用構造無風險組合的方式推導訂價 PDE

用構造無風險組合的方式推導訂價 PDE

推導訂價金融工具的 PDE 是個常考科目,本文記述了一個直觀但不嚴謹的通用方法——構造無風險組合,並展現了 BS 模型、Heston 模型和零息債券單因子模型三個案例。app

BS 模型

從最簡單的童話故事開始。函數

股價 \(S_t\) 的 SDE 以下:工具

\[dS_t = (\mu - q)S_t dt + \sigma S_t dB_t \]

\(S_t\) 和歐式期權 \(E(t,S_t)\) 構造一個組合 \(V\),連續動態調整 \(S\)\(E\) 的比例使得 \(V\) 是一個無風險組合。因爲期權和股票都只在一個隨機變量 \(B_t\) 上有敞口,而且股票是可直接交易的對象,所以無需引入其餘的交易標的就能夠實現徹底對重,進而構造出無風險組合。這一點和後面的兩個例子不一樣。spa

\(w_1\)\(w_2\) 分別是 \(S\)\(E\) 的市值佔比,那麼 \(V\) 的「瞬時」收益率能夠表示爲:3d

\[\frac{dV}{V} = w_1 \frac{dS}{S} + w_2 \frac{dE}{E} +w_1 q dt \tag{1} \]

由 Ito 公式可知,對象

\[\begin{align*} & w_1 \frac{dS}{S} + w_2 \frac{dE}{E} \\ & = w_1 \left((\mu-q)dt + \sigma dB_t \right) + w_2 \frac{1}{E}\left( \frac{\partial E}{\partial t}dt + \frac{\partial E}{\partial S}dS_t + \frac{1}{2} \frac{\partial^2 E}{\partial S^2}d[S_t] \right)\\ & = w_1 \left((\mu-q)dt + \sigma dB_t \right) + w_2 \frac{1}{E}\left( \frac{\partial E}{\partial t}dt + \frac{\partial E}{\partial S}((\mu - q)S dt + \sigma S dB_t) + \frac{1}{2} \frac{\partial^2 E}{\partial S^2}\sigma^2S^2 dt \right) \end{align*} \]

\(V\) 是無風險組合的話數學

\[\frac{dV}{V} = r dt \tag{2} \]

爲了消除 \(V\) 的隨機性,也就是說消除掉 \(dB_t\)\(w_1\)\(w_2\) 要知足一個方程組:class

\[\left\{\begin{matrix} w_1 + w_2 &= 1\\ w_1 + \frac{S}{E} \frac{\partial E}{\partial B} w_2 &= 0 \end{matrix}\right. \]

方程組的解是變量

\[w_1 = \frac{-\frac{S}{E}\frac{\partial E}{\partial S}}{1 - \frac{S}{E}\frac{\partial E}{\partial S}}\\ w_2 = \frac{1}{1 - \frac{S}{E}\frac{\partial E}{\partial S}} \]

丟進(1),同時注意到(2),稍加整理將獲得 PDE(略過邊界條件):lambda

\[\frac{\partial E}{\partial t} + \frac{1}{2} \frac{\partial^2 E}{\partial S^2}\sigma^2S^2 + \frac{\partial E}{\partial S}S(r-q) - Er=0 \]

零息債券的單因子模型

債券的語境下就是另外一個故事了。

短時間利率過程 \(r_t\) 的 SDE 以下:

\[dr_t = \mu(r,t) dt + \sigma(r,t) dB_t \]

和期權的狀況不一樣,因爲短時間利率不是一個能夠直接交易的對象,所以,要構造無風險組合能夠考慮用兩個期限不一樣的債券對衝掉彼此的隨機性敞口。

\(w_1\)\(w_2\) 分別是 \(P(t, T_1)\)\(P(t, T_2)\) 的市值佔比(\(T_2 \neq T_1\)),那麼 \(V\) 的「瞬時」收益率能夠表示爲:

\[\frac{dV}{V} = w_1 \frac{dP_1}{P_1} + w_2 \frac{dP_2}{P_2} \tag{3} \]

