【练习】Option Pricing

基于ECON5020期末考试试题;

Option pricing;

Question:

Consider a multi-period market model $M=(B,S)$ with two assets: the saving account $B$ and the risky asset $S$. For a natural number $T$, consider the following model for the price of the risky asset $S$:

$$S_t = S_0 +Y_1,Y_2+...+Y_t,$$

where the random variables $Y_1,…,Y_T$ are independent and identically distributed under the real-world probability measure $\mathbb{P}$, specifically,

$$P(Y_1=+a)=.25,\ P(Y_1=-a)=.75$$

where $a>0$ is a strictly positive constant. We assume that the risk-free rate $r=0$ so that the savings account equals $B_t=1$ for every $t=0,1,…,T$.

1. Check whether the model $M=(B,S)$ is arbitrage-free and complete.
2. Find the option price process $\pi_t(X)$, $t=0,1,2,3$ and the replicating strategy $\varphi=(\varphi_0,\varphi_1)$

Answer:

Q1

我们可以看到，在给定前一个时刻的股票价格时，每个时间$t$都存在两个状态。分别是$+a$增长一个值，或者$-a$减少一个值，所以这是个$binomial tree$。为了判断模型是不是无套利的，我可以用定义：当$d<1+r<u$ 时无套利。其中$d=\frac{s_t}{s_{t-1}}=\frac{s_{t-1}-a}{s_{t-1}}=1-\frac{a}{s_{t-1}}$,$u=\frac{s_t}{s_{t-1}}=\frac{s_{t-1}+a}{s_{t-1}}=1+\frac{a}{s_{t-1}}$, 所以等式成立。

Q2

主要流程使用 $Backward induction$.

已知$T=3$,可以先画出股价的变动图。后续会往内补充数字：

t=0	t=1	t=2	t=3	value at T
			$S_0+3a$ , $\omega_1$	$\gets x_1$
		$S_0+2a$
	$S_0+a$		$S_0+a$ , $\omega_2$	$\gets x_2$
$S_0$		$S_0$
	$S_0-a$		$S_0-a$ , $\omega_3$	$\gets x_3$
		$S_0-2a$
			$S_0-3a$ , $\omega_4$	$\gets x_4$

因为$X$ 没给定明确的值，我们假设在时间$T$ 有四个状态$\{\omega_1,\omega_2,\omega_3,\omega_4\}$, 同时还有对应的价值$\{x_1,x_2,x_3,x_4\}$。对应$multi-period$ 的模型，需要有$self-financing$ 的过程:

$$\varphi_0^t+\varphi_1^tS_{t+1}=\varphi_0^{t+1}+\varphi_1^{t+1}S_{t+1}$$

对应最后时间$T$我们需要有以下式子:

$$\varphi_0^2+\varphi_1^2S_{t+1}=h(X)$$

然后开始从后向前推导:

From $t:2 \to 3$ where $A_1=\{\omega_1,\omega_2\}$

$$\begin{cases} \varphi_0^2+\varphi_1^2(S_0+3a)=x_1\\ \varphi_0^2+\varphi_1^2(S_0+a)=x_2 \end{cases} \to \begin{cases} \varphi_1^2·2a=x_1-x_2\\ \varphi_0^2=x_1-\varphi_1^2(S_0+3a) \end{cases}$$

we find that $(\varphi_0^2,\varphi_1^2)=(x_1-\frac{(S_0+3a)(x_1 - x_2)}{2a},\frac{x_1-x_2}{2a})$

and $\pi_2(A_1)=x_1-\frac{(S_0+3a)(x_1 - x_2)}{2a}+\frac{(S_0+2a)(x_1-x_2)}{2a}=x_1-\frac{x_1 - x_2}{2}=\frac{x_1 + x_2}{2}$

From $t:2 \to 3$ where $A_2=\{\omega_2,\omega_3\}$

$$\begin{cases} \varphi_0^2+\varphi_1^2(S_0+a)=x_2\\ \varphi_0^2+\varphi_1^2(S_0-a)=x_3 \end{cases} \to \begin{cases} \varphi_1^2·2a=x_2-x_3\\ \varphi_0^2=x_2-\varphi_1^2(S_0+a) \end{cases}$$

we find that $(\varphi_0^2,\varphi_1^2)=(x_2-\frac{(S_0+a)(x_2 - x_3)}{2a},\frac{x_2-x_3}{2a})$

and $\pi_2(A_2)=x_2-\frac{(S_0+a)(x_2 - x_3)}{2a}+\frac{S_0(x_2-x_3)}{2a}=x_2-\frac{x_2 - x_3}{2}=\frac{x_2 + x_3}{2}$

From $t:2 \to 3$ where $A_3=\{\omega_3,\omega_4\}$

$$\begin{cases} \varphi_0^2+\varphi_1^2(S_0-a)=x_3\\ \varphi_0^2+\varphi_1^2(S_0-3a)=x_4 \end{cases} \to \begin{cases} \varphi_1^2·2a=x_3-x_4\\ \varphi_0^2=x_3-\varphi_1^2(S_0-a) \end{cases}$$

we find that $(\varphi_0^2,\varphi_1^2)=(x_3-\frac{(S_0-a)(x_3 - x_4)}{2a},\frac{x_3-x_4}{2a})$

and $\pi_2(A_3)=x_3-\frac{(S_0-a)(x_3 - x_4)}{2a}+\frac{(S_0-2a)(x_3-x_4)}{2a}=x_2-\frac{x_3 - x_4}{2}=\frac{x_3 + x_4}{2}$

