Section 2.5 — Continuous random variables#
This notebook contains all the code examples from Section 2.5 Continuous random variables of the No Bullshit Guide to Statistics.
Topics covered in this notebook:
Definitions of continuous random variables
Examples of random variables
Probability calculations
Computer models for random variables
Overview of
scipy.statsmethods
Real-world example to demo probability applications
Discussion
Bulk and tails of a the normal distribution
Notebook setup#
We’ll start by importing the Python modules we’ll need for this notebook.
# load Python modules
import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
# Figures setup
sns.set_theme(
context="paper",
style="whitegrid",
palette="colorblind",
rc={'figure.figsize': (7,4)},
)
%config InlineBackend.figure_format = 'retina'
%pip install ministats
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Note: you may need to restart the kernel to use updated packages.
from ministats import plot_pdf_and_cdf
# set random seed for repeatability
np.random.seed(3)
Definitions#
Random variables#
random variable \(X\): a quantity that can take on different values.
outcome: a particular value \(\{X = x\}\) or range of values \(\{a \leq X \leq b\}\) that can occur as a result of observing the random variable \(X\).
sample space \(\mathcal{X}\): describes the set of all possible outcomes of the random variable.
\(f_X\): the probability distribution function is a function that assigns probabilities to the different outcome in the sample space of a random variable. The probability distribution function of the random variable \(X\) is a function of the form \(f_X: \mathcal{X} \to \mathbb{R}\).
\(F_X\): the cumulative distribution function (CDF) tells us the probability of an outcome less than or equal to a given value: \(F_X(b) = Pr(\{ X \leq b \})\).
\(F_X^{-1}\): the inverse cumulative distribution function computes contains the information about the quantiles of the probability distribution. The value \(F_X^{-1}(q)=x_q\) tells how far you need to go in the sample space so that the event \(\{ X \leq x_q \}\) contains a proportion \(q\) of the total probability: \(\Pr(\{ X \leq x_q \})=q\).
\(\mathbb{E}_X[w]\): the expected value of the function \(w(X)\) computes the average value of \(w(X)\) computed for all the possible values of the random variable \(X\).
Example 1: Uniform distribution#
The uniform distribution \(\mathcal{U}(0,1)\) is described by the following probability density function:
where \(U\) is the name of the random variable and \(u\) are particular values it can take on.
The above equation describes tells you how likely it is to observe \(\{U=u\}\). For a uniform distribution \(\mathcal{U}(0,1)\), each \(u\) between 0 and 1 is equally likely to occur, and values of \(u\) outside this range have zero probability of occurring.
Computer simulation
The continuous uniform family of distribution \(\mathcal{U}(\alpha,\beta)\), which assigns equal probabilities to all outcomes in the interval \([\alpha,\beta]\). To create a computer model for a continuous uniform distribution, use the code
uniform(alpha,beta), wherealphaandbetaare two floats.
We’ll introduce computer models for random variables is Section 2.1.5 — Computer models for random variables below, but since we’re looking at a notebook, we can show a little preview of the calculations you’ll learn by the end of the section.
# define the computer model `rvU` for the random variable U
from scipy.stats import uniform
rvU = uniform(0, 1)
# use `quad` function to integrate rvU.pdf between 0.2 and 0.5
from scipy.integrate import quad
quad(rvU.pdf, 0.2, 0.5)[0]
0.3
Example 2: Normal distribution#
A random variable \(N\) with a normal distribution \(\mathcal{N}(\mu,\sigma)\) is described by the probability density function:
The mean \(\mu\) (the Greek letter mu) and the standard deviation \(\sigma\) (the Greek letter sigma) are called the parameters of the distribution.
