Jupyter Notebook Binomial_Lattice_Calls_Puts Last Checkpoint: 09/27/2019 (autosaved) Current Kernel Logo Python 3 File Edit View Insert Cell Kernel Widgets Help

Binomial lattice Program ​ ​ import numpy as np from math import exp, sqrt, log s0 = 20; strike = 20;tp = 1.0;rf = 0.06;q= 0.04;sigma = 0.30;nperiods = 3 Ocall = True; Otype = 'E' def make_stuff(tp,nperiods, sigma, rf, q): dt= tp/nperiods u = exp(sigmasqrt(dt)) d = 1/u disc = exp(-rfdt) p = (exp((rf-q)*dt) -d)/(u-d) pbar = 1-p return u, d, p, pbar, disc ​ u, d, p, pbar, disc = make_stuff(tp,nperiods, sigma, rf, q) print(u, d, p, disc) 1.1891099436471448 0.8409651313930471 0.4760197513442101 0.9801986733067553 def make_stock_lattice(s0, u, d, nperiods): S = np.zeros((nperiods+1, nperiods+1)) S[0,0] = s0 for t in range(1, nperiods+1): S[0,t] = S[0, t-1]*u S[t,t] = S[t-1, t-1]*d

for t in range(2, nperiods+1):
    for j in  range(1, t):
        S[j,t] = S[j-1, t]*d**2
return S

S= make_stock_lattice(s0, u, d, nperiods) print(S)
[[20. 23.78219887 28.27964916 33.62761202] [ 0. 16.81930263 20. 23.78219887] [ 0. 0. 14.14444704 16.81930263] [ 0. 0. 0. 11.89498677]] def make_option_lattice(S, strike, nperiods, p, pbar, disc, Otype = 'E', Ocall= True): C = np.zeros((nperiods+1, nperiods+1)) # Get Boundary Values for j in range(0, nperiods+1): C[j, nperiods] = max(0, S[j, nperiods]- strike) if (Ocall == False): C[j, nperiods] = max(0, strike- S[j, nperiods])

for t in range(nperiods-1, -1, -1):
    for j in range(0, t+1):
        C[j,t] = disc*(p*C[j, t+1]+ pbar*C[j+1, t+1])
        if (Otype == 'A' and Ocall == True):
            C[j,t] = max(C[j,t], S[j,t]-strike)
        if (Otype == 'A' and Ocall == False):
                C[j, t] = max(C[j,t], strike-S[j,t])
        
return C

Otype = 'E'; Ocall = True C = make_option_lattice(S, strike, nperiods, p, pbar, disc, Otype, Ocall) print(np.round(C,2))

[[ 2.65 4.78 8.3 13.63] [ 0. 0.82 1.76 3.78] [ 0. 0. 0. 0. ] [ 0. 0. 0. 0. ]] def option_price(s0, strike, tp, rf, q, sigma, nperiods, Otype='E', Ocall= True): #Step1: u, d,p, pbar, disc = make_stuff(tp,nperiods, sigma, rf, q) #Step 2 S = make_stock_lattice(s0, u, d, nperiods) #Step 3 C = make_option_lattice(S, strike, nperiods, p, pbar, disc, Otype, Ocall) return C[0,0] ​ ​ mycall = option_price(s0, strike, tp, rf, q, sigma, nperiods, Otype, Ocall) print(mycall) 2.6530621686003197 def get_all_options(s0, strike, tp, rf, q, sigma, nperiods): EC = option_price(s0, strike, tp, rf, q, sigma, nperiods, 'E', True) AC = option_price(s0, strike, tp, rf, q, sigma, nperiods, 'A', True) EP = option_price(s0, strike, tp, rf, q, sigma, nperiods, 'E', False) AP = option_price(s0, strike, tp, rf, q, sigma, nperiods, 'A' , False) return(EC, AC, EP, AP) ​ ans = get_all_options(s0, strike, tp, rf, q, sigma, nperiods) print('The price of an European call is ', round(ans[0],3)) print('The price of an American call is ', round(ans[1],3),'\n') print('The price of an European put is ', round(ans[2],3)) print('The price of an American put is ', round(ans[3],3)) The price of an European call is 2.653 The price of an American call is 2.653

The price of an European put is 2.273 The price of an American put is 2.328 def sensitivity_nperiods(s0, strike, tp, rf, q, sigma): mylist = np.arange(2,100) n = len(mylist) myoptions = np.zeros(n) for i, nperiods in enumerate(mylist): myoptions[i] = option_price(s0, strike, tp, rf, q, sigma, nperiods) return mylist, myoptions ​ mylist, myoptions = sensitivity_nperiods(s0, strike, tp, rf, q, sigma)
import matplotlib.pyplot as plt %matplotlib inline plt.plot(mylist, myoptions) [<matplotlib.lines.Line2D at 0x1f67314c2b0>]

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