Finance Project

Cox-Ross-Rubinstein Model

Let \(t_{0}=0\), \(t_{N}=T\) and \(t_{i+1}-t{i}=\delta\) with \((t_{i}) _{0 \leqslant i \leqslant N}, \forall i \in \{1,...,N-1\}\) where \(\delta=\frac{T}{N}\). We define for the rest of the subject the following variables :

• The price of the risk-free asset : \(S_{t_{i}}^{0}=(1+r_{N})^{i}, \forall i \in \{0,...,N\}\)
• The price of the risk asset :
  • \(S_{0}^{(N)}=s > 0\)
  • \(S_{t_{i}}^{(N)}=T_{i}^{(N)}S_{t_{i-1}}^{(N)}, \forall i \in \{1,...,N\}\)

Question 1

Let \(\mathbb{Q}\) be the risk-neutral probability and \(\mathcal{F}_{k}=\sigma(S_{t_{i}})\). Define \(q_{N}=\mathbb{Q}[T_{1}^{(N)}=1+\textit{h}_{N}]\) On one side we know that \(S_{t_1}^{(N)}\) is a martingales under \(\mathbb{Q}\) and with the definition \(\tilde{S}_{t_1}^{(N)} = \frac{s\times T_{1}^{(N)}}{1+r_N}\) and \(\mathbb{E}[\tilde{S}_{t_1}^{(N)}|\mathcal{F}_0]=S_0^{(N)}=s\) And on the other side \(T_1^{(N)}(\Omega)=\{1+b_N,1+h_N\}\) Then \[\forall s>0, s=\frac{1}{1+r_N}(s(1+h_N)q_N + s(1+b_N)(1-q_N))\] Finally \[q_N=\frac{r_N - b_N}{h_N - b_N}\]

Soit \(\mathbb{Q}\) tq \(\mathbb{E}[\frac{S_{t_{i+1}}}{S_{t_{i}}}/F_{t_{i}}]=\mathbb{E}[\frac{S^{o}_{t_{i+1}}}{S_{t_{i}}}/F_{t_{i}}]=\mathbb{E}[\frac{T_{i+1}S_{ti}}{S_{ti}}/\mathcal{F}_{i}]\)

Let \(\frac{1}{(1+r_{n})^{N}}\mathbb{E}[f(S_{t_{N}})]\) donc \(S_{tN}=s(1+h_{N})^{X}(1+b_{N})^{N-X}\) donc \(S_{t_{0}}=s\) Ainsi \(X~\mathcal{B}(N,q_{N})\)

Question 2

Let \(p_{(N)}=\frac{1}{(1+r_N)^N}\mathbb{E}[f(S_{t_N}^{(N)})]\) We see that : \(\forall k \in \{1,...,N\}, S_{t_N}^{(N)}= (\prod\limits_{i=1}^{N} T_i^{(N)})\times s\) Moreover: \[ \mathbb{E}_{\mathbb{Q}}[f(S_{t_N}^{(N)}]=\sum\limits_{\alpha\in S_{t_N}^{(N)}(\Omega)}f(\alpha)\mathbb{Q}(S_{t_N}^{(N)}=\alpha)\] But the \((T_i^{(N)})_{i\in\{1,...,N\}}\) are i.i.d and are equals to either \(( 1+h_N)\) or \((1+b_N)\) with z probability of, resp. , \((q_N)\) et \((1-q_N)\) then with the definition we have: \(\forall\alpha\in S_{t_N}^{(N)}(\Omega), \exists (i,j)\in\mathbb{N}^2 | i+j = N\) et \(\alpha=(1+h_N)^i(1+b_N)^j s\), \[\mathbb{E}_{\mathbb{Q}}[f(S_{t_N}^{(N)})]=\sum\limits_{(i,j)\in\mathbb{N}^2}^{i+j=N} f((1+h_N)^i(1+b_N)^j s)q_N^i (1-q_N)^j\] \[= \sum\limits_{i=0}^N \dbinom{N}{i} ((1+h_N)^i(1+b_N)^{N-i}q_N^i(1-q_N)^{N-i}\] Finally: \[p_{(N)}=\frac{1}{(1+r_N)^N}\times\sum\limits_{i=0}^N \dbinom{N}{i} ((1+h_N)^i(1+b_N)^{N-i}q_N^i(1-q_N)^{N-i}\]

