(b=0,c=1,a=\mathbb E[g_t]\) 时平凡成立——于是伪因子系数显著、有用因子不显著、模型完美拟合（Kan-Zhang 1999a,b；Burnside 2016）。Kan et al. (2013)、Gospodinov et al. (2014) 发展了误设下的渐近理论。The classic method is the Fama and MacBeth (1973) two-pass regression. First pass: regress asset returns \(R_{j,t}\) on a constant and the factors to estimate \(\alpha,\beta\) (\(R_{j,t}=\alpha_j+\sum_{k=1}^K\beta_{k,j}f_{k,t}+\epsilon_{j,t}\)); if the model holds, \(\alpha_j\) is common across assets. Second pass: regress \(\hat\mu_j\) (the time-series sample mean of \(R_{j,t}\)) on a constant and the estimated betas to estimate the zero-beta rate \(\gamma_0\) and risk premia \(\gamma_k\); the \(R^2\) of the second pass is viewed as goodness-of-fit. The modern approach uses GMM with the moment condition \(\mathbb E_t[m_{t+1}R_{j,t+1}]=1\) (or \(\mathbb E_t[m_{t+1}(R_{j,t+1}-R_{f,t})]=0\)) and a \(J\) test. Spurious-factor problem: if \(m_t=a-b'f_t\) is approximately true but one includes a spurious \(g_t\) independent of all returns, the moment condition \(\mathbb E_t[(a-b'f_t-cg_t)(R_{j,t}-R_{f,t})]=0\) holds trivially with \(b=0,c=1,a=\mathbb E[g_t]\) — so the spurious factor is significant, the useful factors insignificant, and the model fits perfectly (Kan-Zhang 1999a,b; Burnside 2016). Kan et al. (2013) and Gospodinov et al. (2014) develop the asymptotic theory under misspecification.

设定与期权术语 / Setup and option terminology 作为无套利定价的应用，本节讲 Cox et al. (1979) 的二项期权定价模型。\(T\) 期经济 \(t=0,\dots,T\)，两资产：股票与债券。毛无风险利率常数 \(R\)，股价 \(S_t\) 随机，可涨可跌：\(S_{t+1}=US_t\)（涨）或 \(DS_t\)（跌），\(U>R>D\)。问：执行价 \(K\) 的看涨期权值多少？答案不依赖涨跌概率（乐观者与悲观者对期权定价一致）。术语：执行价 \(K\)、到期 \(T\) 的看涨（看跌）期权赋予持有人在到期前以 \(K\) 买入（卖出）股票的权利（非义务）；按约定价买卖称行权 (exercise)。任意时刻可行权为美式，仅到期可行权为欧式（详见 Shreve 2004）。As an application of no-arbitrage pricing, this section covers the binomial option pricing model of Cox et al. (1979). A \(T\)-period economy \(t=0,\dots,T\), two assets: a stock and a bond. The gross risk-free rate is constant at \(R\), the stock price \(S_t\) is random and can go up or down: \(S_{t+1}=US_t\) (up) or \(DS_t\) (down), with \(U>R>D\). Question: what is the price of a call with strike \(K\)? The answer does not depend on the up/down probability (an optimist and a pessimist agree on the price). Terminology: a call (put) with strike \(K\) and maturity \(T\) gives the holder the right (not the obligation) to buy (sell) the stock at \(K\) until maturity; buying/selling at the specified price is exercising. Exercisable at any time is American, only at maturity is European (see Shreve 2004).

3.1 欧式期权 / 3.1 European options \(T=0\) 时 \(C=\max\{S_0-K,0\}\)。一期：涨态 \(C_u=\max\{US_0-K,0\}\)、跌态 \(C_d=\max\{DS_0-K,0\}\)，由无套利 \(C=p_u C_u+p_d C_d\)。用股票与债券的无套利定 \(p_u,p_d\)：\(1=p_u U+p_d D\)（股票）、\(1=p_u R+p_d R\)（债券）。解得For \(T=0\), \(C=\max\{S_0-K,0\}\). One period: up state \(C_u=\max\{US_0-K,0\}\), down state \(C_d=\max\{DS_0-K,0\}\), and by no-arbitrage \(C=p_u C_u+p_d C_d\). Determine \(p_u,p_d\) from the stock and bond no-arbitrage conditions \(1=p_u U+p_d D\) (stock) and \(1=p_u R+p_d R\) (bond). Solving,

$$\begin{bmatrix}p_u\\p_d\end{bmatrix}=\frac1R\begin{bmatrix}p\\1-p\end{bmatrix},\qquad p=\frac{R-D}{U-D},$$

