Exercises: Variance - Answer Key

By Christopher van Hoecke, Maxwell Margenot, and Delaney Mackenzie

Lecture Link :

https://www.quantopian.com/lectures/variance

IMPORTANT NOTE:

This lecture corresponds to the Variance lecture, which is part of the Quantopian lecture series. This homework expects you to rely heavily on the code presented in the corresponding lecture. Please copy and paste regularly from that lecture when starting to work on the problems, as trying to do them from scratch will likely be too difficult.

Part of the Quantopian Lecture Series:



In [1]:

    
# Useful Libraries
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt

Data:



In [2]:

    
X = np.random.randint(100, size = 100)

Exercise 1:

Using the skills aquired in the lecture series, find the following parameters of the list X above:

Range
Mean Absolute Deviation
Variance and Standard Deviation
Semivariance and Semideviation
Target variance (with B = 60)



In [3]:

    
# Range of X

print 'Range of X: %s' %(np.ptp(X))









    



Range of X: 99



In [4]:

    
# Mean Absolute Deviation
# First calculate the value of mu (the mean)

mu = np.mean(X)

abs_dispersion = [np.abs(mu - x) for x in X]
MAD = np.sum(abs_dispersion)/len(abs_dispersion)

print 'Mean absolute deviation of X:', MAD









    



Mean absolute deviation of X: 24.6858



In [5]:

    
# Variance and standard deviation

print 'Variance of X:', np.var(X)
print 'Standard deviation of X:', np.std(X)









    



Variance of X: 817.7859
Standard deviation of X: 28.5969561317



In [6]:

    
# Semivariance and semideviation

lows = [e for e in X if e <= mu]

semivar = np.sum( (lows - mu) ** 2 ) / len(lows)

print 'Semivariance of X:', semivar
print 'Semideviation of X:', np.sqrt(semivar)









    



Semivariance of X: 804.51664902
Semideviation of X: 28.3640026974



In [7]:

    
# Target variance

B = 60
lows_B = [e for e in X if e <= B]
semivar_B = sum(map(lambda x: (x - B)**2,lows_B))/len(lows_B)

print 'Target semivariance of X:', semivar_B
print 'Target semideviation of X:', np.sqrt(semivar_B)









    



Target semivariance of X: 1182
Target semideviation of X: 34.3802268753

Exercise 2:

Using the skills aquired in the lecture series, find the following parameters of prices for AT&T stock over a year:

30 days rolling variance
15 days rolling Standard Deviation



In [8]:

    
att = get_pricing('T', fields='open_price', start_date='2016-01-01', end_date='2017-01-01')



In [9]:

    
# Rolling mean
variance = att.rolling(window = 30).var()



In [10]:

    
# Rolling standard deviation
std = att.rolling(window = 15).std()

Exercise 3 :

The portfolio variance is calculated as

$$\text{VAR}_p = \text{VAR}_{s1} (w_1^2) + \text{VAR}_{s2}(w_2^2) + \text{COV}_{S_1, S_2} (2 w_1 w_2)$$

Where $w_1$ and $w_2$ are the weights of $S_1$ and $S_2$.

Find values of $w_1$ and $w_2$ to have a portfolio variance of 50.



In [11]:

    
asset1 = get_pricing('AAPL', fields='open_price', start_date='2016-01-01', end_date='2017-01-01')
asset2 = get_pricing('XLF', fields='open_price', start_date='2016-01-01', end_date='2017-01-01')

cov = np.cov(asset1, asset2)[0,1]

w1 = 0.87
w2 = 1 - w1

v1 = np.var(asset1)
v2 = np.var(asset2)

pvariance = (w1**2)*v1+(w2**2)*v2+(2*w1*w2)*cov

print 'Portfolio variance: ', pvariance









    



Portfolio variance:  50.1862438743

Congratulations on completing the Variance answer key!

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