First pricer

Question

[language=Python,caption={Implémentation de Price1}]

from scipy import special

def price1(N,rN,hN,bN,s,f):
    qN = (rN-bN)/(hN-bN)
    return 1.0/((1.0+rN)**N) * sum([special.comb(N,i)*f(s*((1+hN)**i)*((1+bN)**(N-i)))*(qN**i)*((1-qN)**(N-i)) for i in range(N)])
    
def f(x):
    return max(x-90,0)
    
print(price1(30,0.01,0.05,-0.05,100,f))

The output for the following value $s=100$, $h_N=0.05$, $b_N=-0.05$, $r_N=0.01$, $N=30$ et $f(x)=max(x-90,0)$

Second Pricer

Question

[language=Python,caption=Implémentation de Price2]

import numpy as np

def price2(N,rN,hN,bN,s,f):
    cst=1/(1+rN)
    qN=(rN-bN)/(hN-bN)
    M =np.zeros((N+2,N+2),float)
    for i in range(1,N+1):
        M[i,N+1]=f(s*(1+hN)**(N-i+1)*(1+bN)**(i-1)) 
    for j in range(N,0,-1):
        for k in range(1,j+1):
            M[k,j]=cst*(M[k,j+1]*qN+M[k+1,j+1]*(1-qN)) 
    
    return M[1,1]

def f(x):
    return max(x-90,0)
    
print(price2(30,0.01,0.05,-0.05,100,f))

Question

Comparison

Question

[language=Python,caption=Comparaison de Price1 et Price2]

import random

def compare(rN,hN,bN,s,f):
    N=random.randint(5,15)
    print(N)
    print(price1(N,rN,hN,bN,s,f))
    print(price2(N,rN,hN,bN,s,f))
    

The financial cover

On the final date, our portfolio must verify : \[\alpha_{N-1}(S_{t_{N-1}}^{(N)})S_{t_N}^{(N)} +\beta_{N-1}(S_{t_{N-1}}^{(N)})S_{t_N}^{0}=f(S_{t_N}^{(N)})\]

In this section, we suppose that at the \(T\) date, we are two states:

• \(S_{t_N}^{(N)}=(1+h_N)S_{t_{N-1}}^{(N)}\)
• \(S_{t_N}^{(N)}=(1+b_N)S_{t_{N-1}}^{(N)}\)

We have the following system : \[\alpha_{N-1}(S_{t_{N-1}}^{(N)})(1 + h_N)S_{t_{N-1}}^{(N)} +\beta_{N-1}(S_{t_{N-1}}^{(N)})S_{t_N}^{0}=f((1+h_N)S_{t_{N-1}}^{(N)})\]

\[\alpha_{N-1}(S_{t_{N-1}}^{(N)})(1+b_N)S_{t_{N-1}}^{(N)} +\beta_{N-1}(S_{t_{N-1}}^{(N)})S_{t_N}^{0}=f((1+b_N)S_{t_{N-1}}^{(N)})\]

Which gives us :

\[\alpha_{N-1}(S_{t_{N-1}}^{(N)})S_{t_{N-1}}^{(N)}(h_N - b_N)=f((1 + h_N)S_{t_{N-1}}^{(N)}) - f((1 + b_N)S_{t_{N-1}}^{(N)})\] \[\beta_{N-1}(S_{t_{N-1}}^{(N)})S_{t_N}^{0}(b_N - h_N)=f((1+b_N)S_{t_{N-1}}^{(N)})(1+h_N) - f((1+h_N)S_{t_{N-1}}^{(N)})(1+b_N)\]

Finally, we get :

\[\alpha_{N-1}(S_{t_{N-1}}^{(N)}) = \frac{f((1 + h_N)S_{t_{N-1}}^{(N)}) - f((1 + b_N)S_{t_{N-1}}^{(N)})}{S_{t_{N-1}}^{(N)}(h_N - b_N)}\] \[\beta_{N-1}(S_{t_{N-1}}^{(N)}) = \frac{f((1+b_N)S_{t_{N-1}}^{(N)})(1+h_N) - f((1+h_N)S_{t_{N-1}}^{(N)})(1+b_N)}{(1+r_N)^N(b_N - h_N)}\]

Question

\[\alpha_{k-1}(S_{t_{k-1}}^{(N)})(1 + h_N)S_{t_{k-1}}^{(N)} +\beta_{k-1}(S_{t_{k-1}}^{(N)})S_{t_k}^{0}=v_k((1+h_N)S_{t_{k-1}}^{(N)})\]

\[\alpha_{k-1}(S_{t_{k-1}}^{(N)})(1+b_N)S_{t_{k-1}}^{(N)} +\beta_{k-1}(S_{t_{k-1}}^{(N)})S_{t_k}^{0}=f((1+b_N)S_{t_{k-1}}^{(N)})\]