故 \(C=\dfrac1R(pC_u+(1-p)C_d)=\dfrac1{1+r}\tilde{\mathbb E}[C_s]\)（风险中性概率 \(p\)）。一般 \(T\) 期：支付按 \(R^T\) 贴现，\(T\) 时风险中性分布为参数 \(p\) 的二项分布，恰好 \(n\) 次涨的概率为 \(\binom Tn p^n(1-p)^{T-n}\)、终值 \(S_T=U^n D^{T-n}S_0\)，故so \(C=\dfrac1R(pC_u+(1-p)C_d)=\dfrac1{1+r}\tilde{\mathbb E}[C_s]\) (risk-neutral probability \(p\)). General \(T\) periods: payoffs are discounted by \(R^T\), the risk-neutral distribution at \(T\) is binomial with parameter \(p\), the probability of exactly \(n\) up states is \(\binom Tn p^n(1-p)^{T-n}\) with final value \(S_T=U^n D^{T-n}S_0\), so

$$C=\frac1{R^T}\sum_{n=0}^T\binom Tn p^n(1-p)^{T-n}\max\{U^n D^{T-n}S_0-K,0\}.$$

看跌期权类似：\(P=\dfrac1{R^T}\sum_{n=0}^T\binom Tn p^n(1-p)^{T-n}\max\{K-U^n D^{T-n}S_0,0\}\)。The put is analogous: \(P=\dfrac1{R^T}\sum_{n=0}^T\binom Tn p^n(1-p)^{T-n}\max\{K-U^n D^{T-n}S_0,0\}\).

买卖权平价 / Put-call parity 欧式期权的重要性质是买卖权平价：\(C-P=S_0-KR^{-T}\)。证明：因An important property of European options is put-call parity: \(C-P=S_0-KR^{-T}\). Proof: since

$$\max\{S_T-K,0\}-\max\{K-S_T,0\}=S_T-K,$$

买一份看涨、卖一份看跌，其终端支付等于持有股票并在到期日付 \(K\)；该组合的现值恰为 \(S_0-KR^{-T}\)，故买卖权平价成立。（已知看涨价即可由 \(P=C-S_0+KR^{-T}\) 算看跌价。）buying one call and shorting one put has terminal payoff equal to holding the stock and paying \(K\) at maturity; the present value of this portfolio is exactly \(S_0-KR^{-T}\), so put-call parity holds. (Given the call price, the put is \(P=C-S_0+KR^{-T}\).)

3.2 美式期权与命题 2（Merton 1973）/ 3.2 American options and Proposition 2 美式可提前行权，故美式 \(\ge\) 欧式。\(T=0\) 时二者相同。一般地，记 \(C(S,T)\) 为股价 \(S\)、到期 \(T\) 时美式期权价（隐去对 \(K,R\) 的依赖），则American options can be exercised early, so American \(\ge\) European. For \(T=0\) they coincide. In general, letting \(C(S,T)\) be the American price when the stock price is \(S\) and maturity is \(T\) (suppressing dependence on \(K,R\)),

$$C(S,T)=\max\left\{S-K,\ \frac1R(pC(US,T-1)+(1-p)C(DS,T-1))\right\}.\tag{2}$$

命题 2（Merton 1973）：若 \(R\ge1\)，则提前行权看涨期权从不最优，美式看涨与欧式看涨价格相同。Proposition 2 (Merton 1973): if \(R\ge1\), it is never optimal to exercise the call early, and the American and European calls have identical prices.

