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import matplotlib.pyplot as plt
%matplotlib inline
import matplotlib_inline
matplotlib_inline.backend_inline.set_matplotlib_formats('svg')
import seaborn as sns
sns.set_context("paper")
sns.set_style("ticks");
The Uniform Distribution#
The uniform distribution is the most common continuous distribution. It corresponds to a random variable equally likely to take a value within a given interval. We write:
\[
X\sim U([0,1]),
\]
and we read \(X\) follows a uniform distribution taking values in \([0,1]\).
The PDF of the uniform is constant in \([0,1]\) and zero outside it. We have:
\[\begin{split}
f_X(x) := U(x|[0,1]) := \begin{cases}
1,&\;0\le x \le 1,\\
0,&\;\text{otherwise}.
\end{cases}
\end{split}\]
Here is how you can make this random variable in scipy.stats:
import scipy.stats as st
X = st.uniform()
You can evaluate the PDF anywhere like this:
X.pdf(0.5)
1.0
X.pdf(-0.1)
0.0
Here is the plot of the PDF:
import numpy as np
x = np.linspace(-0.1, 1.1, 100)
plt.plot(x, X.pdf(x))
plt.xlabel('$x$')
plt.ylabel('$f_X(x)$')
plt.title('Probability density function of $U([0,1])$')
sns.despine(trim=True);
The CDF of the uniform is for \(x\) in \([0,1]\):
\[
F(x) = p(X \le x) = \int_0^x f_X(u) du = \int_0^x du = x.
\]
Obviously, we have \(F(x) = 0\) for \(x < 0\) and \(F(x) = 1\) for \(x > 1\).
If you have a scipy random variable, you can evaluate the CDF like this:
X.cdf(0.5)
0.5
And here is the plot of the CDF:
fig, ax = plt.subplots()
ax.plot(x, X.cdf(x))
ax.set_xlabel("$x$")
ax.set_ylabel("$F(x)$")
sns.despine(trim=True);
The probability that \(X\) takes values in \([a,b]\) for \(a < b\) in \([0,1]\) is:
\[
p(a \le X \le b) = F(b) - F(a) = b - a.
\]
The expectation of the uniform is:
\[
\mathbb{E}[X] = \int_0^1 xdx = \frac{1}{2}.
\]
The variance of the uniform is:
\[
\mathbb{V}[X] = \mathbb{E}[X^2] - \left(\mathbb{E}[X]\right)^2 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}.
\]
You can get the expectation from scipy like this:
print(f"E[X] = {X.expect():.2f}")
E[X] = 0.50
Similarly for the variance:
# The variance is:
print(f"V[X] = {X.var():.2f}")
V[X] = 0.08
Here is how you can sample from the uniform one hundred times:
X.rvs(size=100)
Show code cell output
array([0.09539232, 0.77524467, 0.1165174 , 0.17251136, 0.40433054,
0.72158 , 0.38796207, 0.06261768, 0.6408136 , 0.18536892,
0.35787089, 0.81968133, 0.03752348, 0.39551494, 0.24574092,
0.75824217, 0.9642853 , 0.06583551, 0.08171757, 0.25362503,
0.72914223, 0.8837704 , 0.02631866, 0.95030773, 0.7060529 ,
0.8930821 , 0.39822756, 0.82410637, 0.18128167, 0.55625658,
0.97856713, 0.06468715, 0.97174219, 0.94518788, 0.8766769 ,
0.02038775, 0.30426117, 0.17927524, 0.13101756, 0.2317204 ,
0.88418449, 0.45675778, 0.69759984, 0.04638995, 0.13877671,
0.68515328, 0.70644869, 0.76041054, 0.09915309, 0.99840908,
0.04972821, 0.58472795, 0.64568846, 0.28645003, 0.28803998,
0.16994 , 0.3151959 , 0.77308367, 0.37186481, 0.26694347,
0.73817172, 0.6150509 , 0.61194572, 0.80847813, 0.93691158,
