2014年4月24日 星期四

[隨機分析] Black-Scholes PDE for European Call option (1)

延續上篇 [隨機分析] Black-Scholes PDE for European Call option (0),這次要來證明 Black-Scholes Formula
\[
{\scriptsize f(t,S_t) = S \Phi \left( \frac{\ln(S_t/K) + (r+\frac{1}{2} \sigma^2) (T-t) }{\sigma \sqrt{T-t}}\right) - K e^{-r (T-t)} \Phi \left( \frac{\ln(S_t/K) + (r - \frac{1}{2}\sigma^2) (T-t) }{\sigma \sqrt{T-t}}\right)}
\]其中 $\Phi (\cdot)$ 為 Standard Normal Cumulative distribution function (CDF)

現在令 $x= S_t$,上述的 Black-Scholes Formula 確實為 Black-Scholes PDE 的解,亦即上式為下列PDE的解:
\[
rf(t,x) = {f_t}(t,x) + rx{f_x}(t,x) + \frac{1}{2}{f_{xx}}(t,x){\sigma ^2}{x^2}
\]且滿足終端邊界條件 $f(T,x) = h(x), \ \forall x \in \mathbb{R}$


我們將分成下列幾個小步驟逐步完成此證明:


步驟1:首先證明下列等式成立:

Claim 1: $K e^{-r(T-t)} \Phi ' (d_2(T-t,x)) = x \Phi' (d_1(T-t,x))$
其中
\[
d_1 (T-t, x) =  \frac{\ln(x/K) + (r+\frac{1}{2} \sigma^2) (T-t) }{\sigma \sqrt{T-t}} \\
d_2 (T-t,x) =  \frac{\ln(x/K) + (r - \frac{1}{2}\sigma^2) (T-t) }{\sigma \sqrt{T-t}} \\
\]
Proof
我們證明
\[
K e^{-r(T-t)} \Phi ' (d_2(T-t,x))- x \Phi' (d_1(T-t,x))=0
\]由於 $\Phi (\cdot)$ 為 Standard Normal Cumulative distribution function (CDF) ,故 $\Phi '(\cdot)$ 即為 Standard Normal density,亦即
 \[\left\{ \begin{array}{l}
\Phi'\left( {{d_1}\left( {T - t,x} \right)} \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{d_1}{{\left( {T - t,x} \right)}^2}}}{2}}}\\
\Phi'\left( {{d_2}\left( {T - t,x} \right)} \right) = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{d_2}{{\left( {T - t,x} \right)}^2}}}{2}}}
\end{array} \right.
\]現在觀察 $d_2(T-t,x) = d_1(T-t,x) - \sigma \sqrt{T-t}$,故直接計算
\[\begin{array}{l}
K{e^{ - r(T - t)}}\Phi '({d_2}(T - t,x)) - x\Phi '({d_1}(T - t,x))\\
 = K{e^{ - r(T - t)}}\frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{d_2}{{\left( {T - t,x} \right)}^2}}}{2}}} - x\frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{d_1}{{\left( {T - t,x} \right)}^2}}}{2}}}\\
 = \frac{1}{{\sqrt {2\pi } }}\left[ {K{e^{ - r(T - t)}}{e^{ - \frac{{{d_2}{{\left( {T - t,x} \right)}^2}}}{2}}} - x{e^{ - \frac{{{d_1}{{\left( {T - t,x} \right)}^2}}}{2}}}} \right]\\
 = \frac{1}{{\sqrt {2\pi } }}\left[ {K{e^{ - r(T - t)}}{e^{ - \frac{{{{\left[ {{d_1}\left( {T - t,x} \right) - \sigma \sqrt {T - t} } \right]}^2}}}{2}}} - x{e^{ - \frac{{{d_1}{{\left( {T - t,x} \right)}^2}}}{2}}}} \right]\\
 = \frac{1}{{\sqrt {2\pi } }}\left[ {K{e^{ - r(T - t)}}{e^{ - \frac{{{d_1}^2\left( {T - t,x} \right) - 2{d_1}\left( {T - t,x} \right)\sigma \sqrt {T - t}  + {\sigma ^2}\left( {T - t} \right)}}{2}}} - x{e^{ - \frac{{{d_1}{{\left( {T - t,x} \right)}^2}}}{2}}}} \right]\\
 = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{d_1}{{\left( {T - t,x} \right)}^2}}}{2}}}\left[ {K{e^{ - r(T - t)}}{e^{\frac{{2{d_1}\left( {T - t,x} \right)\sigma \sqrt {T - t} }}{2}}}{e^{\frac{{ - {\sigma ^2}\left( {T - t} \right)}}{2}}} - x} \right]\\
 = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{d_1}{{\left( {T - t,x} \right)}^2}}}{2}}}\left[ {K{e^{ - r(T - t)}}{e^{\frac{{\ln (x/K) + (r + \frac{1}{2}{\sigma ^2})\left( {T - t} \right)}}{{\sigma \sqrt {T - t} }}\sigma \sqrt {T - t} }}{e^{\frac{{ - {\sigma ^2}\left( {T - t} \right)}}{2}}} - x} \right]\\
 = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{d_1}{{\left( {T - t,x} \right)}^2}}}{2}}}\left[ {K{e^{ - r(T - t)}}\left( {\frac{x}{K}} \right){e^{(r + \frac{1}{2}{\sigma ^2})\left( {T - t} \right)}}{e^{\frac{{ - {\sigma ^2}\left( {T - t} \right)}}{2}}} - x} \right]\\
 = \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{d_1}{{\left( {T - t,x} \right)}^2}}}{2}}}\left[ {{e^{ - r(T - t)}}x{e^{r\left( {T - t} \right)}} - x} \right] = 0. \ \ \ \ \square
\end{array}
\]

