chap08ex.ipynb

Kernel: Python 3

Examples and Exercises from Think Stats, 2nd Edition

http://thinkstats2.com

Copyright 2016 Allen B. Downey

MIT License: https://opensource.org/licenses/MIT

from __future__ import print_function, division

%matplotlib inline

import numpy as np

import brfss

import thinkstats2
import thinkplot

The estimation game

Root mean squared error is one of several ways to summarize the average error of an estimation process.

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.

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()

Experiment 1
rmse xbar 0.3862603150323446
rmse median 0.48032836544952223

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.

Estimating variance

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.

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 $S_{n-1}^2$.

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()

mean error biased -0.1374939469040375
mean error unbiased 0.006257061945289593

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.

The sampling distribution

The following function simulates experiments where we estimate the mean of a population using $\bar{x}$, and returns a list of estimates, one from each experiment.

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.

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 $\mu$.

np.mean(xbars)

89.94056816952832

An interval that contains 90% of the values in the sampling disrtribution is called a 90% confidence interval.

ci = cdf.Percentile(5), cdf.Percentile(95)
ci

(85.87345905535176, 94.11925824713033)

And the RMSE of the sample means is called the standard error.

stderr = RMSE(xbars, 90)
stderr

2.487879588208278

Confidence intervals and standard errors quantify the variability in the estimate due to random sampling.

Estimating rates

The following function simulates experiments where we try to estimate the mean of an exponential distribution using the mean and median of a sample.

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()

rmse L 1.075066354067901
rmse Lm 1.7723826281429338
mean error L 0.315202440018787
mean error Lm 0.464767815400564

The RMSE is smaller for the sample mean than for the sample median.

But neither estimator is unbiased.

Exercises

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.

# Solution goes here

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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.

# Solution goes here

# Solution goes here

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?

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

# Solution goes here

# Solution goes here