From $t:1 \to 2$ where $A_u=\{A_1,A_2\}$

$$\begin{cases} \varphi_0^1+\varphi_1^1(S_0+2a)=\frac{x_1 + x_2}{2}\\ \varphi_0^1+\varphi_1^1S_0=\frac{x_2 + x_3}{2} \end{cases} \to \begin{cases} \varphi_1^1·2a=\frac{x_1 - x_3}{2}\\ \varphi_0^1=\frac{x_1 + x_2}{2}-\varphi_1^1(S_0+2a) \end{cases}$$

we find that $(\varphi_0^1,\varphi_1^1)=(\frac{x_1 + x_2}{2}-\frac{(S_0+2a)(x_1 - x_3)}{4a},\frac{x_1-x_3}{4a})$

and $\pi_1(A_u)=\frac{x_1 + x_2}{2}-\frac{(S_0+2a)(x_1 - x_3)}{4a}+\frac{(S_0+a)(x_1-x_3)}{4a}=\frac{x_1 + x_2}{2}-\frac{x_1 - x_3}{4}=\frac{x_1 + 2x_2-x_3}{4}$

From $t:1 \to 2$ where $A_d=\{A_2,A_3\}$

$$\begin{cases} \varphi_0^1+\varphi_1^1S_0=\frac{x_2 + x_3}{2}\\ \varphi_0^1+\varphi_1^1(S_0-2a)=\frac{x_3 + x_4}{2} \end{cases} \to \begin{cases} \varphi_1^1·2a=\frac{x_2 - x_4}{2}\\ \varphi_0^1=\frac{x_2 + x_3}{2}-\varphi_1^1S_0 \end{cases}$$

we find that $(\varphi_0^1,\varphi_1^1)=(\frac{x_2 + x_3}{2}-\frac{S_0(x_2 - x_4)}{4a},\frac{x_2-x_4}{4a})$

and $\pi_1(A_u)=\frac{x_2 + x_3}{2}-\frac{S_0(x_2 - x_4)}{4a}+\frac{(S_0-a)(x_2-x_4)}{4a}=\frac{x_2 + x_3}{2}-\frac{x_2 - x_4}{4}=\frac{x_2 + 2x_3-x_4}{4}$

From $t:0 \to 1$ where $A_0=\{A_u,A_d\}$

$$\begin{cases} \varphi_0^0+\varphi_1^0(S_0+a)=\frac{x_1 + 2x_2-x_3}{4}\\ \varphi_0^0+\varphi_1^0(S_0-a)=\frac{x_2 + 2x_3-x_4}{4} \end{cases} \to \begin{cases} \varphi_1^0·2a=\frac{x_1 +x_2 -x_3 -x_4}{4}\\ \varphi_0^0=\frac{x_1 + 2x_2-x_3}{4}-\varphi_1^0(S_0+a) \end{cases}$$

we find that $(\varphi_0^0,\varphi_1^0)=(\frac{x_1 + 2x_2-x_3}{4}-\frac{(S_0+a)(x_1 +x_2 -x_3 -x_4)}{8a},\frac{x_1 +x_2 -x_3 -x_4}{8a})$

$$\begin{align}\pi_0(A_0)&=\frac{x_1 + 2x_2-x_3}{4}-\frac{(S_0+a)(x_1 +x_2 -x_3 -x_4)}{8a}+\frac{S_0(x_1 +x_2 -x_3 -x_4)}{8a}\\ &=\frac{x_1 + 2x_2-x_3}{4}-\frac{x_1 +x_2 -x_3 -x_4}{8}=\frac{x_1 + 3x_2-3x_3 - x_4}{8}\end{align}$$

总结

题目的难度不大，主要在计算过程中比较繁杂的化简，选择性化简是比较重要的，我们可以发现$(S_0+a)$通常都作为一个整体出现，所以无需拆分。同时因为这道题特别的对仗工整，我们可以发现在同一时间$t$下的计算特别相似，所以有时可以采用直接替换数字。当计算时，如果出现了不友好的数字，比如一些没有规律的，与上下式子不对应的答案，可以检查过程中是否出现，少除，错减之类的。

Matlab代码

syms s0 a x1 x2 x3 x4
r = 0; u = a; d = -a; S0 = s0;imax = 3;
SS = sym(zeros(imax+1,imax+1));
SS(1,1) = S0;
for t = 2:imax+1 % 对应股价变化
    for n = 1:t
        SS(n,t) = S0+u*(t-n)+d*(n-1);
    end
end
payoff = [x1;x2;x3;x4]; % 在T时的价值
PayOffMatr = sym(zeros(imax+1,imax+1));
PayOffMatr(:,imax+1) = payoff;
phiMatr = sym(zeros(imax,imax,2));
for j=imax:-1:1
    for k = 1:j
        pi1=PayOffMatr(k,j+1);
        pi2=PayOffMatr(k+1,j+1);
        s1=SS(k,j+1);
        s2=SS(k+1,j+1);
        phiMatr(k,j,2)=simplify((pi1-pi2)/(s1-s2)); % phi1
        phiMatr(k,j,1)=simplify((pi1-phiMatr(k,j,2)*s1)/(1+r)); % phi0
        PayOffMatr(k,j)=simplify(phiMatr(k,j,1)+phiMatr(k,j,2)*SS(k,j)); % pi
    end
end