0.67311239, 0.12722363, 0.18093929, 0.53266398, 0.68873768,
0.43517983, 0.69703678, 0.04487793, 0.96773282, 0.39516136,
0.6490977 , 0.06924699, 0.74477372, 0.17175712, 0.00946398,
0.21438731, 0.90366678, 0.22026249, 0.19304039, 0.63719114,
0.02878915, 0.63537944, 0.66062147, 0.64241864, 0.9159541 ,
0.32671907, 0.7661052 , 0.91168297, 0.41423243, 0.07588457,
0.40145792, 0.67562391, 0.77792787, 0.59545402, 0.10586877])
An alternative way is to use the functionality of numpy:
np.random.rand(100)
Show code cell output
array([0.80586055, 0.11824385, 0.16970896, 0.23699974, 0.37712543,
0.95392772, 0.3167166 , 0.29377652, 0.03252983, 0.12010956,
0.50480532, 0.74594687, 0.46786984, 0.23535419, 0.76501325,
0.4868536 , 0.24281247, 0.56834828, 0.73510174, 0.65209475,
0.40019549, 0.51742407, 0.90883814, 0.07134155, 0.81721201,
0.03008084, 0.18048499, 0.60085674, 0.04253718, 0.2604787 ,
0.21244718, 0.97700233, 0.4507617 , 0.5235795 , 0.66534106,
0.89016926, 0.65434185, 0.97440304, 0.31200708, 0.08642 ,
0.82021427, 0.66455954, 0.72986889, 0.60272292, 0.59986421,
0.71106608, 0.02076019, 0.644577 , 0.21984685, 0.83109709,
0.28888499, 0.19897096, 0.35633719, 0.31888049, 0.88103772,
0.61011618, 0.78859155, 0.92172975, 0.09794496, 0.26354898,
0.88442024, 0.14780904, 0.72500108, 0.74646945, 0.64247196,
0.89706545, 0.28715704, 0.46458434, 0.76452549, 0.07750805,
0.93848269, 0.56619284, 0.36794383, 0.37993545, 0.39500912,
0.18791823, 0.96039442, 0.82659114, 0.12148711, 0.99411616,
0.95130704, 0.20910239, 0.27437508, 0.5050475 , 0.93813519,
0.83777349, 0.8593612 , 0.26183294, 0.45411316, 0.13210563,
0.76234838, 0.93534534, 0.46668538, 0.81264633, 0.5485471 ,
0.99254412, 0.54846434, 0.2541332 , 0.7283944 , 0.92028981])
Finally, let’s find the probability that X is between two numbers. In particular, we will find \(p(-1 \le X \le 0.3)\):
a = -1.0
b = 0.3
prob_X_is_in_ab = X.cdf(b) - X.cdf(a)
print(f"p({a:.2f} <= X <= {b:.2f}) = {prob_X_is_in_ab:.2f}")
p(-1.00 <= X <= 0.30) = 0.30
The uniform distribution over an arbitrary interval \([a, b]\)#
One can define a uniform distribution over an arbitrary interval \([a,b]\). We write:
\[
X \sim U([a, b]).
\]
The PDF of this random variable is:
\[\begin{split}
f_X(x) = \begin{cases}
c,&\;x\in[a,b],\\
0,&\;\text{otherwise},
\end{cases}
\end{split}\]
where \(c\) is a positive constant. The formula tells us that the probability density of finding \(X\) in \([a,b]\) is positive and that the probability density of finding outside is zero. We can determine the positive constant \(c\) by imposing the normalization condition:
\[
\int_{-\infty}^{+\infty}f_X(x)dx = 1.
\]
Carrying out the integral:
\[
1 = \int_{-\infty}^{+\infty}p(x)dx = \int_a^bc dx = c \int_a^bdx = c (b-a).
\]
Therefore:
\[
c = \frac{1}{b - a},
\]
and we can now write:
\[\begin{split}
f_X(x) = \begin{cases}
\frac{1}{b-a},&x \in [a, b],\\
0,&\;\text{otherwise},
\end{cases}
\end{split}\]
From the PDF, we can now find the CDF for \(x \in [a,b]\):
\[
F(x) = p(X\le x) = \int_{-\infty}^x f_X(u)du = \int_a^x \frac{1}{b-a}du = \frac{1}{b-a}\int_a^xdu = \frac{x-a}{b-a}.