由 Ito 公式可知,

\[\begin{align*} & w_1 \frac{dP_1}{P_1} + w_2 \frac{dP_2}{P_2} \\ & = w_1 \frac{1}{P_1}\left( \frac{\partial P_1}{\partial t}dt + \frac{\partial P_1}{\partial r}dr_t + \frac{1}{2} \frac{\partial^2 P_1}{\partial r^2}d[r_t] \right) + w_2 \frac{1}{P_2}\left( \frac{\partial P_2}{\partial t}dt + \frac{\partial P_2}{\partial r}dr_t + \frac{1}{2} \frac{\partial^2 P_2}{\partial r^2}d[r_t] \right)\\ & = w_1 \frac{1}{P_1}\left( \frac{\partial P_1}{\partial t}dt + \frac{\partial P_1}{\partial r}(\mu dt + \sigma dB_t) + \frac{1}{2} \frac{\partial^2 P_1}{\partial r^2}\sigma^2 dt \right) + w_2 \frac{1}{P_2}\left( \frac{\partial P_2}{\partial t}dt + \frac{\partial P_2}{\partial r}(\mu dt + \sigma dB_t) + \frac{1}{2} \frac{\partial^2 P_2}{\partial r^2}\sigma^2 dt \right) \end{align*} \]

爲了消除 \(V\) 的隨機性,\(w_1\)\(w_2\) 要知足一個方程組,

\[\left\{\begin{matrix} w_1 + w_2 &= 1\\ \frac{1}{P_1} \frac{\partial P_1}{\partial r} w_1 + \frac{1}{P_2} \frac{\partial P_2}{\partial r} w_2 &= 0 \end{matrix}\right. \]

方程組的解是

\[w_1 = \frac{\frac{1}{P_2} \frac{\partial P_2}{\partial r}}{\frac{1}{P_2} \frac{\partial P_2}{\partial r} - \frac{1}{P_1} \frac{\partial P_1}{\partial r}}\\ w_2 = \frac{-\frac{1}{P_1} \frac{\partial P_1}{\partial r}}{\frac{1}{P_2} \frac{\partial P_2}{\partial r} - \frac{1}{P_1} \frac{\partial P_1}{\partial r}} \]

丟進(3),同時注意到(2)將獲得

\[w_1 \frac{1}{P_1}\left( \frac{\partial P_1}{\partial t} + \frac{1}{2} \frac{\partial^2 P_1}{\partial r^2}\sigma^2 \right) + w_2 \frac{1}{P_2}\left( \frac{\partial P_2}{\partial t} + \frac{1}{2} \frac{\partial^2 P_2}{\partial r^2}\sigma^2 \right) = r_t \]

稍加整理獲得

\[\frac{\frac{\partial P_1}{\partial t} + \frac{1}{2} \frac{\partial^2 P_1}{\partial r^2}\sigma^2 - rP_1}{\frac{\partial P_1}{\partial r}} = \frac{\frac{\partial P_2}{\partial t} + \frac{1}{2} \frac{\partial^2 P_2}{\partial r^2}\sigma^2 - rP_2}{\frac{\partial P_2}{\partial r}} \]

也就是說

\[\frac{\frac{\partial P}{\partial t} + \frac{1}{2} \frac{\partial^2 P}{\partial r^2}\sigma^2 - rP}{\frac{\partial P}{\partial r}} \tag{4} \]

與債券的期限無關,能夠記爲一個 \(t\)\(r\) 的函數 \(\Lambda(t,r)\),進而獲得 PDE(略過邊界條件):

\[\frac{\partial P}{\partial t} + \frac{1}{2} \frac{\partial^2 P}{\partial r^2}\sigma^2 - Pr - \frac{\partial P}{\partial r}\Lambda = 0 \]

\(\Lambda\) 究竟是什麼?

仍是 Ito 公式,

\[\frac{dP}{P} = \frac{1}{P}(\frac{\partial P}{\partial t} + \frac{\partial P}{\partial r}\mu + \frac{1}{2} \frac{\partial^2 P}{\partial r^2}\sigma^2) dt + \frac{1}{P}\frac{\partial P}{\partial r}\sigma dB_t \]

能夠看到,債券瞬時回報分解成爲了肯定性和隨機性的兩部分,套用業績歸因的 Sharp 比的概念,若是把債券的 Sharp 比(並非真正意義上的 Sharp 比,而是一種類比)表示爲:

\[\lambda(t,r) = \frac{\frac{1}{P}(\frac{\partial P}{\partial t} + \frac{\partial P}{\partial r}\mu + \frac{1}{2} \frac{\partial^2 P}{\partial r^2}\sigma^2) - r}{\frac{1}{P}\frac{\partial P}{\partial r}\sigma} \]