The math notation \(\mathcal{N}(\mu, \sigma)\) is used to describe the whole family of normal probability distributions, and \(N \sim \mathcal{N}(1000, 100)\) is a particular instance of the distribution with mean \(\mu = 1000\) and standard deviation \(\sigma = 100\).
plot_pdf_and_cdf(rvU, b=0.7, xlims=[-0.2,1.2], rv_name="U", title="auto");
The code example below shows the calculation of the probability \(\Pr\!\left( \{ 800 \leq N \leq 900 \} \right)\), which corresponds to the integral \(\int_{n=800}^{n=900} f_N(n) dn\).
# define the computer model `rvN` for the random variable N
from scipy.stats import norm
mean = 1000
std = 100
rvN = norm(mean, std)
# use `quad` function to integrate rvN.pdf between 800 and 900
quad(rvN.pdf, 800, 900)[0]
0.13590512198327784
plot_pdf_and_cdf(rvN, b=1100, rv_name="N", title="auto");
Calculations with random variables#
Example 1: mean and variance of the uniform distribution#
Mean#
from sympy import symbols, integrate
u = symbols('u')
integrate(u * 1, (u,0,1))
So the mean is \(\mu_U = \frac{1}{2} = 0.5\).
Variance#
The formula for the variance is
from sympy import symbols, integrate
u = symbols('u')
integrate( (u-1/2)**2 * 1, (u,0,1) )
So the variance of \(U\) is \(\sigma_U^2 = \frac{1}{12} = 0.08\overline{3}\).
We can compute the standard deviation \(\sigma_U\) by taking the square root of the variance.
import numpy as np
np.sqrt(0.0833333333333333)
0.2886751345948128
Example 2: mean and variance of a normal distribution#
import numpy as np
mu = 1000
sigma = 100
def fN(n):
z = (n - mu)/sigma
C = sigma * np.sqrt(2*np.pi)
return 1 / C * np.exp(-1/2 * z**2)
The mean of \(N\) is
from scipy.integrate import quad
def n_times_fN(n):
return n * fN(n)
muN = quad(n_times_fN, 0, 3000)[0]
muN
1000.0000000000002
The standard deviation of \(N\) is
def n_minus_mu_sq_times_fN(n):
return (n-muN)**2 * fN(n)
sigma_sq = quad(n_minus_mu_sq_times_fN, 0, 2000)[0]
sigmaN = np.sqrt(sigma_sq)
sigmaN
100.0
Skewness and kurtosis#
from scipy.stats import norm
mu = 0 # position of the centre
sigma = 1 # dispersion
rvZ = norm(mu, sigma)
mean, var, skew, kurt = rvZ.stats(moments="mvsk")
mean, var, skew, kurt
(0.0, 1.0, 0.0, 0.0)
Computer models for random variables#
<model>: the family of probability distributions<params>: parameters of the model—specific value of the control knobs we choose in the general family of distributions to create a particular distribution<model>(<params>): a particular instance of probability model created by choosing a model family<model>and the model parameters<params>.example 1: uniform family of distribution \(\mathcal{U}(\alpha,\beta)\) with parameters \(\alpha\) and \(\beta\).
example 2: normal family of distribution \(\mathcal{N}(\mu,\sigma)\) with parameters \(\mu\) and \(\sigma\).
\(\sim\): math shorthand symbol that stands for “is distributed according to.” For example \(X \sim \texttt{model}(\theta)\) means the random variable \(X\) is distributed according to model instance \(\texttt{model}\) and parameters \(\theta\).
from scipy.stats import norm
# create a normal random variable with mean 1000 and std 100
rvN = norm(1000, 100)
type(rvN)
scipy.stats._distn_infrastructure.rv_continuous_frozen
## see all attributes and methods:
# [attr for attr in dir(rvN) if "__" not in attr]
Plotting the probability density function#
ns = np.linspace(0, 2000, 1000)
fNs = rvN.pdf(ns)
sns.lineplot(x=ns, y=fNs, label="pdf of $N$")
<Axes: >
The cumulative distribution is the integral of the probability density function:
ns = np.linspace(0, 2000, 1000)
FNs = rvN.cdf(ns)
sns.lineplot(x=ns, y=FNs, label="CDF of $N$")
<Axes: >
Properties of the distribution#
rvN.mean()
1000.0
rvN.std()
100.0
rvN.var()
10000.0
np.sqrt( rvN.var() ) # = rvN.std()
100.0
rvN.median()
1000.0
rvN.support()
(-inf, inf)
Computing probabilities#
Suppose you want to compute the probability of the outcome \(\{ a \leq N \leq b \}\) for the random variable \(N\).
