【FinE】Ito Integral习题(1)「建议收藏」

习题

Q1：几何布朗运动(GBM)的期望

题面：
计算GBM的数学期望
解析：
在概率测度$\mathbb{P}$下，GBM方程为
$$d{S}_t=\mu S_{t}dt+\sigma S_{t}d{w}_t^{\mathbb{P}}$$
在概率测度$\mathbb{Q}$下，GBM方程为
$$d{S}_t=r{S}_{t}dt+\sigma S_{t}d{w}_t^{\mathbb{Q}}$$
在$\mathbb{P}$测度下，令$f(t,x)=\ln x,f_t=0,f_x=\frac{1}{x},f_{xx}=\frac{-1}{x^2}$，根据Ito Formula计算
$$d\ln S_t=\frac{1}{S_t}d{S}_t-\frac{1}{2}\frac{1}{S_t^2}dt$$
代入$d{S}_t=\mu S_{t}dt+\sigma S_{t}d{w}_t^{\mathbb{P}}$，可得
$$\begin{array}{rl}d\ln S_t& =\mu dt+\sigma d{w}_t^{\mathbb{P}}-\frac{1}{2S_t^2}{\sigma }^2{S}_t^{2}dt\\ & =(\mu -\frac{1}{2}{\sigma }^2)dt+\sigma d{w}_t^{\mathbb{P}}\end{array}$$
两边同时积分可得
$$\begin{array}{rl}\ln S_t-\ln S_0& ={\int }_0^t(\mu -\frac{1}{2}{\sigma }^2)ds+\sigma d{w}_s^{\mathbb{P}}\\ & =(\mu -\frac{1}{2}{\sigma }^2)t+\sigma w_t^{\mathbb{P}}\end{array}$$
得到$S_t$表达式
$$S_t=S_0\exp [(\mu -\frac{1}{2}{\sigma }^2)t+\sigma w_t^{\mathbb{P}}]$$
计算期望
$$\mathbb{E}[S_t]=S_0\exp [(\mu -\frac{1}{2}{\sigma }^2)t]\times \mathbb{E}[\exp [\sigma w_t^{\mathbb{P}}]]$$
其中$w_t^{\mathbb{P}}\sim \mathcal{N}(0,t)$，计算$\mathbb{E}[\exp [\sigma w_t^{\mathbb{P}}]]$
$$\begin{array}{rl}\mathbb{E}[\exp [\sigma w_t^{\mathbb{P}}]]& ={\int }_{-\infty }^{\infty }{e}^{\sigma x}\frac{1}{\sqrt{2\pi t}}{e}^{\frac{-x^2}{2t}}dx\\ & =\underset{\mathcal{N}(\sigma t,t)}{\frac{1}{\sqrt{2\pi t}}{\int }_{-\infty }^{\infty }{e}^{-\frac{(x-\sigma t{)}^2}{2t}}dx}\times e^{\frac{{\sigma }^{2}t}{2}}\\ & =e^{\frac{{\sigma }^{2}t}{2}}\end{array}$$
所以
$$\mathbb{E}[S_t]=S_0{e}^{\mu t}$$

