In [3]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
import datetime


The randomly selected asset price:

In [4]:
def generate_brownian_asset_price(S, sigma, r, T):
"""
S     : current stock
sigma : volatility
r     : rate
T     : time
"""
ret = S * np.exp((r - 0.5 * sigma**2) * T + sigma * np.sqrt(T) * np.random.randn())
return ret


The basic call payout:

In [5]:
def call_payout(S_T, K):
"""
S_T : asset price (prediction) at time T
K   : strike price
"""
return max(0, S_T - K)


Now set the price at 120, the volatility at 25%, the rate at 0.21%, and the strike price at 120 on Oct 16, 2015.

In [6]:
S = 120
sigma = 0.25
r = 0.0021
T = (datetime.date(2015,10,16) - datetime.date(2015,9,17)).days / 365.0
K = 120


Now we can run this once. But that doesn't give any confidence. So we'll run it many times.

In [10]:
nsim = 10000
payout = np.zeros((nsim,))
discount = np.exp(-r * T)

In [11]:
for i in range(nsim):
S_T = generate_brownian_asset_price(S, sigma, r, T)
payout[i] = call_payout(S_T, K)

In [12]:
price = discount * payout.sum() / nsim
print("price: %g" % price)

price: 3.35782


Notice this price seem "reasonable". If we had no interest and no volatility, then the price should be \$2. But with interest and with volatility, one would expect this to be more expensive because the payout may be better. So 3.3 or so seems reasonable.

Let's look at how many payouts were at each "value"

In [15]:
_ = plt.hist(payout, bins=100, log=True)
plt.xlabel('payout')
plt.ylabel('frequency')

Out[15]:
<matplotlib.text.Text at 0x10fa56d68>

Notice that

1. There are many zeros. This means that the simulation landed with a price below the strike.
2. There's the potential for a much larger payout than 3.3

It might be easier to look at it here:

In [16]:
plt.plot(np.sort(payout))

Out[16]:
[<matplotlib.lines.Line2D at 0x10f8fadd8>]
In [ ]: