周三上午十点半，一位上海私募的衍生品交易员盯着两块屏幕：一块是 SSE 50ETF（510050）期权链，本周到期、行权价 2.55 的 call 中间价 0.0185；另一块是他自己的 Python 估值脚本，输出 0.0192。差 0.0007——折成单张合约 ¥70（合约乘数 10000），全市场未平仓约 1.2 万张就是 ¥84 万的潜在分歧。屏幕上同一行权价的 put 还报 0.0341，方向和他的脚本对得上，但绝对数能不能对上他没看。在他点开订单簿之前要回答三个问题：(1) 是自己模型用的隐含波动率（implied volatility, IV）输入偏离了市场，还是公式本身错？(2) 把 q q q
一、起点:欧式期权在到期日的现金流
欧式期权（European option）只能在到期日 T T T T T T max ( S T − K , 0 ) \max(S_T - K, 0) max ( S T − K , 0 ) max ( K − S T , 0 ) \max(K - S_T, 0) max ( K − S T , 0 ) S T S_T S T Q Q Q
S T = S 0 exp [ ( r − 1 2 σ 2 ) T + σ W T Q ] , W T Q ∼ N ( 0 , T ) . S_T = S_0 \exp\!\left[\left(r - \tfrac{1}{2}\sigma^2\right) T + \sigma W_T^Q\right], \quad W_T^Q \sim \mathcal{N}(0, T). S T = S 0 exp [ ( r − 2 1 σ 2 ) T + σ W T Q ] , W T Q ∼ N ( 0 , T ) . 也就是说，ln ( S T / S 0 ) \ln(S_T / S_0) ln ( S T / S 0 ) ( r − 1 2 σ 2 ) T (r - \tfrac{1}{2}\sigma^2) T ( r − 2 1 σ 2 ) T σ 2 T \sigma^2 T σ 2 T S T S_T S T Q Q Q
C 0 = e − r T E Q [ max ( S T − K , 0 ) ] . C_0 = e^{-rT}\, E_Q\!\left[\max(S_T - K, 0)\right]. C 0 = e − r T E Q [ max ( S T − K , 0 ) ] . 这一期望可走两条独立的路径计算:(a) 直接用对数正态密度对 S T S_T S T (b) 把上一节推导出来的 Black-Scholes 偏微分方程（partial differential equation, PDE）通过变量替换约化到热方程（heat equation），再套高斯基本解。两条路终点对得上——同一个 d 1 d_1 d 1 d 2 d_2 d 2 C 0 C_0 C 0 d 1 d_1 d 1 d 2 d_2 d 2
二、推导 d 1 d_1 d 1 d 2 d_2 d 2
整个推导分四步:
把期望拆成两块——一块是 S T ⋅ 1 S_T \cdot \mathbb{1} S T ⋅ 1 K ⋅ Q ( S T > K ) K \cdot Q(S_T > K) K ⋅ Q ( S T > K )
算第二块——标的到期落在行权之上的风险中性概率，结果是 N ( d 2 ) N(d_2) N ( d 2 )
算第一块——重新组合后落到 S 0 N ( d 1 ) S_0 N(d_1) S 0 N ( d 1 )
拼回 C 0 C_0 C 0
步骤 1 :把期望拆成两块。
E Q [ max ( S T − K , 0 ) ] = E Q [ S T ⋅ 1 { S T > K } ] − K ⋅ Q ( S T > K ) . E_Q\!\left[\max(S_T - K, 0)\right] = E_Q\!\left[S_T \cdot \mathbb{1}_{\{S_T > K\}}\right] - K \cdot Q(S_T > K). E Q [ max ( S T − K , 0 ) ] = E Q [ S T ⋅ 1 { S T > K } ] − K ⋅ Q ( S T > K ) . 步骤 2 :先算第二项——标的在到期价高于行权价的风险中性概率。令 Z = W T Q / T ∼ N ( 0 , 1 ) Z = W_T^Q / \sqrt{T} \sim \mathcal{N}(0,1) Z = W T Q / T ∼ N ( 0 , 1 ) S T > K S_T > K S T > K