Q2：欧式看涨期权与BSM公式

题面：
对于欧式看涨期权$c(t,x)$，到期时间为$T$，行权价格是$K$，BS公式计算在时刻$t$，如果股票价格为$x$，期权价格为
$$c(t,x)=xN(d_{+})-K{e}^{-r(T-t)}N(d_{-})$$
其中
$$\begin{gathered} d_{+}=\frac{\ln (\frac{x}{K})+(r+\frac{1}{2}{\sigma }^2)(T-t)}{\sigma \sqrt{T-t}} \\ {d}_{-}=\frac{\ln \frac{x}{K}+(r-\frac{1}{2}{\sigma }^2)(T-t)}{\sigma \sqrt{T-t}} \end{gathered}$$
期权价格$c(t,x)$满足BSM偏微分方程
$$c_t(t,x)+rx{c}_x(t,x)+\frac{1}{2}{\sigma }^2{x}^2{c}_{xx}(t,x)=rc(t,x),0\le t\le T,x>0$$
终值条件
lim ⁡ t → T c ( t , x ) = max ⁡ { x − K , 0 } , x > 0 , x ≠ K \lim_{t\to T}c(t, x)=\max\{x-K, 0\}, x>0, x\neq K t→Tlim​c(t,x)=max{
x−K,0},x>0,x​=K
边界条件
$$\begin{array}{rl}& (1)\underset{x\to 0}{\lim }c(t,x)=0\\ & (2)\underset{x\to \infty }{\lim }[c(t,x)-(x-e^{-r(T-t)}K)]=0,0\le t\le K\end{array}$$
(1) 证明方程成立
$$K{e}^{-r(T-t)}{N}^{\prime }(d_{-})=x{N}^{\prime }(d_{+})$$
解析：
移项方程，可知
$$K{e}^{-r(T-t)}=x\frac{N^{\prime }(d_{+})}{N^{\prime }(d_{-})}$$
代入
$$N^{\prime }(x)=({\int }_{-\infty }^x\frac{1}{\sqrt{2\pi }}{e}^{-\frac{s^2}{2}}ds{)}^{\prime }=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$$
得到
$$\begin{array}{rl}& K{e}^{-r(T-t)}=x\frac{e^{-\frac{d_{+}^2}{2}}}{e^{-\frac{d_{-}^2}{2}}}=x{e}^{\frac{d_{-}^2-d_{+}^2}{2}}\\ & \iff \frac{x}{K}{e}^{r(T-t)}=e^{\frac{d_{+}^2-d_{-}^2}{2}}\end{array}$$
可以计算出
$$\begin{array}{rl}{d}_{+}^2-d_{-}^2& =\frac{[\ln (\frac{x}{K})+(r+\frac{1}{2}{\sigma }^2)(T-t){]}^2-[\ln (\frac{x}{K})+(r-\frac{1}{2}{\sigma }^2)(T-t){]}^2}{{\sigma }^2(T-t)}\\ & =\frac{[4\ast (\ln (\frac{x}{K})+r(T-t))]\ast \frac{1}{2}{\sigma }^2(T-t)}{{\sigma }^2(T-t)}\\ & =2\ast (\ln (\frac{x}{K})+r(T-t))\end{array}$$
计算方程R.H.S
$$e^{\frac{d_{-}^2-d_{+}^2}{2}}=e^{\ln (\frac{x}{K})+r(T-t)}=\frac{x}{K}{e}^{r(T-t)}$$
和方程L.H.S相同.$\square$

(2) 证明欧式看涨期权的$\Delta =c_x(t,x)=N(d_{+})$
解析：
复合函数求导得
$$c_x(t,x)=N(d_{+})+x{N}^{\prime }(d_{+})\frac{\partial d_{+}}{\partial x}-K{e}^{-r(T-t)}{N}^{\prime }(d_{-})\frac{\partial d_{-}}{\partial x}$$
计算偏导数部分
$$\frac{\partial d_{+}}{\partial x}=\frac{\partial d_{-}}{\partial x}=\frac{\partial }{\partial x}[\frac{\ln (\frac{x}{K})}{\sigma \sqrt{T-t}}]=\frac{1}{\sigma x\sqrt{T-t}}$$
代入$c_x(t,x)$得到
$$c_x(t,x)=N(d_{+})+(x{N}^{\prime }(d_{+})-K{e}^{-r(T-t)}{N}^{\prime }(d_{-}))\frac{1}{\sigma x\sqrt{T-t}}$$
有(1)中结果可知
$$c_x(t,x)=N(d_{+})$$
所以$\Delta =c_x(t,x)=N(d_{+})$成立.$\square$