它能夠看作是在度量短時間利率敞口的風險溢價水平。對比一下(4)就能夠獲得

\[\Lambda = \sigma \lambda - \mu \]

最終,零息債券的 PDE 寫成:

\[\frac{\partial P}{\partial t} + \frac{1}{2} \frac{\partial^2 P}{\partial r^2}\sigma^2 + \frac{\partial P}{\partial r}(\mu - \sigma \lambda) - Pr = 0 \]

Heston 模型

Heston 模型的故事又深邃了一點,實際上是前述兩個案例的綜合。

\[dS_t = (\mu - q)S dt + \sqrt{v} S dB^1_t\\ dv_t = \kappa(\theta - v)dt + \sigma \sqrt{v}dB^2_t \]

在 Heston 模型中歐式期權 \(E(t,S_t,v_t)\)\(B^1\)\(B^2\) 都有敞口,可是 \(S_t\) 只對 \(B^1\) 有直接的敞口,所以單純 \(E\)\(S\) 的組合沒法徹底對衝掉隨機性敞口。此外,和短時間利率類似,隨機波動率也是沒法直接交易的,能夠考慮引入另外一個不一樣期限的期權來對衝掉對 \(B^2\) 的敞口。

\(w_1\)\(w_2\)\(w_3\) 分別是 \(S\)\(E_1(T_1)\)\(E_2(T_2)\) 的市值佔比(\(T_2 \neq T_1\)),那麼

\[\frac{dV}{V} = w_1 \frac{dS}{S} + w_2 \frac{dE_1}{E_1} + w_3 \frac{dE_2}{E_2} + w_1 q dt \tag{5} \]

由 Ito 公式可知,

\[\begin{align*} & w_1 \frac{dS}{S} + w_2 \frac{dE_1}{E_1} + w_3 \frac{dE_2}{E_2} \\ & = w_1 \left( (\mu - q) dt + \sqrt{v_t} dB^1_t \right)\\ & + w_2 \frac{1}{E_1}\left( \frac{\partial E_1}{\partial t}dt + \frac{\partial E_1}{\partial S}dS_t + \frac{1}{2} \frac{\partial^2 E_1}{\partial S^2}d[S_t] + \frac{\partial E_1}{\partial v}dv_t + \frac{1}{2} \frac{\partial^2 E_1}{\partial v^2}d[v_t] + \frac{\partial^2 E_1}{\partial S \partial v}d[S_t, v_t] \right)\\ & + w_3 \frac{1}{E_2}\left( \frac{\partial E_2}{\partial t}dt + \frac{\partial E_2}{\partial S}dS_t + \frac{1}{2} \frac{\partial^2 E_2}{\partial S^2}d[S_t] + \frac{\partial E_2}{\partial v}dv_t + \frac{1}{2} \frac{\partial^2 E_2}{\partial v^2}d[v_t] + \frac{\partial^2 E_2}{\partial S \partial v}d[S_t, v_t] \right)\\ & = w_1 \left( (\mu - q) dt + \sqrt{v_t} dB^1_t \right)\\ & + w_2 \frac{1}{E_1}\left( \frac{\partial E_1}{\partial t}dt + \frac{\partial E_1}{\partial S}((\mu - q)S dt + \sqrt{v} S dB^1_t) + \frac{1}{2} \frac{\partial^2 E_1}{\partial S^2}vdt + \frac{\partial E_1}{\partial v}(\kappa(\theta - v)dt + \sigma \sqrt{v}dB^2_t) + \frac{1}{2} \frac{\partial^2 E_1}{\partial v^2}\sigma^2vdt + \frac{\partial^2 E_1}{\partial S \partial v}\rho \sigma Sv dt \right)\\ & + w_3 \frac{1}{E_2}\left( \frac{\partial E_2}{\partial t}dt + \frac{\partial E_2}{\partial S}((\mu - q)S dt + \sqrt{v} S dB^1_t) + \frac{1}{2} \frac{\partial^2 E_2}{\partial S^2}vdt + \frac{\partial E_2}{\partial v}(\kappa(\theta - v)dt + \sigma \sqrt{v}dB^2_t) + \frac{1}{2} \frac{\partial^2 E_2}{\partial v^2}\sigma^2vdt + \frac{\partial^2 E_2}{\partial S \partial v}\rho \sigma Sv dt \right)\\ & = w_1(x_1dt + y_1dB^1_t) \\ & + w_2(x_2dt + y_2dB^1_t + z_2dB^2_t) \\ & + w_3(x_3dt + y_3dB^1_t + z_3dB^2_t) \end{align*} \]