步驟2:證明下列Claim:
Claim 2: (Theta of the Option)
\[
{f_t}\left( {t,x} \right) =  - rK{e^{ - r(T - t)}}\Phi ({d_2}(T - t,x)) - \frac{{\sigma x}}{{2 \sqrt {T - t} }}\Phi '({d_1}(T - t,x))
\]
Proof
直接計算 $f(t,x)$ 的對 $t$ 偏導數
\[\begin{array}{l}
{f_t}(t,x) = \frac{\partial }{{\partial t}}f\left( {t,x} \right) = \frac{\partial }{{\partial t}}\left[ {x\Phi \left( {{d_1}(T - t,x)} \right) - K{e^{ - rT}}{e^{rt}}\Phi \left( {{d_2}(T - t,x)} \right)} \right]\\
 \Rightarrow {f_t}(t,x) = x\frac{{\partial \Phi }}{{\partial t}}\left( {{d_1}(T - t,x)} \right) \\
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - K{e^{ - r(T)}}\left[ {r{e^{r(t)}}\Phi \left( {{d_2}(T - t,x)} \right) + {e^{r(t)}}\frac{{\partial \Phi }}{{\partial t}}\left( {{d_2}(T - t,x)} \right)} \right]\\
 \Rightarrow {f_t}(t,x) =  - rK{e^{ - r(T - t)}}\Phi \left( {{d_2}(T - t,x)} \right) + x\frac{{\partial \Phi \left( {{d_1}(T - t,x)} \right)}}{{\partial t}} \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - K{e^{ - r(T - t)}}\frac{{\partial \Phi \left( {{d_2}(T - t,x)} \right)}}{{\partial t}} \ \ \ \ (*)
\end{array}
\]由 Standard Normal CDF $\Phi(\cdot)$ 定義,我們可以分別計算:
\[\begin{array}{l}
\frac{{\partial \Phi \left( {{d_1}(T - t,x)} \right)}}{{\partial t}} = \frac{\partial }{{\partial t}}\left\{ {\frac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^{{d_1}(T - t,x)} {{e^{\frac{{ - {z^2}}}{2}}}} dz} \right\} \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = \frac{1}{{\sqrt {2\pi } }}\left[ {{e^{\frac{{ - {{\left( {{d_1}(T - t,x)} \right)}^2}}}{2}}}\frac{\partial }{{\partial t}}{d_1}(T - t,x)} \right]\\
\frac{{\partial \Phi \left( {{d_2}(T - t,x)} \right)}}{{\partial t}} = \frac{1}{{\sqrt {2\pi } }}\left[ {{e^{\frac{{ - {{\left( {{d_2}(T - t,x)} \right)}^2}}}{2}}}\frac{\partial }{{\partial t}}{d_2}(T - t,x)} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{\sqrt {2\pi } }}\left[ {{e^{\frac{{ - {{\left( {{d_2}(T - t,x)} \right)}^2}}}{2}}}\frac{\partial }{{\partial t}}\left[ {{d_1}(T - t,x) - \sigma \sqrt {T - t} } \right]} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{\sqrt {2\pi } }}{e^{\frac{{ - {{\left( {{d_2}(T - t,x)} \right)}^2}}}{2}}}\left[ {\frac{{\partial {d_1}(T - t,x)}}{{\partial t}} + \frac{\sigma }{{2\sqrt {T - t} }}} \right]
\end{array}
\]將上面計算出來的兩項偏導數式子帶回 $(*)$,並利用 Claim 1 的結果整理可得
\[
 \Rightarrow {f_t}(t,x) =  - rK{e^{ - r(T - t)}}\Phi \left( {{d_2}(T - t,x)} \right) \\
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac{\sigma }{{2\sqrt {T - t} }}K{e^{ - r(T - t)}}\frac{1}{{\sqrt {2\pi } }}{e^{\frac{{ - {{\left( {{d_2}(T - t,x)} \right)}^2}}}{2}}}\\
\]利用 $d_1(T-t,x) = d_2(T-t,x) - \sigma \sqrt{T-t}$,帶入上式
\[\begin{array}{l}
 \Rightarrow {f_t}(t,x) =  - rK{e^{ - r(T - t)}}\Phi \left( {{d_2}(T - t,x)} \right) \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac{\sigma }{{2\sqrt {T - t} }}K{e^{ - r(T - t)}}\frac{1}{{\sqrt {2\pi } }}{e^{\frac{{ - {{\left( {{d_1}(T - t,x) - \sigma \sqrt {T - t} } \right)}^2}}}{2}}}\\
 \Rightarrow {f_t}(t,x) =  - rK{e^{ - r(T - t)}}\Phi \left( {{d_2}(T - t,x)} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}
\end{array} - \frac{\sigma }{{2\sqrt {T - t} }}K{e^{ - r(T - t)}}\frac{1}{{\sqrt {2\pi } }}{e^{\frac{{ - {d_1}^2(T - t,x)}}{2}}}{e^{\frac{{2\sigma {d_1}(T - t,x)\sqrt {T - t} }}{2}}}{e^{\frac{{ - {\sigma ^2}\left( {T - t} \right)}}{2}}}\\
 \Rightarrow {f_t}(t,x) =  - rK{e^{ - r(T - t)}}\Phi \left( {{d_2}(T - t,x)} \right) - \frac{{\sigma x}}{{2\sqrt {T - t} }}\Phi '\left( {{d_1}(T - t,x)} \right)
\end{array}
\] $\square$