命题 2 证明（凸性）/ Proof of Proposition 2 (convexity) 先证 \(C(S,T)\) 关于 \(S\) 凸（归纳）：\(T=0\) 时 \(C(S,0)=\max\{S-K,0\}\) 凸；设 \(C(S,T-1)\) 凸，因 \(S\mapsto US,DS\) 线性故 \(C(US,T-1),C(DS,T-1)\) 凸，正加权和凸，两凸函数取大仍凸，故 (2) 中 \(C(S,T)\) 凸。再由凸性与 \(R\ge1\)：First show \(C(S,T)\) is convex in \(S\) (by induction): for \(T=0\), \(C(S,0)=\max\{S-K,0\}\) is convex; if \(C(S,T-1)\) is convex, then since \(S\mapsto US,DS\) are linear, \(C(US,T-1),C(DS,T-1)\) are convex, their positive weighted sum is convex, and the max of two convex functions is convex, so \(C(S,T)\) in (2) is convex. Then by convexity and \(R\ge1\):

$$\frac1R(pC(US,T-1)+(1-p)C(DS,T-1))\ge\frac1R C((pU+(1-p)D)S,T-1)=\frac1R C(RS,T-1),$$

又 \(\dfrac1R C(RS,T-1)\ge\dfrac1R\max\{RS-K,0\}=\max\{S-K/R,0\}\ge S-K/R\ge S-K\)（末式用 \(R\ge1\)）。故 (2) 右端第一项总不大于第二项，理性主体不提前行权。（看跌期权时 \(R>1\) 使最后一步不成立，美式看跌 $>$ 欧式看跌，提前行权可能最优。）\(\blacksquare\)and \(\dfrac1R C(RS,T-1)\ge\dfrac1R\max\{RS-K,0\}=\max\{S-K/R,0\}\ge S-K/R\ge S-K\) (the last using \(R\ge1\)). So the first term in (2) is always \(\le\) the second, and rational agents do not exercise early. (For puts, \(R>1\) breaks the last step; American puts $>$ European puts, and early exercise may be optimal.) \(\blacksquare\)

3.3 连续时间极限与 Black-Scholes 公式 / 3.3 Continuous-time limit and the Black-Scholes formula 用风险中性定价，看涨价 \(C=e^{-rT}\tilde{\mathbb E}[\max\{S_T-K,0\}]\)。设股价服从几何布朗运动 \(dS_t=\mu S_t dt+\sigma S_t dB_t\)。从风险中性主体看，\(dS_t=rS_t dt+\sigma S_t dB_t\)（否则有套利）。由 Ito 公式 \(d(\log S_t)=(r-\tfrac{\sigma^2}2)dt+\sigma dB_t\)，故 \(\log S_t\sim N((r-\sigma^2/2)t,\sigma^2 t)\)，可写 \(\log S_T=(r-\sigma^2/2)T+\sigma\sqrt T z\)（\(z\sim N(0,1)\)）。代入：By risk-neutral pricing, the call price is \(C=e^{-rT}\tilde{\mathbb E}[\max\{S_T-K,0\}]\). Suppose the stock follows a geometric Brownian motion \(dS_t=\mu S_t dt+\sigma S_t dB_t\). From the risk-neutral agent's view, \(dS_t=rS_t dt+\sigma S_t dB_t\) (otherwise there is arbitrage). By Ito's formula \(d(\log S_t)=(r-\tfrac{\sigma^2}2)dt+\sigma dB_t\), so \(\log S_t\sim N((r-\sigma^2/2)t,\sigma^2 t)\) and \(\log S_T=(r-\sigma^2/2)T+\sigma\sqrt T z\) (\(z\sim N(0,1)\)). Substituting:

$$C=\int_{-\infty}^\infty\max\left\{S_0 e^{-\sigma^2T/2+\sigma\sqrt T z}-Ke^{-rT},0\right\}\frac1{\sqrt{2\pi}}e^{-z^2/2}\,dz.$$