爲了消除 \(V\) 的隨機性,\(w_1\)\(w_2\)\(w_3\) 要知足一個方程組,

\[\left\{\begin{matrix} w_1 + w_2 + w_3 &= 1\\ y_1w_1 + y_2w_2 + y_3w_3 &= 0\\ z_2w_2 + z_3w_3 &= 0\\ \end{matrix}\right. \]

方程組的解是

\[w_1 = \frac{z_2y_3 - z_3y_2}{z_3(y_1-y_2) - z_2(y_1-y_3)}\\ w_2 = \frac{y_1z_3}{z_3(y_1-y_2) - z_2(y_1-y_3)}\\ w_3 = \frac{-y_1z_2}{z_3(y_1-y_2) - z_2(y_1-y_3)} \]

丟進(5),同時注意到(2),那麼

\[w_1(x_1+q) + w_2x_2 + w_3x_3 = r \]

稍加整理將獲得

\[\frac{(x_1-r+q)y_2 - (x_2-r)y_1}{z_2} = \frac{(x_1-r+q)y_3 - (x_3-r)y_1}{z_3} \]

也就是說

\[\frac{(x_1-r+q)y - (x-r)y_1}{z} \]

與期權期限無關,僅僅是 \(t\)\(S\)\(v\) 的函數,能夠記作 \(\Lambda(t,S,v)\),進而獲得 PDE:

\[\frac{\partial E}{\partial t} + \frac{\partial E}{\partial S}S(r-q) + \frac{1}{2}\frac{\partial^2 E}{\partial S^2}S^2v + \frac{\partial E}{\partial v}[\kappa(\theta - v) + \sigma\Lambda] + \frac{1}{2}\frac{\partial^2 E}{\partial v^2}\sigma^2v + \frac{\partial^2 E}{\partial S \partial v}\rho\sigma Sv - Er = 0 \]

\(\Lambda\) 究竟是什麼?

注意到

\[\frac{(x_1-r+q)y - (x-r)y_1}{z} \\= -\frac{yy_1}{z}\left(\frac{x-r}{y}-\frac{x_1-r+q}{y_1}\right) \\= -\frac{S\sqrt{v}}{\sigma}\left(\frac{x-r}{y}-\frac{x_1-r+q}{y_1}\right) \]

\[\lambda = S\sqrt{v}\left(\frac{x-r}{y}-\frac{x_1-r+q}{y_1}\right) \]

它能夠看作期權和股票 Sharp 比(並不是真正意義的 Sharp 比,而是一個類比)的差,再乘以股票價格瞬時變化的標準差。若是把期權看做是交易隨機波動率的工具,而股票僅是隨機波動率的被動承擔者,那麼

\[\frac{x-r}{y}-\frac{x_1-r+q}{y_1} \]

能夠看作是在衡量隨機波動率敞口的風險溢價水平,由於期權在兩個敞口上均有暴露,股票只有一個敞口,二者相減便剩餘隨機波動率敞口。

最終,PDE 變成:

\[\frac{\partial E}{\partial t} + \frac{\partial E}{\partial S}S(r-q) + \frac{1}{2}\frac{\partial^2 E}{\partial S^2}S^2v + \frac{\partial E}{\partial v}[\kappa(\theta - v) - \lambda] + \frac{1}{2}\frac{\partial^2 E}{\partial v^2}\sigma^2v + \frac{\partial^2 E}{\partial S \partial v}\rho\sigma Sv - Er = 0 \]

總結

嚴格的來講,上述推導均不嚴謹。

\(\frac{1}{V}dV\) 爲例,這是一種不規範的寫法,SDE 雖然名字中帶有微分,可是規範的寫法實際上是一組積分表達式,\(dS\) 僅僅是一種「縮寫」。正是由於不規範的寫法,\(\frac{1}{V}dV\) 不能再寫成 \(d\ln V\)。這是在隨機性世界,和肯定性世界有不同的物理規則。

儘管數學表述不規範,但背後的經濟意義卻很是明瞭。

若是把微分視做差分的極限,\(\frac{1}{V}dV\) 能夠看作是瞬時收益率,也就是算數收益率的極限形式的。合約以及資產價值有各自的隨機驅動因子(也就是敞口),消除掉組合的隨機性,也就實現了完美意義上的對衝。

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