步驟3:證明下列Claim:
Claim 3: (Delta of the Option)
\[
{f_x}\left( {t,x} \right) =  \Phi ({d_1}(T - t,x))
\]
Proof
想法同Claim 2,直接計算對 $x$ 偏導數
\[\begin{array}{l}
{f_x}\left( {t,x} \right) = \frac{\partial }{{\partial x}}f(t,x) = \frac{\partial }{{\partial x}}\left[ {x\Phi \left( {{d_1}(T - t,x)} \right) - K{e^{ - r(T - t)}}\Phi \left( {{d_2}(T - t,x)} \right)} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \Phi \left( {{d_1}(T - t,x)} \right) + x\frac{\partial }{{\partial x}}\Phi \left( {{d_1}(T - t,x)} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} - K{e^{ - r(T - t)}}\frac{\partial }{{\partial x}}\Phi \left( {{d_2}(T - t,x)} \right) \ \ \ \ (**)
\end{array}
\]分別計算
\[\begin{array}{l}
\frac{{\partial \Phi \left( {{d_1}(T - t,x)} \right)}}{{\partial x}} = \frac{\partial }{{\partial x}}\left\{ {\frac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^{{d_1}(T - t,x)} {{e^{\frac{{ - {z^2}}}{2}}}} dz} \right\}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{\sqrt {2\pi } }}\left[ {{e^{\frac{{ - {{\left( {{d_1}(T - t,x)} \right)}^2}}}{2}}}\frac{\partial }{{\partial x}}{d_1}(T - t,x)} \right]\\
\frac{{\partial \Phi \left( {{d_2}(T - t,x)} \right)}}{{\partial x}} = \frac{1}{{\sqrt {2\pi } }}\left[ {{e^{\frac{{ - {{\left( {{d_2}(T - t,x)} \right)}^2}}}{2}}}\frac{\partial }{{\partial x}}{d_2}(T - t,x)} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{\sqrt {2\pi } }}\left[ {{e^{\frac{{ - {{\left( {{d_2}(T - t,x)} \right)}^2}}}{2}}}\frac{\partial }{{\partial x}}\left[ {{d_1}(T - t,x) - \sigma \sqrt {T - t} } \right]} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}
\end{array} = \frac{1}{{\sqrt {2\pi } }}{e^{\frac{{ - {{\left( {{d_2}(T - t,x)} \right)}^2}}}{2}}}\left[ {\frac{{\partial {d_1}(T - t,x)}}{{\partial x}}} \right]
\end{array}
\]帶回偏導式 $(**)$ 可得
\[\begin{array}{l}
{f_x}\left( {t,x} \right) = \Phi \left( {{d_1}(T - t,x)} \right) + \frac{1}{{\sqrt {2\pi } }}\left[ {x{e^{\frac{{ - {{\left( {{d_1}(T - t,x)} \right)}^2}}}{2}}}\frac{\partial }{{\partial x}}{d_1}(T - t,x)} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} - K{e^{ - r(T - t)}}\frac{1}{{\sqrt {2\pi } }}{e^{\frac{{ - {{\left( {{d_2}(T - t,x)} \right)}^2}}}{2}}}\left[ {\frac{{\partial {d_1}(T - t,x)}}{{\partial x}}} \right]\\
 \Rightarrow {f_x}\left( {t,x} \right) = \Phi \left( {{d_1}(T - t,x)} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + \frac{1}{{\sqrt {2\pi } }}\left\{ {x{e^{\frac{{ - {{\left( {{d_1}(T - t,x)} \right)}^2}}}{2}}} - K{e^{ - r(T - t)}}{e^{\frac{{ - {{\left( {{d_2}(T - t,x)} \right)}^2}}}{2}}}} \right\}\left[ {\frac{{\partial {d_1}(T - t,x)}}{{\partial x}}} \right]
\end{array}
\]利用Claim 1的結果,我們可知
\[{f_x}\left( {t,x} \right) = \Phi \left( {{d_1}(T - t,x)} \right) \ \ \ \ \square
\]。

