【练习】BS Model To BS PDE

基于B站 up Jerry Xu 的视频《Ito积分练习题(Part 1)》；

从 BS model 反向推符合 PDE

Question

For a European Call option at time T, with strike price K.

$$c(t,x) = xN(d_+)- Ke^{-r(T-t)}N(d_-)$$

where

$$d_+=\frac{ln\frac{x}{K}+(r + \frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}}$$ $$d_-=\frac{ln\frac{x}{K}+(r - \frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}}$$

Show that this function satisfies the BS-PDE

$$rc(t,x)=c_t+\frac12\sigma^2x^2c_{xx}+rxc_x$$

with

$$\lim_{t\to T}c(t,x)=max\{x-K,0\},x>0,x\ne K$$ $$\lim_{x\to 0}c(t,x)=0,\ \lim_{x\to \infty}[c(t,x)-(x-e^{r(T-t)}K)=0,\ 0\leq t<T$$

Step

(1) Show that $Ke^{-r(T-t)}N’(d_-)=xN’(d_+)$;

$$N'(x)=\frac1{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$

原问题可以转换为

$$e^{\frac{d_+^2-d_-^2}{2}}=\frac xKe^{r(T-t)}$$

where

$$d_+^2-d_-^2=\frac{[ln\frac{x}{K}+(r + \frac{1}{2}\sigma^2)(T-t)]^2-[ln\frac{x}{K}+(r - \frac{1}{2}\sigma^2)(T-t)]^2}{\sigma^2(T-t)}$$

Set $A=ln\frac xK+r(T-t)$, $B=\frac12\sigma^2(T-t)$

$$(A+B)^2-(A-B)^2=4AB$$ $$d_+^2-d_-^2=\frac{4(ln\frac xK+r(T-t))\frac12\sigma^2(T-t)}{\sigma^2(T-t)}=2(ln\frac xK+r(T-t))$$ $$e^{\frac{d_+^2-d_-^2}{2}}=e^{ln\frac xK+r(T-t)}=\frac xKe^{r(T-t)}$$

(2) Show that Delta $c_x(t,x)=N(d_+)$

$$c_x(t,x)=N(d_+)+xN'(d_+)\frac{\partial d_+}{\partial x}-ke^{-r(T-t)}N'(d_-)\frac{\partial d_-}{\partial x}$$ $$\frac{\partial d_+}{\partial x}=\frac{\partial d_-}{\partial x}=\frac{1}{\sigma \sqrt{T-t}}\frac{K}{x}\frac1K=\frac{1}{x\sigma \sqrt{T-t}}$$ $$c_x(t,x)=N(d_+)+\frac{xN'(d_+)-ke^{-r(T-t)}N'(d_-)}{x\sigma \sqrt{T-t}}=N(d_+)+0$$