\]
The expectation is:
\[
\mathbb{E}[X] = \frac{1}{2}(a+b),
\]
and the variance is:
\[
\mathbb{V}[X] = \frac{1}{12}(b-a)^2.
\]
Here is how you can do this using scipy.stats for \(a=-2\) and \(b=5\):
a = -2.0
b = 5.0
X = st.uniform(loc=a, scale=(b-a))
The PDF is:
Show code cell source
fig, ax = plt.subplots()
xs = np.linspace(a - 0.1, b + 0.1, 100)
ax.plot(xs, X.pdf(xs))
ax.set_xlabel("$x$")
ax.set_ylabel("$f_X(x)$")
sns.despine(trim=True);
The CDF is:
Show code cell source
fig, ax = plt.subplots()
xs = np.linspace(a - 0.1, b + 0.1, 100)
ax.plot(xs, X.cdf(xs))
ax.set_xlabel("$x$")
ax.set_ylabel("$F(x)$")
sns.despine(trim=True);
The expectation is:
print(f"E[X] = {X.expect():.2f}")
E[X] = 1.50
And the variance:
# The variance is:
print(f"V[X] = {X.var():.2f}")
V[X] = 4.08
And here are a few random samples:
X.rvs(size=100)
Show code cell output
array([ 1.44828788, 3.41171026, 2.87684174, 4.8983561 , -0.1312805 ,
-1.70245574, 2.26979439, 1.6466782 , 4.34652482, 0.35543454,
4.53818962, -1.33718623, 0.04536362, 0.41322555, 3.78249496,
4.68216223, 3.28923988, 2.60780789, 4.18933089, 4.00611092,
2.57420565, 4.94090822, 1.98921428, -1.49104842, 1.08427018,
0.77995007, -0.18844224, 0.4527731 , -0.52566889, 1.92956296,
4.02993221, 4.41533211, 2.00825063, 3.67496822, -1.74089817,
4.37543734, 3.18883896, 2.05146579, -1.54908546, 3.57873337,
-1.53078455, 4.10284175, -1.25869879, 4.39609846, 4.08372762,
1.59054529, -1.88254861, -1.31531624, 2.72709832, -1.72031435,
4.16180122, 1.2298621 , 0.90993575, -1.44662087, 4.11556168,
-0.02735795, 2.34190311, 0.95310295, 3.94416729, -0.37474025,
2.89650481, 0.32614986, 1.87890769, 2.30208277, 1.70514826,
-1.2640151 , 3.98274136, -0.51666423, 0.38175858, 4.50485523,
2.59398841, 4.56469552, 1.21358654, 2.88875989, 4.69242704,
-0.4574542 , 0.90836318, -1.53365789, -0.98508496, 2.78899496,
-1.44765717, -1.48304314, 4.30368088, 0.29395015, 3.02711301,
4.31544499, -1.99617301, 2.12661733, 1.95365674, -0.28018737,
3.70934999, -0.09200282, 1.98819365, 3.75772142, 4.32743015,
3.77284658, -0.83542659, 0.56102761, 2.29900879, -0.36813604])
Alternative way to get \(U([a,b])\)#
There is another way to obtain samples from \(U([a,b])\) that uses only samples from \(U([0,1])\). Here is how. Let \(Z\) be a standard uniform random variable:
\[
Z\sim U([0,1]).
\]
Then define the random variable:
\[
X = a + (b-a) Z.
\]
Then, \(X\sim U([a,b])\). Why? Well, let’s just show that the CDF of \(X\) has the right form:
\[
p(X \le x) = p(a + (b-a)Z \le x) = p((b-a)Z \le x - a) = p\left(Z \le \frac{x-a}{b-a}\right) = \frac{x-a}{b-a},
\]
where the last step follows from the fact that the CDF of \(Z\) is simply: \(p(Z \le z) = z\).