# Pr({800 < N < 1200}) = integral of f_N between 800 and 1200
quad(rvN.pdf, 800, 1200)[0]
0.9544997361036417
# Pr({800 < N < 1200}) = F_N(1200) - F_N(800)
rvN.cdf(1200) - rvN.cdf(800)
0.9544997361036416
Computing quantiles#
The inverse question is to find the interval \((-\infty, n_q]\) that contains proportion \(q\) of the total probability.
For example the \(q=0.25\) quantile is located at
# first quartile
rvN.ppf(0.25)
932.5510249803918
# verify that Pr({N<=932.5510249803918)}) == 0.25
rvN.cdf(932.5510249803918)
0.25
# second quartile == median
rvN.ppf(0.5)
1000.0
# third quartile
rvN.ppf(0.75)
1067.4489750196083
Left tail#
rvN.ppf(0.05)
835.5146373048527
Right tail#
rvN.ppf(0.95)
1164.4853626951472
Computing confidence intervals#
To compute a 90% confidence interval for the random variable \(N\),
we can use the rvN.interval() method.
rvN.interval(0.90)
(835.5146373048527, 1164.4853626951472)
Note the method rvN.interval(0.90) is just a shortcut for computing (rvN.ppf(0.05),rvN.ppf(0.95)).
Generating random observations#
Let’s say you want to generate \(n=10\) observations from the random variable \(N\).
You can do this by calling the method rvN.rvs(n).
ns = rvN.rvs(10)
ns
array([1178.86284734, 1043.65098505, 1009.64974681, 813.65072966,
972.26117975, 964.52410207, 991.72585185, 937.29993232,
995.6181831 , 952.27819696])
ns_mean = sum(ns) / len(ns)
ns_mean
985.9521754922469
Computing expectations#
Suppose the distributor accepts only bottles contain between 800 ml and 1200 ml, and you’ll receive a receive payment of \(\$2\) for each bottle. Bottles outside that range get rejected and you don’t get paid for them.
def payment(n):
if 800 <= n and n <= 1200:
return 2
else:
return 0
# get paid if in spec
payment(1050)
2
# don't get paid if out of spec
payment(1250)
0
# expected value of payment
rvN.expect(payment, lb=0, ub=2000)
1.9089994721789976
Visually speaking, only parts of the probability mass of the random variable “count” towards the payment, the subset of the values inside the yellow region shown below.
xs = np.linspace(0, 2000, 1000)
ys = [payment(x)/500 for x in xs]
sns.lineplot(x=xs, y=rvN.pdf(xs))
sns.lineplot(x=xs, y=ys, label="payment(x)")
<Axes: >
Multiple random variables#
Joint probability density functions#
TODO: insert figure as attachment here.
See notebook ../figure_generation/continuous_RVs.ipynb for the code to generate the figure in the book.
Marginal density functions#
TODO: insert figure as attachment here.
See notebook ../figure_generation/continuous_RVs.ipynb for the code to generate the figure in the book.