\(\max\) 内关于 \(z\) 严格递增，零点为 \(\bar z=\dfrac{\sigma\sqrt T}2-\dfrac1{\sigma\sqrt T}\log\!\left(\dfrac{S_0}{Ke^{-rT}}\right)\)。分两项积分（第二项 \(Ke^{-rT}\Phi(-\bar z)\)；第一项配方化为 \(S_0\Phi(-\bar z+\sigma\sqrt T)\)，\(\Phi\) 为标准正态 CDF），得The expression inside \(\max\) is strictly increasing in \(z\), with zero at \(\bar z=\dfrac{\sigma\sqrt T}2-\dfrac1{\sigma\sqrt T}\log\!\left(\dfrac{S_0}{Ke^{-rT}}\right)\). Splitting the integral (the second term is \(Ke^{-rT}\Phi(-\bar z)\); the first, by completing the square, is \(S_0\Phi(-\bar z+\sigma\sqrt T)\), with \(\Phi\) the standard normal CDF) gives

$$C=S_0\Phi(x)-Ke^{-rT}\Phi(x-\sigma\sqrt T),\qquad x=-\bar z+\sigma\sqrt T=\frac1{\sigma\sqrt T}\log\!\left(\frac{S_0}{Ke^{-rT}}\right)+\frac{\sigma\sqrt T}2.$$

这就是 Black-Scholes 公式。This is the Black-Scholes formula.

习题 / Exercises 1. (a) 设 \(\{C_i\}_{i\in I}\subset\mathbb R^N\) 为一族凸集，证明 \(\bigcap_{i\in I}C_i\) 凸。(b) 对任意集合 \(A\)，证明存在包含 \(A\) 的最小凸集（\(A\) 的凸包）。2. 证明引理 3。3. (a) 设 \(0\neq a\in\mathbb R^N\)、\(c\in\mathbb R\)，证明超平面 \(H=\{x\mid\langle a,x\rangle=c\}\) 与半空间 \(H^+=\{x\mid\langle a,x\rangle\ge c\}\) 是凸集。(b) 设 \(A\) 为 \(M\times N\) 矩阵、\(b\in\mathbb R^M\)，证明多面体 \(P=\{x\mid Ax\le b\}\) 凸。4. (a) 证明平行四边形法则 \(\|a+b\|^2+\|a-b\|^2=2\|a\|^2+2\|b\|^2\)。(b) 用平行四边形法则证明 (3)。5. 设 \(A=\{(x,y)\mid y>x^3\}\)、\(B=\{(x,y)\mid x\ge1,y\le1\}\)：作图；问 \(A,B\) 能否分离（若能给出分离直线方程，否则说明原因）。6. 设 \(C=\{(x,y)\mid y>e^x\}\)、\(D=\{(x,y)\mid y\le0\}\)：作图；给出分离 \(C,D\) 的直线方程；\(C,D\) 能否严格分离（是/否并解释）。7. 证明 Stiemke 引理：设 \(A\) 为 \(M\times N\) 矩阵，则以下恰有一者成立：(1) 存在 \(x\in\mathbb R_{++}^N\) 使 \(Ax=0\)；(2) 存在 \(y\in\mathbb R^M\) 使 \(A'y>0\)。8. 线性规划 \(\min\langle c,x\rangle\) s.t. \(Ax\ge b\)（单纯形法）：(a) 证明单纯形法有限步终止；(b) 证明终止时即为原问题的解。1. (a) Let \(\{C_i\}_{i\in I}\subset\mathbb R^N\) be convex sets; prove \(\bigcap_{i\in I}C_i\) is convex. (b) For any set \(A\), prove there is a smallest convex set containing \(A\) (the convex hull of \(A\)). 2. Prove Lemma 3. 3. (a) For \(0\neq a\in\mathbb R^N\), \(c\in\mathbb R\), show the hyperplane \(H=\{x\mid\langle a,x\rangle=c\}\) and half space \(H^+=\{x\mid\langle a,x\rangle\ge c\}\) are convex. (b) For an \(M\times N\) matrix \(A\) and \(b\in\mathbb R^M\), show the polytope \(P=\{x\mid Ax\le b\}\) is convex. 4. (a) Prove the parallelogram law \(\|a+b\|^2+\|a-b\|^2=2\|a\|^2+2\|b\|^2\). (b) Use it to prove (3). 5. Let \(A=\{(x,y)\mid y>x^3\}\), \(B=\{(x,y)\mid x\ge1,y\le1\}\): draw them; can \(A,B\) be separated (give a separating line, or explain why not)? 6. Let \(C=\{(x,y)\mid y>e^x\}\), \(D=\{(x,y)\mid y\le0\}\): draw them; give a separating line; can \(C,D\) be strictly separated (yes/no and explain)? 7. Prove Stiemke's Lemma: for an \(M\times N\) matrix \(A\), exactly one holds: (1) there is \(x\in\mathbb R_{++}^N\) with \(Ax=0\); (2) there is \(y\in\mathbb R^M\) with \(A'y>0\). 8. Linear program \(\min\langle c,x\rangle\) s.t. \(Ax\ge b\) (simplex method): (a) prove it terminates in finitely many steps; (b) prove that on termination you are at a solution.

