This exercise notebook refers to this lecture. Please use the lecture for explanations and sample code.
https://www.quantopian.com/lectures#Instability-of-Estimates
Part of the Quantopian Lecture Series:
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from statsmodels.stats.stattools import jarque_bera
# Set a seed so we can play with the data without generating new random numbers every time
np.random.seed(123)
Using the below normal distribution with mean 100 and standard deviation 50, find the means and standard deviations of samples of size 5, 25, 100, and 500.
POPULATION_MU = 100
POPULATION_SIGMA = 25
sample_sizes = [5, 25, 100, 500]
#Your code goes here
for i in range(len(sample_sizes)):
sample = np.random.normal(POPULATION_MU, POPULATION_SIGMA, sample_sizes[i])
print 'Mean', (i+1),':', np.mean(sample),'Std',(i+1),':',np.std(sample)
print "\nAs sample size increases, the mean and standard deviation approach those of the population. However, even at the 500 sample level the sample mean is not the same as the population mean."
X = [ 31., 6., 21., 32., 41., 4., 48., 38., 43., 36., 50., 20., 46., 33., 8., 27., 17., 44., 16., 39., 3., 37.,
35., 13., 49., 2., 18., 42., 22., 25., 15., 24., 11., 19., 5., 40., 12., 10., 1., 45., 26., 29., 7., 30.,
14., 23., 28., 0., 34., 9., 47.]
Y = [ 15., 41., 33., 29., 3., 28., 28., 8., 15., 22., 39., 38., 22., 10., 39., 40., 24., 15., 21., 25., 17., 33.,
40., 32., 42., 5., 39., 8., 15., 25., 37., 33., 14., 25., 1., 31., 45., 5., 6., 19., 13., 39., 18., 49.,
13., 38., 8., 25., 32., 40., 17.]
Z = [ 38., 23., 16., 35., 48., 18., 48., 38., 24., 27., 24., 35., 37., 28., 11., 12., 31., -1., 9., 19., 20., 0.,
23., 33., 34., 24., 14., 28., 12., 25., 53., 19., 42., 21., 15., 36., 47., 20., 26., 41., 33., 50., 26., 22.,
-1., 35., 10., 25., 23., 24., 6.]
#Your code goes here
print "Mean X:", np.mean(X)
print "Mean Y:", np.mean(Y)
print "Mean Z:", np.mean(Z)
Use the jarque_bera
function to conduct a Jarque-Bera test on $X$, $Y$, and $Z$ to determine whether their distributions are normal.
from statsmodels.stats.stattools import jarque_bera
print 'X jarque_bera:', jarque_bera(X)[1]
if jarque_bera(X)[1] < 0.05:
print 'The distribution of X is likely normal. \n'
else:
print 'The distribution of X is likely not normal. \n'
print 'Y jarque_bera:', jarque_bera(Y)[1]
if jarque_bera(Y)[1] < 0.05:
print 'The distribution of Y is likely normal. \n'
else:
print 'The distribution of Y is likely not normal. \n'
print 'Z jarque_bera:', jarque_bera(Z)[1]
if jarque_bera(Z)[1] < 0.05:
print 'The distribution of Z is likely normal. \n'
else:
print 'The distribution of Z is likely not normal. \n'
Create a histogram of the sample distributions of $X$, $Y$, and $Z$ along with the best estimate/mean based on the sample.
plt.hist([X, Y, Z], normed=1, histtype='bar', stacked=False, alpha = 0.7);
plt.ylabel('Frequency')
plt.xlabel('Value');
plt.axvline(np.mean(X));
plt.axvline(np.mean(Y), c='r');
plt.axvline(np.mean(Z), c='g');
print "All three datasets have a similar mean, but have very different distributions. Mean alone is very non-informative about what is going on in data, and should not be used alone as an estimator."
Just as in the lecture, find the mean and standard deviation of the running sharpe ratio for THO, this time testing for multiple window lengths: 300, 150, and 50. Restrict your mean and standard deviation calculation to pricing data up to 200 days away from the end.
def sharpe_ratio(asset, riskfree):
return np.mean(asset - riskfree)/np.std(asset - riskfree)
start = '2010-01-01'
end = '2015-01-01'
treasury_ret = get_pricing('BIL', fields='price', start_date=start, end_date=end).pct_change()[1:]
pricing = get_pricing('THO', fields='price', start_date=start, end_date=end)
returns = pricing.pct_change()[1:]
#Your code goes here
def sharpe_ratio(asset, riskfree):
return np.mean(asset - riskfree)/np.std(asset - riskfree)
for window in [50, 150, 300]:
# Compute the running Sharpe ratio
running_sharpe = [sharpe_ratio(returns[i-window+10:i], treasury_ret[i-window+10:i]) for i in range(window-10, len(returns))]
# Compute the mean and std of the running Sharpe ratios up to 200 days before the end
mean_rs = np.mean(running_sharpe[:-200])
std_rs = np.std(running_sharpe[:-200])
print (window),'Mean of running Sharpe ratio:', mean_rs
print (window),'std of running Sharpe ratio: ', std_rs, '\n'
Plot the running sharpe ratio of all three window lengths, as well as their in-sample mean and standard deviation bars.