現在,由上述 Claim 2 的結果: ${f_x}\left( {t,x} \right) = \Phi \left( {{d_1}(T - t,x)} \right)$ 我們可立刻求得 $f_x(t,x)$ 對 $x$ 的二階偏導數 (Gamma of the Option):
\[{f_{xx}}\left( {t,x} \right) = \frac{{\partial \Phi \left( {{d_1}(T - t,x)} \right)}}{{\partial x}} = \frac{1}{{\sqrt {2\pi } }}\left[ {{e^{\frac{{ - {{\left( {{d_1}(T - t,x)} \right)}^2}}}{2}}}\frac{\partial }{{\partial x}}{d_1}(T - t,x)} \right]
\]由於
\[\frac{\partial }{{\partial x}}\left[ {\frac{{\ln (x/K) + (r + \frac{1}{2}{\sigma ^2})(T - t)}}{{\sigma \sqrt {T - t} }}} \right] = \frac{1}{{\sigma \sqrt {T - t} }}\frac{1}{x}
\]故
\[{f_{xx}}\left( {t,x} \right) = \frac{1}{{\sqrt {2\pi } }}\left[ {{e^{\frac{{ - {{\left( {{d_1}(T - t,x)} \right)}^2}}}{2}}}\frac{1}{{\sigma x\sqrt {T - t} }}} \right]\]

步驟4:證明下列 Claim:
Claim 4: Black-Scholes Formula satisfies Black-Scholes PDE

Proof
回憶 Black-Scholes PDE:
\[
rf(t,x) = {f_t}(t,x) + rx{f_x}(t,x) + \frac{1}{2}{f_{xx}}(t,x){\sigma ^2}{x^2}
\]現在將 Claim 2, 3 計算出來的結果 $f_t(t,x), f_x(t,x)$ 與 二階偏導數 $f_{xx}(t,x)$ 帶入PDE中
\[
rf(t,x) = {f_t}(t,x) + rx{f_x}(t,x) + \frac{1}{2}{f_{xx}}(t,x){\sigma ^2}{x^2}
\],我們得到
\[\begin{array}{l}
 \Rightarrow rf(t,x) =  - rK{e^{ - r(T - t)}}\Phi ({d_2}(T - t,x)) - \frac{{\sigma x}}{{2\sqrt {T - t} }}\Phi '({d_1}(T - t,x))\\
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + rx\Phi \left( {{d_1}(T - t,x)} \right) + \frac{1}{2}\frac{1}{{\sqrt {2\pi } }}\left[ {{e^{\frac{{ - {{\left( {{d_1}(T - t,x)} \right)}^2}}}{2}}}\frac{1}{{\sigma x\sqrt {T - t} }}} \right]{\sigma ^2}{x^2}\\
 \Rightarrow f(t,x) = x\Phi \left( {{d_1}(T - t,x)} \right) - K{e^{ - r(T - t)}}\Phi ({d_2}(T - t,x))
\end{array}
\]$ \square$


步驟5:證明符合 Terminal condition:
Claim 5: $f(T,x) = h(x)$

Proof
考慮 $x>K$,我們得到
\[
\mathop {\lim }\limits_{t \to T} {d_1}\left( {T - t,x} \right) =  + \infty  \Rightarrow \mathop {\lim }\limits_{t \to T} {d_2}\left( {T - t,x} \right) =  + \infty
\]故
\[ \Rightarrow f(T,x) = S - K \ \ \ \ (1)
\]

考慮 $0<x<K$,
\[\mathop {\lim }\limits_{t \to T} {d_1}\left( {T - t,x} \right) =  - \infty  \Rightarrow \mathop {\lim }\limits_{t \to T} {d_2}\left( {T - t,x} \right) =  - \infty
\]故
\[ \Rightarrow f(T,x) = 0 \ \ \ \ (2)
\] 合併 $(1) + (2)$ 得到Terminal condition:
\[
f(T,x) = (x - K)_+ = h(x)
\]