凸集、凸包、引理 3 / Convex set, convex hull, Lemma 3 集合 \(C\subset\mathbb R^N\) 凸，若 \(x,y\in C\Rightarrow(1-\alpha)x+\alpha y\in C\)（\(\forall\alpha\in[0,1]\)）——任两点连线段全含于 \(C\)。圆、三角形、正方形凸，星形不凸。包含 \(A\) 的最小凸集称 \(A\) 的凸包 \(\mathrm{co}\,A\)。\(\{x_k\}_{k=1}^K\) 的凸组合为 \(x=\sum_{k=1}^K\alpha_k x_k\)（\(\alpha_k\ge0\)，\(\sum_k\alpha_k=1\)）。引理 3：\(\mathrm{co}\,A\) 恰由 \(A\) 中点的所有凸组合构成。（Toda 的一个玩笑：汉字"凸"本身并不凸。）A set \(C\subset\mathbb R^N\) is convex if \(x,y\in C\Rightarrow(1-\alpha)x+\alpha y\in C\) (\(\forall\alpha\in[0,1]\)) — the segment between any two points lies in \(C\). Circles, triangles, squares are convex; a star is not. The smallest convex set containing \(A\) is its convex hull \(\mathrm{co}\,A\). A convex combination of \(\{x_k\}_{k=1}^K\) is \(x=\sum_{k=1}^K\alpha_k x_k\) (\(\alpha_k\ge0\), \(\sum_k\alpha_k=1\)). Lemma 3: \(\mathrm{co}\,A\) consists of all convex combinations of points of \(A\). (Toda's joke: the Chinese character for "convex," 凸, is itself not convex.)

超平面与半空间 / Hyperplanes and half spaces \(\langle a,x\rangle=a_1x_1+\dots+a_Nx_N\) 为内积。集合 \(\{x\in\mathbb R^N\mid\langle a,x\rangle=c\}\)（\(a\neq0\)）称超平面，\(a\) 是其法向量（取超平面上 \(x_0\)，由 \(\langle a,x-x_0\rangle=0\) 知 \(a\perp(x-x_0)\)）。半空间 \(H^+=\{x\mid\langle a,x\rangle\ge c\}\)、\(H^-=\{x\mid\langle a,x\rangle\le c\}\) 是超平面沿 \(a\)（或 \(-a\)）方向分出的两部分。超平面与半空间都是凸集。\(\langle a,x\rangle=a_1x_1+\dots+a_Nx_N\) is the inner product. The set \(\{x\in\mathbb R^N\mid\langle a,x\rangle=c\}\) (\(a\neq0\)) is a hyperplane, with \(a\) its normal vector (taking \(x_0\) on it, \(\langle a,x-x_0\rangle=0\) shows \(a\perp(x-x_0)\)). The half spaces \(H^+=\{x\mid\langle a,x\rangle\ge c\}\), \(H^-=\{x\mid\langle a,x\rangle\le c\}\) are the two parts split off by the hyperplane toward \(a\) (or \(-a\)). Hyperplanes and half spaces are convex.