for window in [50, 150, 300]:
running_sharpe = [sharpe_ratio(returns[i-window+10:i], treasury_ret[i-window+10:i]) for i in range(window-10, len(returns))]
mean_rs = np.mean(running_sharpe[:-200])
std_rs = np.std(running_sharpe[:-200])
_, ax2 = plt.subplots()
ax2.plot(range(window-10, len(returns)), running_sharpe)
ticks = ax2.get_xticks()
ax2.set_xticklabels([pricing.index[i].date() for i in ticks[:-1]])
ax2.axhline(mean_rs)
ax2.axhline(mean_rs + std_rs, linestyle='--')
ax2.axhline(mean_rs - std_rs, linestyle='--')
ax2.axvline(len(returns) - 200, color='pink');
plt.title(window, fontsize = 20)
plt.xlabel('Date')
plt.ylabel('Sharpe Ratio')
plt.legend(['Sharpe Ratio', 'Mean', '+/- 1 Standard Deviation'])
print "Despite the longer window Sharpe ratios having less variability, they are still unpredictable with repect to just the mean. But within the context of the standard deviation the mean has more predictive value, as we see that even in the out-of-sample periods the ratios of all window lengths stay mainly within 1 standard deviation of the mean."
b15_df = pd.DataFrame([ 29., 22., 19., 17., 19., 19., 15., 16., 18., 25., 21.,
25., 29., 27., 36., 38., 40., 44., 49., 50., 58., 61.,
67., 69., 74., 72., 76., 81., 81., 80., 83., 82., 80.,
79., 79., 80., 74., 72., 68., 68., 65., 61., 57., 50.,
46., 42., 41., 35., 30., 27., 28., 28.],
columns = ['Weekly Avg Temp'],
index = pd.date_range('1/1/2012', periods=52, freq='W') )
#Your code goes here
b15_mean = np.mean(b15_df['Weekly Avg Temp'])
b15_std = np.std(b15_df['Weekly Avg Temp'])
print "Boston Weekly Temp Mean: ", b15_mean
print "Boston Weekly Temp Std: ", b15_std
Find the mean and standard deviation of Palo Alto weekly average temperature data for the year of 2015 stored in p15_df
.
p15_df = pd.DataFrame([ 49., 53., 51., 47., 50., 46., 49., 51., 49., 45., 52.,
54., 54., 55., 55., 57., 56., 56., 57., 63., 63., 65.,
65., 69., 67., 70., 67., 67., 68., 68., 70., 72., 72.,
70., 72., 70., 66., 66., 68., 68., 65., 66., 62., 61.,
63., 57., 55., 55., 55., 55., 55., 48.],
columns = ['Weekly Avg Temp'],
index = pd.date_range('1/1/2012', periods=52, freq='W'))
#Your code goes here
p15_mean = np.mean(p15_df['Weekly Avg Temp'])
p15_std = np.std(p15_df['Weekly Avg Temp'])
print "Palo Alto Weekly Temp Mean: ", p15_mean
print "Palo Alto Weekly Temp Std: ", p15_std
Use the means you found in parts a and b to attempt to predict 2016 temperature data for both cities. Do this by creating two histograms for the 2016 temperature data in b16_df
and p16_df
with a vertical line where the 2015 means were to represent your prediction.
b16_df = pd.DataFrame([ 26., 22., 20., 19., 18., 19., 17., 17., 19., 20., 23., 22., 28., 28., 35., 38., 42., 47., 49., 56., 59., 61.,
61., 70., 73., 73., 73., 77., 78., 82., 80., 80., 81., 78., 82., 78., 76., 71., 69., 66., 60., 63., 56., 50.,
44., 43., 34., 33., 31., 28., 27., 20.],
columns = ['Weekly Avg Temp'],
index = pd.date_range('1/1/2012', periods=52, freq='W'))
p16_df = pd.DataFrame([ 50., 50., 51., 48., 48., 49., 50., 45., 52., 50., 51., 52., 50., 56., 58., 55., 61., 56., 61., 62., 62., 64.,
64., 69., 71., 66., 69., 70., 68., 71., 70., 69., 72., 71., 66., 69., 70., 70., 66., 67., 64., 64., 65., 61.,
61., 59., 56., 53., 55., 52., 52., 51.],
columns = ['Weekly Avg Temp'],
index = pd.date_range('1/1/2012', periods=52, freq='W'))
#Your code goes here
b16_df.plot.hist(title = "Boston 2016 Temperature vs. Prediction");
plt.axvline(b15_mean);
p16_df.plot.hist(title = "Palo Alto 2016 Temperature vs. Prediction");
plt.axvline(p15_mean);
b_avg_error = np.mean(abs(b16_df['Weekly Avg Temp'] - b15_mean))
p_avg_error = np.mean(abs(p16_df['Weekly Avg Temp'] - p15_mean))
print "Avg of Absolute Value of Prediction Error in Boston:", b_avg_error
print "Avg of Absolute Value of Prediction Error in Palo Alto:", p_avg_error
print "\nWe know from parts a and b that the weather in Boston is much more variable than that of Palo Alto. As a result, we can predict that an estimate based on a sample mean in Boston will be less accurate than an estimate based on a sample from Palo Alto, which is confirmed by this test. The Palo Alto predictions had a much lower error than those of Boston. With mean alone we would not have been able to make any conclusions about the accuracy of our predictions."
Congratulations on completing the instability of parameter estimates exercises!
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