from __future__ import print_function, division %matplotlib inline %precision 6 import warnings warnings.filterwarnings('ignore') from thinkbayes2 import Pmf, Cdf import thinkplot import numpy as np from numpy.fft import fft, ifft
from inspect import getsourcelines def show_code(func): lines, _ = getsourcelines(func) for line in lines: print(line, end='')
One of the recurring themes of my books is the use of object-oriented programming to explore mathematical ideas. Many mathematical entities are hard to define because they are so abstract. Representing them in Python puts the focus on what operations each entity supports -- that is, what the objects can do -- rather than on what they are.
In this notebook, I explore the idea of a probability distribution, which is one of the most important ideas in statistics, but also one of the hardest to explain.
To keep things concrete, I'll start with one of the usual examples: rolling dice. When you roll a standard six-sided die, there are six possible outcomes -- numbers 1 through 6 -- and all outcomes are equally likely.
If you roll two dice and add up the total, there are 11 possible outcomes -- numbers 2 through 12 -- but they are not equally likely. The least likely outcomes, 2 and 12, only happen once in 36 tries; the most likely outcome happens 1 times in 6.
And if you roll three dice and add them up, you get a different set of possible outcomes with a different set of probabilities.
What I've just described are three random number generators, which are also called random processes. The output from a random process is a random variable, or more generally a set of random variables. And each random variable has probability distribution, which is the set of possible outcomes and the corresponding set of probabilities.
There are many ways to represent a probability distribution. The most obvious is a probability mass function, or PMF, which is a function that maps from each possible outcome to its probability. And in Python, the most obvious way to represent a PMF is a dictionary that maps from outcomes to probabilities.
thinkbayes2 provides a class called
Pmf that represents a probability mass function. Each
Pmf contains a dictionary named
d that contains the values and probabilities. To show how this class is used, I'll create a
Pmf that represents a six-sided die:
d6 = Pmf() for x in range(1, 7): d6[x] = 1 d6.Print()
1 1 2 1 3 1 4 1 5 1 6 1
Initially the "probabilities" are all 1, so the total probability in the
Pmf is 6, which doesn't make a lot of sense. In a proper, meaningful, PMF, the probabilities add up to 1, which implies that one outcome, and only one outcome, will occur (for any given roll of the die).
We can take this "unnormalized" distribution and make it a proper
Pmf using the
Normalize method. Here's what the method looks like:
def Normalize(self, fraction=1): """Normalizes this PMF so the sum of all probs is fraction. Args: fraction: what the total should be after normalization Returns: the total probability before normalizing """ if self.log: raise ValueError("Normalize: Pmf is under a log transform") total = self.Total() if total == 0: raise ValueError('Normalize: total probability is zero.') factor = fraction / total for x in self.d: self.d[x] *= factor return total
Normalize adds up the probabilities in the PMF and divides through by the total. The result is a
Pmf with probabilities that add to 1.
Here's how it's used:
1 0.16666666666666666 2 0.16666666666666666 3 0.16666666666666666 4 0.16666666666666666 5 0.16666666666666666 6 0.16666666666666666
The fundamental operation provided by a
Pmf is a "lookup"; that is, we can look up an outcome and get the corresponding probability.
__getitem__, so we can use bracket notation to look up an outcome:
And if you look up a value that's not in the
Pmf, the probability is 0.
Exerise: Create a
Pmf that represents a six-sided die that is red on two sides and blue on the other four.
# Solution die = Pmf(dict(red=2, blue=4)) die.Normalize() die.Print()
blue 0.6666666666666666 red 0.3333333333333333
So is a
Pmf a distribution? No. At least in this framework, a
Pmf is one of several representations of a distribution. Other representations include the cumulative distribution function, or CDF, and the characteristic function.
These representations are equivalent in the sense that they all contain the same informaton; if I give you any one of them, you can figure out the others (and we'll see how soon).
So why would we want different representations of the same information? The fundamental reason is that there are many different operations we would like to perform with distributions; that is, questions we would like to answer. Some representations are better for some operations, but none of them is the best for all operations.
So what are the questions we would like a distribution to answer? They include:
Each of these questions corresponds to a method we would like a distribution to provide. But as I said, there is no one representation that answers all of them easily and efficiently. So let's look at the different representations and see what they can do.
Getting back to the
Pmf, we've already seen how to look up the probability of a given outcome. Here's the code:
def __getitem__(self, value): return self.d.get(value, 0)
Python dictionaries are implemented using hash tables, so we expect
__getitem__ to be fast. In terms of algorithmic complexity, it is constant time, or $O(1)$.
Pmf representation is also good for computing mean, variance, and other moments. Here's the implementation of
def Mean(self): """Computes the mean of a PMF. Returns: float mean """ return sum(p * x for x, p in self.Items())
This implementation is efficient, in the sense that it is $O(n)$, and because it uses a comprehension to traverse the outcomes, the overhead is low.