Conditional probability density functions#
Examples#
Example 1: Multivariable uniform#
xmin = 0
xmax = 100
ymin = 0
ymax = 10
# joint pdf of = uniform(0,100) x uniform(0,10)
def fUV(x,y):
A = (xmax-xmin) * (ymax-ymin)
if xmin <= x and x <= xmax and ymin <= y and y <= ymax:
return 1/A
else:
return 0.0
from scipy.stats import uniform
rvU = uniform(0,100)
rvV = uniform(0,10)
# joint pdf of = uniform(0,100) x uniform(0,10)
def fUV(u,v):
return rvU.pdf(u) * rvV.pdf(v)
fUV(70,10)
0.001
from scipy.integrate import dblquad
dblquad(fUV, 0, 20, 0, 300)[0]
1.0000000000000004
from mpl_toolkits.mplot3d import axes3d
fig1a = plt.figure(figsize=(12,6))
ax1a = fig1a.add_subplot(111, projection='3d')
us = np.arange(-10,110,1)
vs = np.arange(-3,16,0.1)
Us,Vs = np.meshgrid(us,vs)
fUVuv = np.vectorize(fUV)(Us,Vs)
# Plot a basic wireframe
# ax.plot_wireframe(X, Y, fUxy, rstride=10, cstride=20)
ax1a.plot_surface(Us, Vs, fUVuv, cmap="Greys")# rstride=10, cstride=20)
ax1a.set_box_aspect((10,4,2))
ax1a.set_zlim(0,0.002)
(0.0, 0.002)
fig1b = plt.figure()
ax1b = fig1b.add_subplot(111)
ax1b.contourf(Us, Vs, fUVuv, cmap="Greys")
<matplotlib.contour.QuadContourSet at 0x7f56d04cc430>
Example 2: Kombucha volume increasing with temperature#
Consider now the joint sample space \((N,T)\), where \(T\) describes the temperature of the Kombucha that is going into the bottles, and \(N\) describes the volume that goes into each bottle.
Suppose the temperature random variable is normally distributed with standard deviation \(\sigma_T = 2\) around the mean of \(\mu_T = 20\), which is written mathematically as \(T \sim \mathcal{N}(20,2)\). The variability of the volume of kombucha depends on the temperature \(t\), and is described by the random variable \(N \sim \mathcal{N}(\mu_N,75)\), where \(\mu_N = 1000 + 35(t-20)\). In other words, the conditional distribution \(f_{N|T}(n|t)\) is distributed according to:
By studying the dependence between the bottling temperature and the variation in the volume, you might be able to improve the reliability of the kombucha bottling process. Recall that your distributor only pays for bottles that are within “spec” (mean of 1000 +/- 2 times std = \([800,1200]\)).
from scipy.stats import norm
rvT = norm(20,2)
# joint pdf f_{NT}(n,t) = f_{N|T}(n|t) * f_T(t)
def fNT2(n,t):
fNgivent = norm(1000+35*(t-20), 75) # = f_{N|T=t}
return fNgivent.pdf(n) * rvT.pdf(t)
ts = np.arange(15,25,0.2)
ns = np.arange(700,1300,10)
Ts, Ns = np.meshgrid(ts,ns)
fNT2nt = np.vectorize(fNT2)(Ns,Ts)
fig2 = plt.figure()
ax2 = fig2.add_subplot(111)
ax2.contourf(Ts, Ns, fNT2nt, cmap="Greys")
<matplotlib.contour.QuadContourSet at 0x7f56c09fa2c0>
Example 3: Temperature-dependent variability#
Consider now an alternative scenario in which the variance of the volume of kombucha \(N\) varies with temperature. The temperature random variable is \(T \sim \mathcal{N}(20,2)\). And the volume of kombucha is described by the random variable \(N \sim \mathcal{N}(1000,\sigma_N)\), where \(\sigma_N = 100 + 5(t-20)\). In other words, the conditional distribution \(f_{N|T}(n|t)\) is distributed according to:
from scipy.stats import norm
rvT = norm(20,2)
infos = {}
# joint pdf f_{NT}(n,t) = f_{N|T}(n|t) * f_T(t)
def fNT3(n,t):
fNgivent = norm(1000, 100 + 14*(t-20))
return fNgivent.pdf(n) * rvT.pdf(t)
ts = np.arange(15,25,0.2)
ns = np.arange(700,1300,10)
Ts, Ns = np.meshgrid(ts,ns)
fNT3nt = np.vectorize(fNT3)(Ns,Ts)
fig3 = plt.figure()
ax3 = fig3.add_subplot(111)
ax3.contourf(Ts, Ns, fNT3nt, cmap="Greys")
<matplotlib.contour.QuadContourSet at 0x7f56c09751b0>
Useful probability formulas#
Multivariable expectation#
Independent, identically distributed random variabls#
Discussion#
Bulk of the normal distribution#
How much of the total probability “weight” lies within \(k\) standard deviations of the mean?