(3) 证明欧式看涨期权的$\Theta$为
$$c_t(t,x)=-r{e}^{-r(T-t)}N(d_{-})-\frac{\sigma x}{2\sqrt{T-t}}{N}^{\prime }(d_{+})$$
解析：
根据(3)中的结果，可以预处理$\Gamma =c_{xx}(t,x)$
$$\Gamma =c_{xx}(t,x)=\frac{\partial }{\partial x}(N(d_{+}))=N^{\prime }(d_{+})\frac{\partial d_{+}}{\partial x}=\frac{N^{\prime }(d_{+})}{\sigma x\sqrt{T-t}}$$
计算
$$\begin{array}{rl}\frac{\partial }{\partial t}N(d_{+})& =N^{\prime }(d_{+})\frac{\partial d_{+}}{\partial t}\\ & =N^{\prime }(d_{+})\frac{1}{{\sigma }^2(T-t)}[-\frac{1}{2}(r+\frac{1}{2}{\sigma }^2)\sigma \sqrt{T-t}+\ln (x/K)\frac{\sigma }{2\sqrt{T-t}}]\\ & =\frac{N^{\prime }(d_{+})}{2\sigma \sqrt{T-t}}[\frac{\ln (x/K)}{T-t}-(r+\frac{1}{2}{\sigma }^2)]\\ \frac{\partial }{\partial t}N(d_{-})& =\frac{N^{\prime }(d_{-})}{2\sigma \sqrt{T-t}}[\frac{\ln (x/K)}{T-t}-(r-\frac{1}{2}{\sigma }^2)]\end{array}\text{\hspace{1em}(3.1)}$$
计算$c(t,x)$对$t$的偏导数，代入方程$(3.1)$中的结果
$$\begin{array}{rl}{c}_t(t,x)& =\frac{\partial }{\partial t}[xN(d_{+})-K{e}^{-r(T-t)}N(d_{-})]\\ & =-\frac{x{N}^{\prime }(d_{+})}{2\sigma \sqrt{T-t}}{\sigma }^2-Kr{e}^{-r(T-t)}N(d_{-})\\ & =-\frac{\sigma x{N}^{\prime }(d_{+})}{2\sqrt{T-t}}-Kr{e}^{-r(T-t)}N(d_{-})\end{array}$$
$\Theta$方程成立.$\square$

(4) 使用(1), (2), (3)中的结果证明$c(t,x)$满足BSM偏微分方程
解析：
BSM偏微分方程L.H.S为
$$\begin{array}{rl}& c_t(t,x)+rx{c}_x(t,x)+\frac{1}{2}{\sigma }^2{x}^2{c}_{xx}(t,x)\\ & =\Theta +rx\Delta +\frac{1}{2}{\sigma }^2{x}^2\Gamma \\ & =r[xN(d_{+})-K{e}^{-r(T-t)}N(d_{-})]\\ & =rc(t,x)\end{array}$$
R.H.S为$rc(t,x)$.$\square$

(5) 计算${\lim }_{t\to T}{d}_{+}$和${\lim }_{t\to T}{d}_{-}$，对于$x>0,x\ne K$，推导终值条件
解析：
计算极限
$$\underset{t\to T}{\lim }{d}_{+}=\underset{t\to T}{\lim }\frac{\ln (x/K)+(r+\frac{1}{2}{\sigma }^2)(T-t)}{\sigma \sqrt{T-t}}$$
分类讨论：
(1). 当$x>K$时，${\lim }_{t\to T}\to +\infty$，终值为
$$\underset{t\to T}{\lim }c(t,x)=\underset{t\to T}{\lim }[xN(d_{+})-K{e}^{-r(T-t)}N(d_{-})]=x-K$$
(2). 当$x<K$时，${\lim }_{t\to 0}xN(d_{+})-K{e}^{-r(T-t)}N(d_{-})=0$
综上，边界条件为 max ⁡ { x − K , 0 } \max\{x-K, 0\} max{
x−K,0}.$\square$