行权价值附近，斜率最大，Delta 约等于 1/2

$$c_{xx}=\frac{\partial N(d_+)}{\partial x}=\frac{N'(d_+)}{x\sigma \sqrt{T-t}}$$

(3) Show $c_t(t,x)=-rKe^{-r(T-t)}N(d_-)-\frac{\sigma x}{2\sqrt{T-t}}N’(d_+)$

$$C_t(t,x)=xN'(d_+)\frac{\partial d_+}{\partial t}- rKe^{-r(T-t)}N(d_-)- Ke^{-r(T-t)}N'(d_-)\frac{\partial d_-}{\partial t}$$ $$\begin{align} \frac{\partial d_+}{\partial t}&=\frac{-(r + \frac{1}{2}\sigma^2)\sigma \sqrt{T-t}+\frac12\sigma (T-t)^{-1/2}(ln\frac{x}{K}+(r + \frac{1}{2}\sigma^2)(T-t))}{\sigma^2(T-t)}\\ &=\frac{-\frac12(r + \frac{1}{2}\sigma^2)\sigma \sqrt{T-t}+\frac12\sigma (T-t)^{-1/2}ln\frac{x}{K}}{\sigma^2(T-t)}\\ &=-\frac{(r + \frac{1}{2}\sigma^2)}{2\sigma \sqrt{T-t}}+\frac{ln\frac{x}{K}}{2\sigma(T-t)^{3/2}}\\ &=\frac{1}{2\sigma\sqrt{T-t}}[\frac{ln\frac{x}{K}}{T-t}-(r + \frac{1}{2}\sigma^2)] \end{align}$$ $$\frac{\partial d_-}{\partial t}=\frac{1}{2\sigma\sqrt{T-t}}[\frac{ln\frac{x}{K}}{T-t}-(r - \frac{1}{2}\sigma^2)]$$ $$\begin{align} C_t(t,x)&=- rKe^{-r(T-t)}N(d_-)+x\frac{N'(d_+)}{2\sigma\sqrt{T-t}}[\frac{ln\frac{x}{K}}{T-t}-(r + \frac{1}{2}\sigma^2)]- Ke^{-r(T-t)}\frac{N'(d_-)}{2\sigma\sqrt{T-t}}[\frac{ln\frac{x}{K}}{T-t}-(r - \frac{1}{2}\sigma^2)]\\ &=- rKe^{-r(T-t)}N(d_-)-x\frac{N'(d_+)}{2\sigma\sqrt{T-t}}\frac{1}{2}\sigma^2- Ke^{-r(T-t)}\frac{N'(d_-)}{2\sigma\sqrt{T-t}}\frac{1}{2}\sigma^2\\ &=- rKe^{-r(T-t)}N(d_-)-\frac{\sigma x N'(d_+)}{2\sqrt{T-t}} \end{align}$$

(4) Prove that satisfies

$$\begin{align} &c_t+\frac12\sigma^2x^2c_{xx}+rxc_x\\ =&- rKe^{-r(T-t)}N(d_-)-\frac{\sigma xN'(d_+)}{2\sqrt{T-t}}+\frac{\sigma xN'(d_+)}{2\sqrt{T-t}}+rxN(d_+)\\ =& rxN(d_+)- rKe^{-r(T-t)}N(d_-)=rc(t,x) \end{align}$$

(5) Calculate $\lim_{t\to T}d_+$ and $\lim_{t\to T}d_-$ for $x>0$, $x\ne K$.

$$\lim_{t\to T}d_+=\lim_{t\to T}d_-=\lim_{t\to T}\frac{ln\frac{x}{K}+(r + \frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}}=\frac{ln\frac{x}{K}}0$$ $$\lim_{t\to T}c(t,x)=\lim_{t\to T}[xN(d_+)- Ke^{-r(T-t)}N(d_-)]$$

If $x>K$：

$$\lim_{t\to T}[xN(d_+)- Ke^{-r(T-t)}N(d_-)]=xN(+\infty)- KN(+\infty)=x-K$$

If $x<K$：

$$\lim_{t\to T}[xN(d_+)- Ke^{-r(T-t)}N(d_-)]=xN(-\infty)- KN(-\infty)=0$$

(6) Calculate $\lim_{x\to 0}d_+$ and $\lim_{x\to 0}d_-$ for $0\leq t<T$.

$$\lim_{x\to 0}d_+=\lim_{x\to 0}d_-=\lim_{x\to 0}\frac{ln\frac{x}{K}+(r + \frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}}=\frac{-\infty}{\sigma \sqrt{T-t}}$$ $$\lim_{x\to 0}c(t,x)=xN(-\infty)- Ke^{-r(T-t)}N(-\infty)=0$$

(7) Calculate $\lim_{x\to \infty}d_+$ and $\lim_{x\to \infty}d_-$ for $0\leq t<T$.

$$\lim_{x\to \infty}d_+=\lim_{x\to \infty}d_-=\frac{+\infty}{\sigma \sqrt{T-t}}$$ $$\lim_{x\to 0}c(t,x)=xN(+\infty)- Ke^{-r(T-t)}N(+\infty)=x- Ke^{-r(T-t)}$$

Hint for (7)：From the formula of $d_+$, we can obtain that

$$x=Kexp(\sigma\sqrt{T-t}d_+-(T-t)(r+\frac12\sigma^2))$$