S T > K ⟺ ln ( S 0 / K ) + ( r − 1 2 σ 2 ) T + σ T Z > 0 ⟺ Z > − d 2 , S_T > K \iff \ln(S_0/K) + \left(r - \tfrac{1}{2}\sigma^2\right) T + \sigma\sqrt{T}\, Z > 0 \iff Z > -d_2, S T > K ⟺ ln ( S 0 / K ) + ( r − 2 1 σ 2 ) T + σ T Z > 0 ⟺ Z > − d 2 , 其中
d 2 = ln ( S 0 / K ) + ( r − 1 2 σ 2 ) T σ T . d_2 = \frac{\ln(S_0/K) + \left(r - \tfrac{1}{2}\sigma^2\right) T}{\sigma\sqrt{T}}. d 2 = σ T ln ( S 0 / K ) + ( r − 2 1 σ 2 ) T . 由标准正态对称性，Q ( S T > K ) = Q ( Z > − d 2 ) = N ( d 2 ) Q(S_T > K) = Q(Z > -d_2) = N(d_2) Q ( S T > K ) = Q ( Z > − d 2 ) = N ( d 2 ) N ( ⋅ ) N(\cdot) N ( ⋅ )
步骤 3 :再算第一项。把 S T S_T S T ϕ ( z ) = ( 1 / 2 π ) e − z 2 / 2 \phi(z) = (1/\sqrt{2\pi}) e^{-z^2/2} ϕ ( z ) = ( 1/ 2 π ) e − z 2 /2
E Q [ S T ⋅ 1 { Z > − d 2 } ] = S 0 e r T ∫ − d 2 ∞ e − 1 2 σ 2 T + σ T z ϕ ( z ) d z . \begin{aligned}
E_Q\!\left[S_T \cdot \mathbb{1}_{\{Z > -d_2\}}\right]
&= S_0 e^{rT} \int_{-d_2}^{\infty} e^{-\tfrac{1}{2}\sigma^2 T + \sigma\sqrt{T} z}\, \phi(z)\, dz.
\end{aligned} E Q [ S T ⋅ 1 { Z > − d 2 } ] = S 0 e r T ∫ − d 2 ∞ e − 2 1 σ 2 T + σ T z ϕ ( z ) d z . 指数里凑配方:
− 1 2 σ 2 T + σ T z − 1 2 z 2 = − 1 2 ( z − σ T ) 2 , -\tfrac{1}{2}\sigma^2 T + \sigma\sqrt{T} z - \tfrac{1}{2} z^2 = -\tfrac{1}{2}(z - \sigma\sqrt{T})^2, − 2 1 σ 2 T + σ T z − 2 1 z 2 = − 2 1 ( z − σ T ) 2 , 被积函数即 ϕ ( z − σ T ) \phi(z - \sigma\sqrt{T}) ϕ ( z − σ T ) u = z − σ T u = z - \sigma\sqrt{T} u = z − σ T
∫ − d 2 ∞ ϕ ( z − σ T ) d z = ∫ − d 2 − σ T ∞ ϕ ( u ) d u = N ( d 2 + σ T ) . \int_{-d_2}^{\infty} \phi(z - \sigma\sqrt{T})\, dz = \int_{-d_2 - \sigma\sqrt{T}}^{\infty} \phi(u)\, du = N\!\left(d_2 + \sigma\sqrt{T}\right). ∫ − d 2 ∞ ϕ ( z − σ T ) d z = ∫ − d 2 − σ T ∞ ϕ ( u ) d u = N ( d 2 + σ T ) . 令 d 1 = d 2 + σ T d_1 = d_2 + \sigma\sqrt{T} d 1 = d 2 + σ T S 0 e r T N ( d 1 ) S_0 e^{rT}\, N(d_1) S 0 e r T N ( d 1 )
步骤 4 :把两项装回 C 0 = e − r T ⋅ ( ⋅ ) C_0 = e^{-rT} \cdot (\,\cdot\,) C 0 = e − r T ⋅ ( ⋅ )
C 0 = S 0 N ( d 1 ) − K e − r T N ( d 2 ) , d 1 = ln ( S 0 / K ) + ( r + 1 2 σ 2 ) T σ T , d 2 = d 1 − σ T . \boxed{\;
\begin{aligned}
C_0 &= S_0\, N(d_1) - K\, e^{-rT}\, N(d_2), \\
d_1 &= \frac{\ln(S_0/K) + \left(r + \tfrac{1}{2}\sigma^2\right) T}{\sigma\sqrt{T}}, \quad d_2 = d_1 - \sigma\sqrt{T}.