(6) 计算${\lim }_{x\to 0}{d}_{+}$和${\lim }_{x\to 0}{d}_{-}$，对于$0\le t<T$，推导边界条件$(1)$.
解析：
计算极限
$$\begin{gathered} \underset{x\to 0}{\lim }{d}_{+}=\underset{x\to 0}{\lim }\frac{\ln (x/K)+(r+\frac{1}{2}{\sigma }^2)(T-t))}{\sigma \sqrt{T-t}}=-\infty \\ \underset{x\to 0}{\lim }{d}_{-}=\underset{x\to 0}{\lim }\frac{\ln (x/K)+(r-\frac{1}{2}{\sigma }^2)(T-t))}{\sigma \sqrt{T-t}}=-\infty \\ \underset{x\to 0}{\lim }c(t,x)=\underset{x\to 0}{\lim }xN(d_{+})-K{e}^{-r(T-t)}N(d_{-})=0 \end{gathered}$$

(7) 计算${\lim }_{x\to \infty }{d}_{+}$和${\lim }_{x\to \infty }{d}_{-}$，对于$0\le t<T$，推导边界条件$(2)$.
解析：
从$d_{+}$反解出$x$为
x = K exp ⁡ { σ T − t d + − ( T − t ) ( r + 1 2 σ 2 ) } x=K\exp\{\sigma\sqrt{T-t}d_+-(T-t)(r+\frac{1}{2}\sigma^2)\} x=Kexp{
σT−t
​d+​−(T−t)(r+21​σ2)}
计算极限
$$\begin{gathered} \underset{x\to \infty }{\lim }{d}_{+}=\underset{x\to \infty }{\lim }\frac{\ln (x/K)+(r+\frac{1}{2}{\sigma }^2)(T-t))}{\sigma \sqrt{T-t}}=\infty \\ \underset{x\to \infty }{\lim }{d}_{-}=\underset{x\to \infty }{\lim }\frac{\ln (x/K)+(r-\frac{1}{2}{\sigma }^2)(T-t))}{\sigma \sqrt{T-t}}=\infty \\ \underset{x\to \infty }{\lim }c(t,x)=\underset{x\to \infty }{\lim }xN(d_{+})-K{e}^{-r(T-t)}N(d_{-})=\underset{\text{远期合约的价值}}{x-K{e}^{-r(T-t)}}=\text{R.H.S} \end{gathered}$$

Q3：标准Brown运动

题面：
设 { B t , t ≥ 0 } \{B_t, t\geq 0\} {
Bt​,t≥0}是标准Brown运动，计算$P(B_2\le 0)$和$P(B_t\le 0,t=0,1,2)$
解析：
$\because B_2$是标准布朗运动. $\therefore B_2\sim (0,2),P(B_2\le 0)=\frac{1}{2}$
$P(B_t\le 0,t=0,1,2)=P(B_1\le 0,B_2\le 0)$
由Brown运动性质：
(1) $B(t)\sim \mathcal{N}(0,t)$
(2) $B(t)-B(s)\sim \mathcal{N}(0,t-s),0\le s\le t$
(3) 独立增量性：$\forall 0=t_0<t_1<\cdots <t_m，B(t_1)-B(t_0)=B(t_1),B(t_2)-B(t_1),B(t_3)-B(t_2),\dots ,B(t_m)-B(t_{m-1})$增量两两独立且$B(t_{j+1})-B(t_j)\sim \mathcal{N}(0,t_{j+1}-t_j)$.
计算协方差，对于$0\le s<t$
$$cov(B(s),B(t))=\mathbb{E}[B(s)B(t)]-\mathbb{E}(B(s))\mathbb{E}(B(t))=\mathbb{E}[B(s)B(t)]$$
运用独立增量性求解
$$\begin{array}{rl}cov(B(s),B(t))& =\mathbb{E}[B(s)B(t)]=\mathbb{E}[B(s)(B(t)-B(s))+B^2(s)]\\ & =\mathbb{E}[B^2(s)]=\mathbb{V}[B^2(s)]\\ & =s\end{array}$$
计算$B(t_1),B(t_2),\dots ,B(t_m)$的方差-协方差矩阵
$$\begin{array}{rl}& \mathbb{E}[B(t_i)B(t_j)]=t_i\\ & \mathbb{E}[B^2(t_i)]=t_i\end{array}$$
方差-协方差矩阵$\Sigma$为
$$\Sigma ={\left(\begin{array}{cccc}{t}_1& t_1& \dots & t_1\\ t_1& t_2& \dots & t_2\\ ⋮& ⋮& ⋮& ⋮\\ t_1& t_2& \dots & t_m\end{array}\right)}_{m\times m}$$
$[B(t_1),B(t_2),\dots ,B(t_m){]}^T$联合pdf是联合正态分布(jointly normal distribution)，即
$$\left(\begin{array}{c}B(t_1)\\ B(t_2)\\ ⋮\\ B(t_m)\end{array}\right)\sim \mathcal{N}(0,\Sigma )$$
$m$维正态分布的pdf函数为
1 ( 2 π ) m 2 1 ∣ Σ ∣ 1 2 exp ⁡ { − 1 2 ( x − μ ) T Σ − 1 ( x − μ ) } \frac{1}{(2\pi)^\frac{m}{2}}\frac{1}{|\Sigma|^\frac{1}{2}}\exp\{-\frac{1}{2}(\mathbf{x}-\pmb{\mu})^T\Sigma^{-1}(\mathbf{x}-\pmb{\mu)}\} (2π)2m​1​∣Σ∣21​1​exp{
−21​(x−μ​μ​​μ)TΣ−1(x−μ)​μ)​​μ)}
考虑问题中的$B_1和B_2$，根据上述公式得到
$$\left(\begin{array}{c}{B}_1\\ B_2\end{array}\right)\sim \mathcal{N}(0,\Sigma )$$
其中
$$\Sigma =\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$$
得到联合pdf为
f ( b 1 , b 2 ) = 1 2 π exp ⁡ { − 1 2 [ b 1 , b 2 ] Σ − 1 [ b 1 , b 2 ] T } = 1 2 π exp ⁡ { − 1 2 ( 2 b 1 2 − 2 b 1 b 2 + b 2 2 ) } \begin{aligned} f(b_1, b_2)&=\frac{1}{2\pi}\exp\{-\frac{1}{2}[b1, b2]\Sigma^{-1}[b1, b2]^T\}\\ &=\frac{1}{2\pi}\exp\{-\frac{1}{2}(2b_1^2-2b_1b_2+b_2^2)\} \end{aligned} f(b1​,b2​)​=2π1​exp{
−21​[b1,b2]Σ−1[b1,b2]T}=2π1​exp{
−21​(2b12​−2b1​b2​+b22​)}​
计算概率
$$\begin{array}{rl}\mathbb{P}(B_1\le 0,B_2\le 0)& ={\int }_{-\infty }^0{\int }_{-\infty }^{0}f(b_1,b_2)d{b}_1{b}_2\\ & ={\int }_{-\infty }^0\frac{1}{\sqrt{2\pi }}{e}^{-\frac{b_1^2}{2}}d{b}_1{\int }_{-\infty }^0\frac{1}{\sqrt{2\pi }}{e}^{\frac{1}{2}(b_1-b_2{)}^2}d{b}_2\\ & ={\int }_{-\infty }^0\phi (b_1)(1-\Phi (b_1))d{b}_1\\ & ={\int }_{-\infty }^{0}1-\Phi (b_1)d\Phi (b_1)\\ & \stackrel{t=\Phi (b_1)}{=}{\int }_0^{1/2}1-tdt\\ & =\frac{3}{8}\end{array}$$

参考资料

Exercise For Ito Integral Jerry Xu

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