Copyright 2016 Allen B. Downey
MIT License: https://opensource.org/licenses/MIT
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from __future__ import print_function, division
%matplotlib inline
import numpy as np
import brfss
import thinkstats2
import thinkplot
Root mean squared error is one of several ways to summarize the average error of an estimation process.
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def RMSE(estimates, actual):
"""Computes the root mean squared error of a sequence of estimates.
estimate: sequence of numbers
actual: actual value
returns: float RMSE
"""
e2 = [(estimate-actual)**2 for estimate in estimates]
mse = np.mean(e2)
return np.sqrt(mse)
The following function simulates experiments where we try to estimate the mean of a population based on a sample with size n=7. We run iters=1000 experiments and collect the mean and median of each sample.
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import random
def Estimate1(n=7, iters=1000):
"""Evaluates RMSE of sample mean and median as estimators.
n: sample size
iters: number of iterations
"""
mu = 0
sigma = 1
means = []
medians = []
for _ in range(iters):
xs = [random.gauss(mu, sigma) for _ in range(n)]
xbar = np.mean(xs)
median = np.median(xs)
means.append(xbar)
medians.append(median)
print('Experiment 1')
print('rmse xbar', RMSE(means, mu))
print('rmse median', RMSE(medians, mu))
Estimate1()
Using $\bar{x}$ to estimate the mean works a little better than using the median; in the long run, it minimizes RMSE. But using the median is more robust in the presence of outliers or large errors.
The obvious way to estimate the variance of a population is to compute the variance of the sample, $S^2$, but that turns out to be a biased estimator; that is, in the long run, the average error doesn't converge to 0.
The following function computes the mean error for a collection of estimates.
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def MeanError(estimates, actual):
"""Computes the mean error of a sequence of estimates.
estimate: sequence of numbers
actual: actual value
returns: float mean error
"""
errors = [estimate-actual for estimate in estimates]
return np.mean(errors)
The following function simulates experiments where we try to estimate the variance of a population based on a sample with size n=7. We run iters=1000 experiments and two estimates for each sample, $S^2$ and $S_{n-1}^2$.
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def Estimate2(n=7, iters=1000):
mu = 0
sigma = 1
estimates1 = []
estimates2 = []
for _ in range(iters):
xs = [random.gauss(mu, sigma) for i in range(n)]
biased = np.var(xs)
unbiased = np.var(xs, ddof=1)
estimates1.append(biased)
estimates2.append(unbiased)
print('mean error biased', MeanError(estimates1, sigma**2))
print('mean error unbiased', MeanError(estimates2, sigma**2))
Estimate2()
The mean error for $S^2$ is non-zero, which suggests that it is biased. The mean error for $S_{n-1}^2$ is close to zero, and gets even smaller if we increase iters.
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def SimulateSample(mu=90, sigma=7.5, n=9, iters=1000):
xbars = []
for j in range(iters):
xs = np.random.normal(mu, sigma, n)
xbar = np.mean(xs)
xbars.append(xbar)
return xbars
xbars = SimulateSample()
Here's the "sampling distribution of the mean" which shows how much we should expect $\bar{x}$ to vary from one experiment to the next.
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cdf = thinkstats2.Cdf(xbars)
thinkplot.Cdf(cdf)
thinkplot.Config(xlabel='Sample mean',
ylabel='CDF')
The mean of the sample means is close to the actual value of $\mu$.
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np.mean(xbars)
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An interval that contains 90% of the values in the sampling disrtribution is called a 90% confidence interval.
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ci = cdf.Percentile(5), cdf.Percentile(95)
ci
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And the RMSE of the sample means is called the standard error.
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stderr = RMSE(xbars, 90)
stderr
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Confidence intervals and standard errors quantify the variability in the estimate due to random sampling.
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def Estimate3(n=7, iters=1000):
lam = 2
means = []
medians = []
for _ in range(iters):
xs = np.random.exponential(1.0/lam, n)
L = 1 / np.mean(xs)
Lm = np.log(2) / thinkstats2.Median(xs)
means.append(L)
medians.append(Lm)
print('rmse L', RMSE(means, lam))
print('rmse Lm', RMSE(medians, lam))
print('mean error L', MeanError(means, lam))
print('mean error Lm', MeanError(medians, lam))
Estimate3()
The RMSE is smaller for the sample mean than for the sample median.
But neither estimator is unbiased.
Exercise: In this chapter we used $\bar{x}$ and median to estimate µ, and found that $\bar{x}$ yields lower MSE. Also, we used $S^2$ and $S_{n-1}^2$ to estimate σ, and found that $S^2$ is biased and $S_{n-1}^2$ unbiased. Run similar experiments to see if $\bar{x}$ and median are biased estimates of µ. Also check whether $S^2$ or $S_{n-1}^2$ yields a lower MSE.