定理 4（分离超平面定理）/ Theorem 4 (Separating Hyperplane Theorem) 超平面 \(\langle a,x\rangle=c\) 分离 \(A,B\)，若 \(A\subset H^-\)、\(B\subset H^+\)，即 \(x\in A\Rightarrow\langle a,x\rangle\le c\)、\(x\in B\Rightarrow\langle a,x\rangle\ge c\)。\(A,B\) 可分离 \(\iff\sup_{x\in A}\langle a,x\rangle\le\inf_{x\in B}\langle a,x\rangle\)；不等式严格则严格分离。定理 4：设 \(C,D\subset\mathbb R^N\) 非空凸且 \(C\cap D=\emptyset\)，则存在分离 \(C,D\) 的超平面；若二者闭且其一紧，则可严格分离。A hyperplane \(\langle a,x\rangle=c\) separates \(A,B\) if \(A\subset H^-\), \(B\subset H^+\), i.e. \(x\in A\Rightarrow\langle a,x\rangle\le c\), \(x\in B\Rightarrow\langle a,x\rangle\ge c\). \(A,B\) are separable \(\iff\sup_{x\in A}\langle a,x\rangle\le\inf_{x\in B}\langle a,x\rangle\); strict inequality gives strict separation. Theorem 4: if \(C,D\subset\mathbb R^N\) are nonempty convex with \(C\cap D=\emptyset\), there is a separating hyperplane; if both are closed and one is compact, they can be strictly separated.

引理 5（投影）与命题 6 / Lemma 5 (projection) and Proposition 6 引理 5：设 \(C\) 非空凸，则任意 \(x\in\mathbb R^N\) 有唯一最近点 \(P_C(x)\in\mathrm{cl}\,C\)（\(x\) 在 \(\mathrm{cl}\,C\) 上的投影），且对任意 \(z\in C\)，\(\langle x-P_C(x),z-P_C(x)\rangle\le0\)。命题 6：设 \(C\) 非空凸、\(\bar x\notin\mathrm{int}\,C\)，则存在分离 \(\bar x\) 与 \(C\) 的超平面 \(\langle a,x\rangle=c\)，即 \(\langle a,\bar x\rangle\ge c\ge\langle a,z\rangle\)（\(\forall z\in C\)）；若 \(\bar x\notin\mathrm{cl}\,C\) 则可取严格不等式。Lemma 5: if \(C\) is nonempty convex, every \(x\in\mathbb R^N\) has a unique closest point \(P_C(x)\in\mathrm{cl}\,C\) (the projection of \(x\) onto \(\mathrm{cl}\,C\)), and for any \(z\in C\), \(\langle x-P_C(x),z-P_C(x)\rangle\le0\). Proposition 6: if \(C\) is nonempty convex and \(\bar x\notin\mathrm{int}\,C\), there is a hyperplane \(\langle a,x\rangle=c\) separating \(\bar x\) and \(C\), i.e. \(\langle a,\bar x\rangle\ge c\ge\langle a,z\rangle\) (\(\forall z\in C\)); if \(\bar x\notin\mathrm{cl}\,C\) the inequalities can be made strict.

