Solutions for http://quant-econ.net/py/python_by_example.html
In [1]:
def factorial(n):
k = 1
for i in range(n):
k = k * (i + 1)
return k
factorial(4)
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In [2]:
from random import uniform
def binomial_rv(n, p):
count = 0
for i in range(n):
U = uniform(0, 1)
if U < p:
count = count + 1 # Or count += 1
return count
binomial_rv(10, 0.5)
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Consider the circle of diameter 1 embedded in the unit square
Let $A$ be its area and let $r=1/2$ be its radius
If we know $\pi$ then we can compute $A$ via $A = \pi r^2$
But here the point is to compute $\pi$, which we can do by $\pi = A / r^2$
Summary: If we can estimate the area of the unit circle, then dividing by $r^2 = (1/2)^2 = 1/4$ gives an estimate of $\pi$
We estimate the area by sampling bivariate uniforms and looking at the fraction that fall into the unit circle
In [3]:
from __future__ import division # Omit if using Python 3.x
from math import sqrt
n = 100000
count = 0
for i in range(n):
u, v = uniform(0, 1), uniform(0, 1)
d = sqrt((u - 0.5)**2 + (v - 0.5)**2)
if d < 0.5:
count += 1
area_estimate = count / n
print(area_estimate * 4) # dividing by radius**2
In [4]:
payoff = 0
count = 0
for i in range(10):
U = uniform(0, 1)
count = count + 1 if U < 0.5 else 0
if count == 3:
payoff = 1
print(payoff)
The next line embeds all subsequent figures in the browser itself
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%matplotlib inline
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import matplotlib.pyplot as plt
from random import normalvariate
alpha = 0.9
ts_length = 200
current_x = 0
x_values = []
for i in range(ts_length + 1):
x_values.append(current_x)
current_x = alpha * current_x + normalvariate(0, 1)
plt.plot(x_values, 'b-')
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In [7]:
alphas = [0.0, 0.8, 0.98]
ts_length = 200
for alpha in alphas:
x_values = []
current_x = 0
for i in range(ts_length):
x_values.append(current_x)
current_x = alpha * current_x + normalvariate(0, 1)
plt.plot(x_values, label='alpha = ' + str(alpha))
plt.legend()
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