from scipy.integrate import quad
from scipy.stats import norm
rvN = norm(1000, 100)
mu = rvN.mean() # mean of the random variable
sigma = rvN.std() # standard deviaiton of the random variable
for k in [1, 2, 3]:
I_k = [mu - k*sigma, mu + k*sigma]
p_k = quad(rvN.pdf, *I_k)[0]
print(f"p_{k} = Pr( N in {I_k} ) = {p_k:.3f}")
p_1 = Pr( N in [900.0, 1100.0] ) = 0.683
p_2 = Pr( N in [800.0, 1200.0] ) = 0.954
p_3 = Pr( N in [700.0, 1300.0] ) = 0.997
The code below highlights the interval \(I_k\) and computes the probability \(p_k\). Change the value of \(k\) to get different plots.
from ministats import calc_prob_and_plot
k = 2 # number of standard deviations around the mean
# values of x in the interval 𝜇 ± k𝜎 = [𝜇-k𝜎, 𝜇+k𝜎]
I_k = [mu-k*sigma, mu+k*sigma]
p_k, _ = calc_prob_and_plot(rvN, *I_k)
p_k
0.9544997361036417
Try changing the value of the variable k to 1 or 3 in the above code cell.
Tails of the normal distribution#
We’re often interested in tail ends of the distribution, which contain the unlikely events.
mu = rvN.mean() # mean of the random variable
sigma = rvN.std() # standard deviaiton of the random variable
for k in [1, 2, 3]:
# compute the probability in the left tail (-∞,𝜇-k𝜎]
x_l = mu - k*sigma
p_l = quad(rvN.pdf, rvN.ppf(0.0000000001), x_l)[0]
# compute the probability in the right tail [𝜇+k𝜎,∞)
x_r = mu + k*sigma
p_r = quad(rvN.pdf, x_r, rvN.ppf(0.9999999999))[0]
# add together to get total probability in the tails
p_tails = p_l + p_r
print(f"Pr( N<{x_l} or N>{x_r} ) = {p_tails:.4f}")
Pr( N<900.0 or N>1100.0 ) = 0.3173
Pr( N<800.0 or N>1200.0 ) = 0.0455
Pr( N<700.0 or N>1300.0 ) = 0.0027
The code below highlights the tails of the distribution and computes the sum of their probability.
from ministats import calc_prob_and_plot_tails
mu = rvN.mean() # mean of the random variable
sigma = rvN.std() # standard deviaiton of the random variable
k = 2 # number of standard deviations around the mean
# the distribution's left tail (-∞,𝜇-k𝜎]
x_l = mu - k*sigma
# the distribution's right tail [𝜇+k𝜎,∞)
x_r = mu + k*sigma
p_tails, _ = calc_prob_and_plot_tails(rvN, x_l, x_r, xlims=[600, 1400])
print(f"Pr( {{N<{x_l}}} ∪ {{N>{x_r}}} ) = {p_tails:.4f}")
Pr( {N<800.0} ∪ {N>1200.0} ) = 0.0455
Try changing the value of the variable k in the above code cell.
The above calculations leads us to an important rule of thumb: the values of the 5% tail of the distribution are \(k=2\) standard deviations away from the mean (more precisely, we should use \(k=1.96\) to get exactly 5%). We’ll use this facts later in STATS to define a cutoff value for events that are unlikely to occur by chance.