By the same method we obtain :

\[\alpha_{k-1}(S_{t_{N-1}}^{(N)}) = \frac{v_k((1 + h_N)S_{t_{k-1}}^{(N)}) - v((1 + b_N)S_{t_{k-1}}^{(N)})}{S_{t_{k-1}}^{(N)}(h_N - b_N)}\]

And,

\[\beta_{k-1}(S_{t_{k-1}}^{(N)}) = \frac{v((1+b_N)S_{t_{k-1}}^{(N)})(1+h_N) - v((1+h_N)S_{t_{k-1}}^{(N)})(1+b_N)}{(1+r_N)^N(b_N - h_N)}\]

Question

[language=Python,caption=Comparison of Price1 and Price2]

def couverture(N,s,rN,hN,bN,f):
    Sh = price2(N,rN,hN,bN,(1+hN)*s,f);
    Sb = price2(N,rN,hN,bN,(1+bN)*s,f);
    alpha =  (Sh-Sb)/(s*(hN-bN));
    beta = (Sh(1+bN)-Sb(1+hN))/((1+rN)**N *(bN-hN));
    print(alpha)
    print(beta)

Black-Scholes’ Model

The modal

Îto’s formula is given by :

Let g be a function in $C^2$. Then the differential of the processus $g(S_t)$ satisfies:
$$dg(S_t)=g^{'}(S_t)dS_t + \frac{|\sigma S_t|^2}{2}g^{''}(S_t)dt$$

Question

We apply the previous lemma with \(S_t \longrightarrow ln(S_t)\). Then, we have : \[dln(S_t) = \frac{1}{S_t}dS_t + \frac{|\sigma S_t|^2}{2}(-\frac{1}{S_t^2})dt\] But: $ dS_t = S_t(rdt+dB_t)$ Then : \[ dln(S_t) = rdt + \sigma dB_t - \frac{\sigma^2}{2}dt\] Then : \[ \frac{dln(S_t)}{dt} = r + \sigma\frac{B_t}{dt} - \frac{\sigma^2}{2}\] An indefinite integration on \(t\) and an evaluation in \(0\) gives us : \[ln(S_t) = rt +\sigma B_t - \frac{\sigma^2}{2}t + c^{te} \text{ with } c^{te}= ln(s)\] by applying the exponential function (continous), we finally get : \[ S_t = \exp(rt + \sigma B_t - \frac{\sigma^2}{2}t ) \times s\]

A pricer with Monte-Carlo Method

The price formula of an asset is given by : \(p:=\mathbb{E}[e^{-rT}f(S_t)]\)

Let \((\xi_i)_{1\leqslant i \leqslant n}\) be a sequence of random variables on some probability space \((\Omega,\mathcal{F},\mathbb{P})\). They are supposed \(i.i.d\) and they follow a \(normal\) \(distribution\) with parameter \(m=0\), \(\sigma=1\). We define : \[\hat{p}_{(N)}:=\frac{1}{n}\sum_{i=1}^{n}e^{-rT}f(s\times\exp((r-\frac{\sigma^2}{2})T+\sigma\sqrt{T}\xi_i))\]

Question

[language=Python,caption=Price3]

def price3(n, s, r, o, T, f):
    return 1.0 / n * exp(-r * T) * sum([f(s * exp((r - o*o/2.0)*T + o * sqrt(T) * ksi)) for ksi in rnorm(0.0, 1.0, n)])

Question

{Plot of Price3}

Let start with two results quite useful for what’s to come. Let \(X_1,X_2,...\) be a sequence of random variables on some probability space \((\Omega,\mathcal{F},\mathbb{P})\). They are supposed \(identically\) \(distributed\). Define \(S_n=X_1+ ... +X_n\). The \(Strong\) \(Law\) \(of\) \(Large\) \(Numbers\) specifies :

If the $X_n$ are independent and identically distributed and have finite mean then 
$$ \mathbb{P}[\lim_{n}n^{-1}S_n=\mathbb{E}[X_1]]=1$$

If the \(X_n\) are independent and let \(f\) be a measurable function then the \(f(X_n)\) are independent.