Equipped with this result, we can sample \(X\) by sampling \(Z\) and then scaling it appropriately (by the way, this is what scipy.stats is doing internally). Here it is using numpy.random.rand to sample in \([0,1]\):
x_samples = a + (b - a) * np.random.rand(1000)
print(x_samples)
Show code cell output
[-1.17968315e+00 -9.91679588e-01 1.34411789e+00 1.14593570e+00
6.37846308e-01 -3.55504159e-01 -1.33171178e+00 1.18303712e+00
3.20876200e+00 3.19667735e+00 -1.27515305e+00 2.51508002e+00
6.84081866e-01 2.44487554e+00 4.91615360e+00 3.09205003e+00
1.44013657e+00 2.15047448e+00 -1.30341195e+00 -4.41178353e-01
2.59401696e+00 1.98714760e+00 2.01425439e+00 4.96415606e+00
4.24264777e+00 4.88138233e+00 -1.55005597e+00 1.62766182e+00
-5.59193830e-01 4.59853007e-01 -1.79581400e+00 2.14846678e+00
3.30474256e+00 -1.42148694e+00 1.61251214e+00 -1.14294509e+00
-9.76485259e-01 3.14474495e-01 -1.58248222e+00 1.78018802e-01
4.13978083e+00 -7.66317089e-01 -2.41124441e-01 -1.14919158e+00
5.44683158e-01 -1.54367814e-01 3.16869882e+00 2.55781126e+00
4.81829577e+00 4.12587892e+00 1.03292936e+00 1.59955571e+00
3.25948809e+00 2.64536298e+00 4.23719619e+00 3.33727502e+00
-1.00696879e-01 3.57550671e+00 1.51327194e+00 1.04445845e+00
-1.02823256e+00 3.63510969e+00 1.79652898e+00 1.56104435e+00
2.50746351e+00 2.67487158e+00 2.20292496e+00 3.90029262e+00
-1.17324854e+00 -1.18876164e+00 1.19950558e+00 -1.65240941e-01
2.32275869e+00 2.47379659e-01 8.92987479e-01 7.37173769e-01
2.77864400e+00 4.97192317e+00 3.08384800e-02 1.23544520e+00
3.64543947e+00 -3.69787503e-01 4.38784321e+00 1.12058327e-01
2.48687455e+00 -1.36717106e+00 1.25885672e+00 1.03121999e+00
3.91622677e+00 1.48368014e+00 4.42145309e+00 3.61980557e+00
4.61402820e+00 3.61519450e+00 2.16269649e-01 1.64432456e+00
1.24830833e+00 1.73655333e+00 3.05957195e+00 3.58732430e+00
-6.99010100e-01 3.50998068e+00 4.81228913e+00 1.82948753e+00
2.02008991e+00 -1.40290818e+00 4.75843655e+00 -1.27289899e+00
-4.84665557e-02 3.09671948e+00 4.13372571e+00 -1.96571316e+00
-1.59265249e+00 -8.50618471e-02 -5.48769674e-01 4.23469323e+00
4.88158664e+00 -1.34303456e+00 3.50968782e+00 4.62135837e+00
4.56286817e+00 3.52011956e+00 9.30428299e-01 1.66177549e+00
2.26345132e+00 4.71694438e+00 -8.64846133e-01 2.74455730e+00
4.13924095e-01 5.93697461e-01 2.48922515e+00 7.29692294e-01
6.92201518e-01 -8.76923231e-01 3.84216815e+00 3.68478361e+00
3.99199047e+00 -7.10897975e-01 -1.10915113e+00 4.84954354e+00
2.50872947e+00 -1.12522238e-02 1.85080108e+00 -1.17286785e+00
-7.02283619e-01 1.15886604e+00 3.95732753e+00 1.89822021e+00
-1.28627619e+00 -1.46130076e+00 3.79115282e+00 -2.90722078e-01
-1.48284716e+00 4.45962731e+00 2.28789043e-01 3.12628672e+00
2.65737999e+00 2.31442143e+00 4.18292470e+00 2.52586745e+00
-1.93987682e+00 3.76118238e+00 -1.79062515e+00 2.88817016e+00
-8.49007671e-01 3.46449930e+00 -1.23173168e+00 -4.03804837e-01
1.81314008e+00 4.54591476e+00 1.44679810e+00 1.61688406e+00
-1.64820579e+00 3.99238562e+00 2.87893082e-01 -9.99431046e-01
2.91072152e+00 -1.50827027e-01 9.63256245e-01 1.31340762e+00