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# Solution
def Estimate4(n=7, iters=100000):
"""Mean error for xbar and median as estimators of population mean.
n: sample size
iters: number of iterations
"""
mu = 0
sigma = 1
means = []
medians = []
for _ in range(iters):
xs = [random.gauss(mu, sigma) for i in range(n)]
xbar = np.mean(xs)
median = np.median(xs)
means.append(xbar)
medians.append(median)
print('Experiment 1')
print('mean error xbar', MeanError(means, mu))
print('mean error median', MeanError(medians, mu))
Estimate4()
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# Solution
def Estimate5(n=7, iters=100000):
"""RMSE for biased and unbiased estimators of population variance.
n: sample size
iters: number of iterations
"""
mu = 0
sigma = 1
estimates1 = []
estimates2 = []
for _ in range(iters):
xs = [random.gauss(mu, sigma) for i in range(n)]
biased = np.var(xs)
unbiased = np.var(xs, ddof=1)
estimates1.append(biased)
estimates2.append(unbiased)
print('Experiment 2')
print('RMSE biased', RMSE(estimates1, sigma**2))
print('RMSE unbiased', RMSE(estimates2, sigma**2))
Estimate5()
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# Solution
# My conclusions:
# 1) xbar and median yield lower mean error as m increases, so neither
# one is obviously biased, as far as we can tell from the experiment.
# 2) The biased estimator of variance yields lower RMSE than the unbiased
# estimator, by about 10%. And the difference holds up as m increases.
Exercise: Suppose you draw a sample with size n=10 from an exponential distribution with λ=2. Simulate this experiment 1000 times and plot the sampling distribution of the estimate L. Compute the standard error of the estimate and the 90% confidence interval.
Repeat the experiment with a few different values of n and make a plot of standard error versus n.
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# Solution
def SimulateSample(lam=2, n=10, iters=1000):
"""Sampling distribution of L as an estimator of exponential parameter.
lam: parameter of an exponential distribution
n: sample size
iters: number of iterations
"""
def VertLine(x, y=1):
thinkplot.Plot([x, x], [0, y], color='0.8', linewidth=3)
estimates = []
for _ in range(iters):
xs = np.random.exponential(1.0/lam, n)
lamhat = 1.0 / np.mean(xs)
estimates.append(lamhat)
stderr = RMSE(estimates, lam)
print('standard error', stderr)
cdf = thinkstats2.Cdf(estimates)
ci = cdf.Percentile(5), cdf.Percentile(95)
print('confidence interval', ci)
VertLine(ci[0])
VertLine(ci[1])
# plot the CDF
thinkplot.Cdf(cdf)
thinkplot.Config(xlabel='estimate',
ylabel='CDF',
title='Sampling distribution')
return stderr
SimulateSample()
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# Solution
# My conclusions:
# 1) With sample size 10:
# standard error 0.762510819389
# confidence interval (1.2674054394352277, 3.5377353792673705)
# 2) As sample size increases, standard error and the width of
# the CI decrease:
# 10 0.90 (1.3, 3.9)
# 100 0.21 (1.7, 2.4)
# 1000 0.06 (1.9, 2.1)
# All three confidence intervals contain the actual value, 2.
Exercise: In games like hockey and soccer, the time between goals is roughly exponential. So you could estimate a team’s goal-scoring rate by observing the number of goals they score in a game. This estimation process is a little different from sampling the time between goals, so let’s see how it works.
Write a function that takes a goal-scoring rate, lam, in goals per game, and simulates a game by generating the time between goals until the total time exceeds 1 game, then returns the number of goals scored.
Write another function that simulates many games, stores the estimates of lam, then computes their mean error and RMSE.
Is this way of making an estimate biased?
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def SimulateGame(lam):
"""Simulates a game and returns the estimated goal-scoring rate.
lam: actual goal scoring rate in goals per game
"""
goals = 0
t = 0
while True:
time_between_goals = random.expovariate(lam)
t += time_between_goals
if t > 1:
break
goals += 1
# estimated goal-scoring rate is the actual number of goals scored
L = goals
return L
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# Solution
# The following function simulates many games, then uses the
# number of goals scored as an estimate of the true long-term
# goal-scoring rate.
def Estimate6(lam=2, m=1000000):
estimates = []
for i in range(m):
L = SimulateGame(lam)
estimates.append(L)
print('Experiment 4')
print('rmse L', RMSE(estimates, lam))
print('mean error L', MeanError(estimates, lam))
pmf = thinkstats2.Pmf(estimates)
thinkplot.Hist(pmf)
thinkplot.Config(xlabel='Goals scored', ylabel='PMF')
Estimate6()
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# Solution
# My conclusions:
# 1) RMSE for this way of estimating lambda is 1.4
# 2) The mean error is small and decreases with m, so this estimator
# appears to be unbiased.
# One note: If the time between goals is exponential, the distribution
# of goals scored in a game is Poisson.
# See https://en.wikipedia.org/wiki/Poisson_distribution
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