Exercises: Instability of Parameter Estimates

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:

In [2]:
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)

Exercise 1: Sample Size vs. Standard Deviation

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.

In [35]:
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."
Mean 1 : 106.118892326 Std 1 : 10.0745513103
Mean 2 : 108.246344326 Std 2 : 23.8941921401
Mean 3 : 102.203295956 Std 3 : 25.8210851427
Mean 4 : 100.962874773 Std 4 : 23.4893196055

As 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.

Exercise 2: Instability of Predictions on Mean Alone

a. Finding Means

Find the means of the following three data sets $X$, $Y$, and $Z$.

In [40]:
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)
Mean X: 25.0
Mean Y: 24.6862745098
Mean Z: 25.6078431373

b. Checking for Normality

Use the jarque_bera function to conduct a Jarque-Bera test on $X$, $Y$, and $Z$ to determine whether their distributions are normal.

In [52]:
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'
X jarque_bera: 0.216026379492
The distribution of X is likely not normal. 

Y jarque_bera: 0.25028131217
The distribution of Y is likely not normal. 

Z jarque_bera: 0.866907001763
The distribution of Z is likely not normal. 

c. Instability of Estimates

Create a histogram of the sample distributions of $X$, $Y$, and $Z$ along with the best estimate/mean based on the sample.

In [58]:
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." 
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.

Exercise 3: Sharpe Ratio Window Adjustment

a. Effect on Variability

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.

In [70]:
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'
50 Mean of running Sharpe ratio: 0.0415782438805
50 std of running Sharpe ratio:  0.152672986316 

150 Mean of running Sharpe ratio: 0.0450634036062
150 std of running Sharpe ratio:  0.0708117281461 

300 Mean of running Sharpe ratio: 0.0424167320946
300 std of running Sharpe ratio:  0.041788831424 

b. Out-of-Sample Instability

Plot the running sharpe ratio of all three window lengths, as well as their in-sample mean and standard deviation bars.

In [73]:
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."
/usr/local/lib/python2.7/dist-packages/pandas/tseries/base.py:192: VisibleDeprecationWarning: using a non-integer number instead of an integer will result in an error in the future
  val = getitem(key)
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.

Exercise 4: Weather

a. Temperature in Boston

Find the mean and standard deviation of Boston weekly average temperature data for the year of 2015 stored in b15_df.

In [90]:
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
Boston Weekly Temp Mean:  49.0769230769
Boston Weekly Temp Std:   22.983979499

b. Temperature in Palo Alto

Find the mean and standard deviation of Palo Alto weekly average temperature data for the year of 2015 stored in p15_df.

In [91]:
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
Palo Alto Weekly Temp Mean:  59.7884615385
Palo Alto Weekly Temp Std:   7.97432548018

c. Predicting 2016 Temperatures

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.

In [92]:
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."
Avg of Absolute Value of Prediction Error in Boston: 20.8106508876
Avg of Absolute Value of Prediction Error in Palo Alto: 7.20857988166

We 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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