Question

Let \(X_1,X_2,...\) be a sequence of random variables define by \((e^{-rT}f(s\times\exp((r+\frac{\sigma^2}{2})T + \sigma\sqrt{T}\xi_i)))_{i\in\{1,...,n\}}\). We have with the hypotheses that \((\xi_i)_{i\in\{1,...,\}}\) are \(i.i.d\). Therefore the \(X_n\) are i.i.d.
Moreover we should check if \(\mathbb{E}[X_1]\) exits but if we do obtain p in the end then everything will be demonstrated. We can apply the \(Strong\) \(Law\) \(of\) \(Large\) \(Number\) which gives us : \[\mathbb{P}[\lim_{n}\frac{1}{n}\sum_{i=0}^{n}X_i = \mathbb{E}[X_1]]=1 \text{ i.e } \frac{1}{n}\sum_{i=0}^{n}X_i \underset{p.s}{\longrightarrow} \mathbb{E}[X_1]\] We shall now show that \(\mathbb{E}[X_1]=p\). Let start from \(\mathbb{E}[X_1]\) We know that \(B\) is a brownien movement so \(B_t - B_s \sim \mathcal{N}(0,t-s)\). Then with \(s=0\) comes \(B_T - B_s \sim \mathcal{N}(0,T-s)\). But \(B_0=0\) and we divide by \(\sqrt{T}\) to obtain \(\frac{B_T}{\sqrt{T}}\sim\mathcal{N}(0,1)\). Therefore we define \(\xi_1=\frac{B_T}{\sqrt{T}}\). And finally we do have by applying question 11. result (and \(\mathbb{E}[X_1]\) does exist) \[\mathbb{E}[X_1]= \mathbb{E}[e^{-rT}f(s\times\exp((r-\frac{\sigma^2}{2})T + \sigma B_T] = \mathbb{E}[e^{-rT}f(S_T)]=p\] To conclude : \[\hat{p}_{(N)} \underset{p.s}{\longrightarrow} p\]

Pricer by a close formula

Question

[language=Python,caption=Put function]

def put(s,r,sigma,T,K):
    d = 1/(sigma*sqrt(T))*(log(s/K)+(r+sigma**2/2)*T);
    va = norm()
    F1 = va.cdf(-d+sigma*sqrt(T))
    F2 = va.cdf(-d)
    return -s*F2+K*exp(-r*T)*F1

Question

Question

Question

Question

We see that \(Price2\) seems to converge to the value of \(put\).

Black-Scholes’ PDE

All along this section, we will consider the Black-Scholes’ equation given by : \[ \frac{\partial P}{\partial t} + rS \frac{\partial P}{\partial S} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 P}{\partial S^2} = rP \]

A few notation :

Finite difference explicite method

\[\frac{\partial P_i^n}{\partial T} \approx \frac{P_i^{n+1}-P_i^n}{\Delta T}\] \[\frac{\partial P_i^n}{\partial s} \approx \frac{P_{i}^{n}-P_{i-1}^n}{\Delta s}\] \[\frac{\partial^2 P_i^n}{\partial s^2} \approx \frac{P_{i+1}^{n}-2P_i^n+P_{i-1}^n}{\Delta s^2}\]

Therefore, we have : \[ P_i^{n+1} = rP_i^{n+1}\Delta T + P_i^n - rS\frac{\Delta T}{2\Delta s}(P_{i}^n - P_{i-1}^n) - \frac{1}{2}\sigma^2S^2\frac{\Delta T}{\Delta s^2}(P_{i+1}^n-2P_i^n+P_{i-1}^n)\] \[ (1-r\Delta T)P_i^{n+1} = (\frac{rS\Delta T}{\Delta s}-\frac{\sigma^2S^2\Delta T}{2\Delta s^2})P_{i-1}^n (-rS\Delta T +1+ \frac{\sigma^2S^2\Delta T}{\Delta s^2})P_i^n +( - \frac{\sigma^2S^2\Delta T}{2\Delta s^2})P_{i+1}^n\]

Finite difference implicite method

\[\frac{\partial P_i^{n}}{\partial T} \approx \frac{P_i^{n+1}-P_i^n}{\Delta T}\] \[\frac{\partial P_i^{n}}{\partial s} \approx \frac{P_{i}^{n+1}-P_{i-1}^{n+1}}{\Delta s}\] \[\frac{\partial^2 P_i^{n}}{\partial s^2} \approx \frac{P_{i+1}^{n+1}-2P_i^{n+1}+P_{i-1}^{n+1}}{\Delta s^2}\]

Therefore, we have:

\[P_i^n=(\frac{-rS\Delta T }{\Delta s}+\frac{\sigma^2S^2\Delta T}{2\Delta s^2})P_{i-1}^{n+1} + (-r\Delta T + 1 - \frac{\sigma^2S^2\Delta T}{\Delta s^2}-\frac{rS\Delta T}{\Delta S})P_i^{n+1} +(\frac{\sigma^2S^2\Delta T}{2\Delta s^2})P_{i+1}^{n+1}\]