引理 5、命题 6、定理 4 证明梗概 / Proof outline of Lemma 5, Proposition 6, Theorem 4 引理 5：设 \(\delta=\inf\{\|x-y\|\mid y\in C\}\)，取 \(\{y_k\}\subset C\) 使 \(\|x-y_k\|\to\delta\)。由平行四边形法则 \(\|y_k-y_l\|^2=2\|x-y_k\|^2+2\|x-y_l\|^2-4\|x-\tfrac12(y_k+y_l)\|^2\) (3)，又 \(\tfrac12(y_k+y_l)\in C\)，故 \(\|y_k-y_l\|^2\le2\|x-y_k\|^2+2\|x-y_l\|^2-4\delta^2\to0\)，\(\{y_k\}\) Cauchy 收敛到 \(y\in\mathrm{cl}\,C\)，且 \(\|x-y\|=\delta\)。唯一性同理。对 \(z\in C\)，由 \(C\) 凸，\((1-\alpha)y_k+\alpha z\in C\)，\(\delta^2\le\|x-(1-\alpha)y_k-\alpha z\|^2\)，令 \(k\to\infty\)、除以 \(\alpha>0\)、令 \(\alpha\to0\) 得 \(\langle x-y,z-y\rangle\le0\)。命题 6：若 \(\bar x\notin\mathrm{cl}\,C\)，令 \(y=P_C(\bar x)\)、\(a=\bar x-y\neq0\)、\(c=\langle a,y\rangle+\tfrac12\|a\|^2\)，则由引理 5 对 \(z\in C\) 有 \(\langle a,z\rangle\le\langle a,y\rangle0\)，严格分离。\(\bar x\in\mathrm{cl}\,C\) 但 \(\notin\mathrm{int}\,C\) 时取 \(x_k\notin\mathrm{cl}\,C\)、\(x_k\to\bar x\)，对每个 \(x_k\) 得 \(\|a_k\|=1\) 的分离向量，取收敛子列取极限。定理 4：令 \(E=C-D=\{x-y\mid x\in C,y\in D\}\) 非空凸，\(C\cap D=\emptyset\Rightarrow0\notin E\)。由命题 6 存在 \(a\neq0\) 使 \(0\ge\langle a,z\rangle\)（\(\forall z\in E\)），即 \(\langle a,x\rangle\le\langle a,y\rangle\)（\(x\in C,y\in D\)），取 \(\sup_C\langle a,x\rangle\le c\le\inf_D\langle a,y\rangle\) 即分离。若 \(C\) 闭、\(D\) 紧，可证 \(E=C-D\) 闭且 \(0\notin E\)，由命题 6 得严格分离。\(\blacksquare\)Lemma 5: let \(\delta=\inf\{\|x-y\|\mid y\in C\}\) and take \(\{y_k\}\subset C\) with \(\|x-y_k\|\to\delta\). By the parallelogram law \(\|y_k-y_l\|^2=2\|x-y_k\|^2+2\|x-y_l\|^2-4\|x-\tfrac12(y_k+y_l)\|^2\) (3), and \(\tfrac12(y_k+y_l)\in C\), so \(\|y_k-y_l\|^2\le2\|x-y_k\|^2+2\|x-y_l\|^2-4\delta^2\to0\); \(\{y_k\}\) is Cauchy, converging to \(y\in\mathrm{cl}\,C\) with \(\|x-y\|=\delta\). Uniqueness is similar. For \(z\in C\), by convexity \((1-\alpha)y_k+\alpha z\in C\), so \(\delta^2\le\|x-(1-\alpha)y_k-\alpha z\|^2\); letting \(k\to\infty\), dividing by \(\alpha>0\), and letting \(\alpha\to0\) gives \(\langle x-y,z-y\rangle\le0\). Proposition 6: if \(\bar x\notin\mathrm{cl}\,C\), let \(y=P_C(\bar x)\), \(a=\bar x-y\neq0\), \(c=\langle a,y\rangle+\tfrac12\|a\|^2\); by Lemma 5, for \(z\in C\), \(\langle a,z\rangle\le\langle a,y\rangle0\), giving strict separation. If \(\bar x\in\mathrm{cl}\,C\) but \(\notin\mathrm{int}\,C\), take \(x_k\notin\mathrm{cl}\,C\), \(x_k\to\bar x\), get separating vectors with \(\|a_k\|=1\), and pass to a convergent subsequence. Theorem 4: let \(E=C-D=\{x-y\mid x\in C,y\in D\}\), nonempty convex; \(C\cap D=\emptyset\Rightarrow0\notin E\). By Proposition 6 there is \(a\neq0\) with \(0\ge\langle a,z\rangle\) (\(\forall z\in E\)), i.e. \(\langle a,x\rangle\le\langle a,y\rangle\) (\(x\in C,y\in D\)); taking \(\sup_C\langle a,x\rangle\le c\le\inf_D\langle a,y\rangle\) separates them. If \(C\) is closed and \(D\) compact, one shows \(E=C-D\) is closed with \(0\notin E\), and Proposition 6 gives strict separation. \(\blacksquare\)