\end{aligned}
\;} C 0 d 1 = S 0 N ( d 1 ) − K e − r T N ( d 2 ) , = σ T ln ( S 0 / K ) + ( r + 2 1 σ 2 ) T , d 2 = d 1 − σ T . N ( d 2 ) N(d_2) N ( d 2 ) N ( d 1 ) N(d_1) N ( d 1 ) S t S_t S t K K K S T S_T S T K K K
路径 (b) 一句话指引 :令 τ = T − t \tau = T - t τ = T − t x = ln ( S / K ) x = \ln(S/K) x = ln ( S / K ) V V V V = K e − α x − β τ u ( x , τ ) V = K e^{-\alpha x - \beta \tau} u(x, \tau) V = K e − α x − β τ u ( x , τ ) α , β \alpha, \beta α , β u τ = 1 2 σ 2 u x x u_\tau = \tfrac{1}{2}\sigma^2 u_{xx} u τ = 2 1 σ 2 u xx u ( x , 0 ) = ( e x − 1 ) + u(x, 0) = (e^x - 1)^+ u ( x , 0 ) = ( e x − 1 ) +
三、连续股息率 q q q
ETF、指数现货、外汇等标的会付一笔连续股息或类股息。把 S t S_t S t Q Q Q r r r r − q r - q r − q S 0 S_0 S 0 S 0 e − q T S_0 e^{-qT} S 0 e − q T d 1 d_1 d 1 d 2 d_2 d 2 r r r r − q r - q r − q
C 0 = S 0 e − q T N ( d 1 ) − K e − r T N ( d 2 ) , d 1 = ln ( S 0 / K ) + ( r − q + 1 2 σ 2 ) T σ T . \begin{aligned}
C_0 &= S_0 e^{-qT} N(d_1) - K e^{-rT} N(d_2), \\
d_1 &= \frac{\ln(S_0/K) + \left(r - q + \tfrac{1}{2}\sigma^2\right) T}{\sigma\sqrt{T}}.
\end{aligned} C 0 d 1 = S 0 e − q T N ( d 1 ) − K e − r T N ( d 2 ) , = σ T ln ( S 0 / K ) + ( r − q + 2 1 σ 2 ) T . 等价地，把远期价 F = S 0 e ( r − q ) T F = S_0 e^{(r - q) T} F = S 0 e ( r − q ) T
C 0 = e − r T [ F N ( d 1 ) − K N ( d 2 ) ] , d 1 = ln ( F / K ) + 1 2 σ 2 T σ T . C_0 = e^{-rT}\!\left[F\, N(d_1) - K\, N(d_2)\right], \quad d_1 = \frac{\ln(F/K) + \tfrac{1}{2}\sigma^2 T}{\sigma\sqrt{T}}. C 0 = e − r T [ F N ( d 1 ) − K N ( d 2 ) ] , d 1 = σ T ln ( F / K ) + 2 1 σ 2 T . 期货期权直接用 Black-76;现货标的把 F F F S 0 e ( r − q ) T S_0 e^{(r - q) T} S 0 e ( r − q ) T q q q r r r e − r T e^{-rT} e − r T
四、看跌看涨平价:闭式公式的自检
同一行权价、同一到期日的欧式 call 与 put 永远满足
C 0 − P 0 = S 0 e − q T − K e − r T . C_0 - P_0 = S_0 e^{-qT} - K e^{-rT}. C 0 − P 0 = S 0 e − q T − K e − r T . 这一式不依赖任何分布假设——只需要无套利与欧式行权约束。把第二、三节的 call 闭式公式代入，反求 put:
P 0 = K e − r T N ( − d 2 ) − S 0 e − q T N ( − d 1 ) . P_0 = K e^{-rT} N(-d_2) - S_0 e^{-qT} N(-d_1). P 0 = K e − r T N ( − d 2 ) − S 0 e − q T N ( − d 1 ) . 实务上这是任何 BS 实现的第一道自检:分别用闭式公式算 call 与 put，再用平价反算一次 put，三者数值差应该在 10 − 10 10^{-10} 1 0 − 10 d 2 = d 1 − σ T d_2 = d_1 - \sigma\sqrt{T} d 2 = d 1 − σ T N ( ⋅ ) N(\cdot) N ( ⋅ ) erf \text{erf} erf erfc \text{erfc} erfc
五、用 510300 期权做一次校准
取 2025 年 6 月某交易日盘中:沪深300 ETF（510300）现价 S 0 = 3.85 S_0 = 3.85 S 0 = 3.85 T = 28 / 365 = 0.0767 T = 28/365 = 0.0767 T = 28/365 = 0.0767 K = 3.85 K = 3.85 K = 3.85 σ = 0.22 \sigma = 0.22 σ = 0.22 r = 0.018 r = 0.018 r = 0.018 q = 0.022 q = 0.022 q = 0.022
d 1 = ln ( 3.85 / 3.85 ) + ( 0.018 − 0.022 + 1 2 ⋅ 0.22 2 ) ⋅ 0.0767 0.22 ⋅ 0.0767 = 0 + 0.02022 ⋅ 0.0767 0.0609 = 0.001551 0.0609 = 0.0255 , d 2 = 0.0255 − 0.0609 = − 0.0354. \begin{aligned}
d_1 &= \frac{\ln(3.85/3.85) + (0.018 - 0.022 + \tfrac{1}{2} \cdot 0.22^2) \cdot 0.0767}{0.22 \cdot \sqrt{0.0767}} \\
&= \frac{0 + 0.02022 \cdot 0.0767}{0.0609} = \frac{0.001551}{0.0609} = 0.0255, \\
d_2 &= 0.0255 - 0.0609 = -0.0354.
\end{aligned} d 1 d 2 = 0.22 ⋅ 0.0767 ln ( 3.85/3.85 ) + ( 0.018 − 0.022 + 2 1 ⋅ 0.2 2 2 ) ⋅ 0.0767 = 0.0609 0 + 0.02022 ⋅ 0.0767 = 0.0609 0.001551 = 0.0255 , = 0.0255 − 0.0609 = − 0.0354. 查表:N ( 0.0255 ) ≈ 0.5102 N(0.0255) \approx 0.5102 N ( 0.0255 ) ≈ 0.5102 N ( − 0.0354 ) ≈ 0.4859 N(-0.0354) \approx 0.4859 N ( − 0.0354 ) ≈ 0.4859
C 0 = 3.85 ⋅ e − 0.022 ⋅ 0.0767 ⋅ 0.5102 − 3.85 ⋅ e − 0.018 ⋅ 0.0767 ⋅ 0.4859 ≈ 0.0635. C_0 = 3.85 \cdot e^{-0.022 \cdot 0.0767} \cdot 0.5102 - 3.85 \cdot e^{-0.018 \cdot 0.0767} \cdot 0.4859 \approx 0.0635. C 0 = 3.85 ⋅ e − 0.022 ⋅ 0.0767 ⋅ 0.5102 − 3.85 ⋅ e − 0.018 ⋅ 0.0767 ⋅ 0.4859 ≈ 0.0635. 合约乘数 10000，单张 call 理论值约 ¥635。屏幕上同一合约报价 0.0648 时差 ¥13，折算成 IV 约高 0.5 个 vol——可能是市场对下周公布的经济数据多付了一点事件溢价。这种「价格 ↔ IV」的双向映射不在本节范围内，1.4.4 模块会专门展开。
把闭式公式拖动看响应（以 ATM 处 N ( d 1 ) ≈ N ( d 2 ) ≈ 0.5 N(d_1) \approx N(d_2) \approx 0.5 N ( d 1 ) ≈ N ( d 2 ) ≈ 0.5 S 0 S_0 S 0 K K K r r r q q q T T T
Formula Explorer
S0 * exp(-q*T) - K * exp(-r*T)直觉检查:(1) r > q r > q r > q T → 0 T \to 0 T → 0 → S 0 − K \to S_0 - K → S 0 − K q > r q > r q > r
六、内在价值与时间价值
把 call 价拆成两部分:
内在价值（intrinsic value） = max ( S 0 − K , 0 ) = \max(S_0 - K, 0) = max ( S 0 − K , 0 ) 时间价值（time value） = C 0 − = C_0 - = C 0 − ≥ 0 \geq 0 ≥ 0
ATM 与 OTM call 的价值全部是时间价值;深度 ITM call 的时间价值会被贴现压扁，逼近 S 0 e − q T − K e − r T S_0 e^{-qT} - K e^{-rT} S 0 e − q T − K e − r T
七、练习
Exercise
某 510300 ETF 欧式 call:S 0 = 4.00 S_0 = 4.00 S 0 = 4.00 K = 4.10 K = 4.10 K = 4.10 T = 0.25 T = 0.25 T = 0.25 σ = 0.20 \sigma = 0.20 σ = 0.20 r = 0.020 r = 0.020 r = 0.020 q = 0.025 q = 0.025 q = 0.025
(a) 求 d 1 d_1 d 1 d 2 d_2 d 2 N ( − 0.21 ) ≈ 0.417 N(-0.21) \approx 0.417 N ( − 0.21 ) ≈ 0.417 N ( − 0.31 ) ≈ 0.378 N(-0.31) \approx 0.378 N ( − 0.31 ) ≈ 0.378 C 0 C_0 C 0 P 0 P_0 P 0 C 0 − P 0 C_0 - P_0 C 0 − P 0 S 0 e − q T − K e − r T S_0 e^{-qT} - K e^{-rT} S 0 e − q T − K e − r T
提示 d 1 d_1 d 1 分子是
ln ( 4.00 / 4.10 ) + ( r − q + 1 2 σ 2 ) T \ln(4.00/4.10) + (r - q + \tfrac{1}{2}\sigma^2) T ln ( 4.00/4.10 ) + ( r − q + 2 1 σ 2 ) T :把
ln ( 4.00 / 4.10 ) ≈ − 0.0247 \ln(4.00/4.10) \approx -0.0247 ln ( 4.00/4.10 ) ≈ − 0.0247 与
( 0.020 − 0.025 + 0.020 ) ⋅ 0.25 = 0.00375 (0.020 - 0.025 + 0.020) \cdot 0.25 = 0.00375 ( 0.020 − 0.025 + 0.020 ) ⋅ 0.25 = 0.00375 相加;分母
σ T = 0.10 \sigma\sqrt{T} = 0.10 σ T = 0.10 ;
d 2 = d 1 − 0.10 d_2 = d_1 - 0.10 d 2 = d 1 − 0.10 。先把这两个数算出来再做下一步。
提示 d 1 ≈ − 0.210 d_1 \approx -0.210 d 1 ≈ − 0.210 、
d 2 ≈ − 0.310 d_2 \approx -0.310 d 2 ≈ − 0.310 。代入:
C 0 ≈ 4.00 ⋅ e − 0.025 ⋅ 0.25 ⋅ 0.417 − 4.10 ⋅ e − 0.020 ⋅ 0.25 ⋅ 0.378 ≈ 0.117 C_0 \approx 4.00 \cdot e^{-0.025 \cdot 0.25} \cdot 0.417 - 4.10 \cdot e^{-0.020 \cdot 0.25} \cdot 0.378 \approx 0.117 C 0 ≈ 4.00 ⋅ e − 0.025 ⋅ 0.25 ⋅ 0.417 − 4.10 ⋅ e − 0.020 ⋅ 0.25 ⋅ 0.378 ≈ 0.117 。平价反推:
P 0 = C 0 − S 0 e − q T + K e − r T ≈ 0.220 P_0 = C_0 - S_0 e^{-qT} + K e^{-rT} \approx 0.220 P 0 = C 0 − S 0 e − q T + K e − r T ≈ 0.220 。两侧差应小于
10 − 3 10^{-3} 1 0 − 3 。
八、通往下一节
到这里你能写出 C 0 C_0 C 0 P 0 P_0 P 0 d 1 d_1 d 1 d 2 d_2 d 2 q q q S 0 S_0 S 0 σ \sigma σ r r r T T T N ( d 1 ) N(d_1) N ( d 1 ) N ( d 1 ) − 1 N(d_1) - 1 N ( d 1 ) − 1 S 0 e − q T N ′ ( d 1 ) T S_0 e^{-qT}\, N'(d_1)\, \sqrt{T} S 0 e − q T N ′ ( d 1 ) T