Crank-Nicholson method

The Crank-Nicholson method result of taking the mean of the two previous interpretation (implicite and explicite). It gives us : \[(I_d + \Delta T B)P^{n+1}=(I_d +\Delta T A)P^n +\frac{1}{2}(Cond_1 + Cond_2)\]

Let assume that \((I_d + \Delta T B)\) is inversible then we should come with the following equation : \[P^{n+1}=(I_d + \Delta T B)^{-1}[(I_d +\Delta T A)P^n +\frac{1}{2}(Cond_1 + Cond_2)]\]

Linear algebra Redefinition

We now rewrite this equation using linear algebra:

\[BP^{n+1} = AP^n + Cond_i,\]

\[ \text{ where } A = \left(\begin{array}{cccccc}A_{0,0} & A_{0,1} & 0 & .. & .. & 0 \\A_{1,0} & .. & .. & .. & & : \\0 & .. & .. & .. & .. & 0 \\: & .. & .. & .. & .. & 0 \\: & & .. & .. & .. & A_{M,M+1} \\0 & .. & .. & 0 & A_{M+1,M} & A_{M+1,M+1}\end{array}\right)\] \[\text{ and } B = \left(\begin{array}{cccccc}B_{0,0} & B_{0,1} & 0 & .. & .. & 0 \\B_{1,0} & .. & .. & .. & & : \\0 & .. & .. & .. & .. & : \\: & .. & .. & .. & .. & 0 \\: & & .. & .. & .. & A_{M,M+1} \\0 & .. & .. & 0 & B_{M+1,M} & B_{M+1,M+1}\end{array}\right) \] \[\text{ and } Cond_i=\left(\begin{array}{c}P(t_i,s_0) \\0 \\: \\: \\0 \\P(t_i,s_{M+1})\end{array}\right)\]

For the explicite version, we get: \(\forall j\in \{1,...,M\},\) \[A_{j,j+1}=\frac{\sigma^2s_j^2}{2\Delta s^2} \text{, } A_{j,j}=\frac{1}{\Delta T} + \frac{rs_j}{\Delta s} + \frac{\sigma^2s_j^2}{\Delta s^2} \text{ and } A_{j,j+1}=-\frac{rs_j}{\Delta s} - \frac{\sigma^2s_j^2}{2\Delta s^2}\] and \[ B_{j,j-1}=0 \text{, } B_{j,j}=-r +\frac{1}{\Delta T} \text{ and } B_{j,j+1}=0\]

For the implicite version, it is: \(\forall j\in \{1,...,M\},\) \[A_{j,j-1}=0 \text{, } A_{j,j}=\frac{1}{\Delta T} \text{ and } A_{j,j+1}=0\] and \[ B_{j,j-1}=\frac{\sigma^2 s_j^2}{2\Delta s^2} \text{, } B_{j,j}=-r + \frac{1}{\Delta T} - \frac{rs_j}{\Delta s} - \frac{\sigma^2s_j^2}{\Delta s^2} \text{ and } B_{j,j+1}=\frac{rs_j}{\Delta s} + \frac{\sigma^2s_j^2}{\Delta s^2}\]

And finally, for Crank-Nicholson: \(\forall j\in \{1,...,M\},\) \[A_{j,j-1}=-\frac{\sigma^2s_j^2}{4\Delta s^2} \text{, } A_{j,j}=\frac{1}{\Delta T} + \frac{rs_j}{2\Delta s} +\frac{\sigma^2s_j^2}{2\Delta s^2} \text{ and } A_{j,j+1}= -\frac{rs_j}{2\Delta s} - \frac{\sigma^2s_j^2}{4\Delta s^2}\] and \[B_{j,j-1}= \frac{\sigma^2s_j^2}{4\Delta s^2} \text{, } B_{j,j} = -r + \frac{1}{\Delta T} - \frac{rs_j}{2\Delta s} -\frac{\sigma^2 s_j^2}{2\Delta s^2} \text{ and } B_{j,j+1}=\frac{rs_j}{2\Delta s} + \frac{\sigma^2s_j^2}{4\Delta s^2}\]

For all three preceding situation, we initialize : \[A_{0,0}=A_{0,1}=A_{M+1,M}=A_{M+1,M+1}=B_{0,0}=B_{0,1}=B_{M+1,M}=B_{M+1,M+1}=0\]

CFL condition

Pour la méthode explicit, le schéma est stable si le pas de temps est de l’ordre du carré du pas d’espace. Autrement dit, si : \[\Delta t\approx \Delta s^2\]