4.10862356e+00 4.74900253e+00 -9.96871664e-01 -1.66698853e+00
2.45847924e+00 -4.10866461e-01 3.02009409e+00 3.95800337e+00
2.91496150e+00 -1.64498094e+00 -2.96170832e-01 3.68000716e-01
4.33827706e+00 3.75519415e+00 1.39413566e+00 -6.70876230e-01
6.64632082e-01 4.46970515e+00 3.07917656e+00 8.60269017e-01
4.68654139e+00 3.40511540e+00 -1.03833445e+00 -1.62963673e-01
-4.64769991e-01 5.27086568e-01 1.62677553e+00 2.09679050e+00
1.99516774e+00 3.83995224e+00 2.48795436e-01 4.60076713e+00
5.86476646e-01 -7.01901275e-01 4.94013081e+00 1.89350669e+00
2.51934322e+00 4.56646324e+00 -3.14754974e-01 -9.72362524e-01
1.85137701e+00 2.57042174e+00 -1.10693823e+00 -6.29945840e-01
2.62747590e+00 4.03724437e+00 -3.35904085e-01 4.92277153e+00
3.77833210e+00 2.99047110e+00 2.85609328e+00 2.60528917e+00
3.33690449e+00 -1.77260483e+00 -7.36628168e-01 -3.46722992e-01
4.94538524e+00 3.44638989e+00 -1.10465184e+00 1.92122119e+00
3.27221482e+00 2.98461427e+00 3.68561931e+00 6.45502620e-01
2.63160190e+00 2.82349001e+00 3.62051805e+00 -1.58135309e+00
2.35474995e+00 -1.73016667e+00 1.54126125e+00 -9.38997653e-01
2.59732064e+00 1.10902090e+00 1.03304785e+00 3.48388935e+00
-1.14124501e+00 2.14887295e+00 -1.63822445e+00 2.57059233e+00
3.32360432e-02 -1.43902543e+00 5.52638193e-01 -5.12670429e-01
2.82026753e+00 -9.91218462e-02 2.43082384e+00 -2.45619298e-01
-1.70300116e+00 3.34293495e+00 1.95664920e+00 2.92331577e+00
4.56635306e+00 -4.98279946e-01 -1.04653454e+00 4.74972290e-01
6.65034502e-01 2.65243000e+00 2.43681424e+00 1.75301147e+00
3.57944551e+00 4.19371942e+00 4.83178144e+00 -8.96606131e-01
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1.14470014e+00 1.81623020e+00 2.75448351e+00 1.89109281e+00
3.09898592e+00 -3.01292075e-01 3.25455867e+00 4.81881825e+00
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3.76885391e+00 1.55656277e+00 2.69930222e+00 -1.63759912e-01
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4.99473288e+00 8.55607816e-01 1.30541738e+00 2.30127458e+00
3.06510865e+00 -1.51599947e-01 3.49814358e-01 -1.67084304e+00
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-1.45816387e-01 3.96240617e+00 2.62051307e+00 1.31791778e+00
3.09028980e+00 9.01557115e-01 2.58879839e+00 -3.29452981e-01
-9.07002460e-01 1.35883064e+00 -1.66821428e+00 2.47880300e+00
2.79033856e+00 4.02906211e+00 2.05679809e+00 -3.05864192e-01
3.71065325e+00 3.70183402e+00 3.04348528e+00 9.99156148e-01
2.77439554e+00 -9.24842652e-01 1.08952793e+00 3.61516544e+00
1.71853445e+00 4.65637095e+00 3.07562353e+00 -1.15923567e+00
1.13099281e+00 1.84052542e+00 3.73088838e+00 1.72587936e-01
-1.86799416e+00 3.95129583e+00 3.23722970e+00 2.81583500e+00]
Let’s also do the histogram of x_samples to make sure they are distributed the right way:
fig, ax = plt.subplots()
ax.plot(xs, X.pdf(xs), label="True PDF")
ax.hist(x_samples, density=True, alpha=0.25, label="Histogram",
color=sns.color_palette()[0])
ax.set_xlabel("$x$")
ax.set_ylabel("$f_X(x)$")
plt.legend(loc="best")
sns.despine(trim=True);
Questions#
Rerun the code above so that the random variable is \(U([1, 10])\).