The implementation of
Pmf.Var is similar:
def Var(self, mu=None): """Computes the variance of a PMF. mu: the point around which the variance is computed; if omitted, computes the mean returns: float variance """ if mu is None: mu = self.Mean() return sum(p * (x-mu)**2 for x, p in self.Items())
And here's how they are used:
The structure of
Var is the same: they traverse the outcomes and their probabilities,
p, and add up the product of
p and some function of
We can generalize this structure to compute the expectation of any function of
x, which is defined as
$E[f] = \sum_x p(x) f(x)$
Expect, which takes a function object,
func, and returns the expectation of
def Expect(self, func): """Computes the expectation of func(x). Returns: expectation """ return np.sum(p * func(x) for x, p in self.Items())
As an example, we can use
Expect to compute the third central moment of the distribution:
mu = d6.Mean() d6.Expect(lambda x: (x-mu)**3)
Because the distribution is symmetric, the third central moment is 0.
The next question we'll answer is the last one on the list: if we run two random processes and add up the results, what is the distribution of the sum? In other words, if the result of the first process is a random variable, $X$, and the result of the second is $Y$, what is the distribution of $X+Y$?
Pmf representation of the distribution can answer this question pretty well, but we'll see later that the characteristic function is even better.
Here's the implementation:
def AddPmf(self, other): """Computes the Pmf of the sum of values drawn from self and other. other: another Pmf returns: new Pmf """ pmf = Pmf() for v1, p1 in self.Items(): for v2, p2 in other.Items(): pmf.Incr(v1 + v2, p1 * p2) return pmf
The outer loop traverses the outcomes and probabilities of the first
Pmf; the inner loop traverses the second
Pmf. Each time through the loop, we compute the sum of the outcome pair,
v2, and the probability that the pair occurs.
Note that this method implicitly assumes that the two processes are independent; that is, the outcome from one does not affect the other. That's why we can compute the probability of the pair by multiplying the probabilities of the outcomes.
To demonstrate this method, we'll start with
d6 again. Here's what it looks like:
When we use the
+ operator, Python invokes
__add__, which invokes
AddPmf, which returns a new
Pmf object. Here's the
Pmf that represents the sum of two dice:
twice = d6 + d6 thinkplot.Pdf(twice, color='green')
And here's the
Pmf that represents the sum of three dice.
thrice = twice + d6 thinkplot.Pdf(d6) thinkplot.Pdf(twice, color='green') thinkplot.Pdf(thrice, color='red')
As we add up more dice, the result converges to the bell shape of the Gaussian distribution.
Exercise: If you did the previous exercise, you have a
Pmf that represents a die with red on 2 sides and blue on the other 4. Use the
+ operator to compute the outcomes of rolling two of these dice and the probabilities of the outcomes.
Note: if you represent the outcomes as strings,
AddPmf concatenates them instead of adding, which actually works.
# Solution dice = die + die dice.Print()
blueblue 0.4444444444444444 bluered 0.2222222222222222 redblue 0.2222222222222222 redred 0.1111111111111111
The next few questions on the list are related to the median and other percentiles. They are harder to answer with the
Pmf representation, but easier with a cumulative distribution function (CDF).
A CDF is a map from an outcome, $x$, to its cumulative probability, which is the probability that the outcome is less than or equal to $x$. In math notation:
$CDF(x) = Prob(X \le x)$
where $X$ is the outcome of a random process, and $x$ is the threshold we are interested in. For example, if $CDF$ is the cumulative distribution for the sum of three dice, the probability of getting 5 or less is $CDF(5)$, and the probability of getting 6 or more is $1 - CDF(5)$.
thinkbayes2 provides a class called Cdf that represents a cumulative distribution function. It uses a sorted list of outcomes and the corresponding list of cumulative probabilities. The
__init__ method is complicated because it accepts a lot of different parameters. The important part is the last 4 lines.
def __init__(self, obj=None, ps=None, label=None): """Initializes. If ps is provided, obj must be the corresponding list of values. obj: Hist, Pmf, Cdf, Pdf, dict, pandas Series, list of pairs ps: list of cumulative probabilities label: string label """ self.label = label if label is not None else DEFAULT_LABEL if isinstance(obj, (_DictWrapper, Cdf, Pdf)): if not label: self.label = label if label is not None else obj.label if obj is None: # caller does not provide obj, make an empty Cdf self.xs = np.asarray() self.ps = np.asarray() if ps is not None: logging.warning("Cdf: can't pass ps without also passing xs.") return else: # if the caller provides xs and ps, just store them if ps is not None: if isinstance(ps, str): logging.warning("Cdf: ps can't be a string") self.xs = np.asarray(obj) self.ps = np.asarray(ps) return # caller has provided just obj, not ps if isinstance(obj, Cdf): self.xs = copy.copy(obj.xs) self.ps = copy.copy(obj.ps) return if isinstance(obj, _DictWrapper): dw = obj else: dw = Hist(obj) if len(dw) == 0: self.xs = np.asarray() self.ps = np.asarray() return xs, freqs = zip(*sorted(dw.Items())) self.xs = np.asarray(xs) self.ps = np.cumsum(freqs, dtype=np.float) self.ps /= self.ps[-1]
xs is the sorted list of values, and
freqs are their frequencies or probabilities.
ps is the list of cumulative frequencies or probabilities, which we normalize by dividing through by the last element.
Here's how we use it to create a
Cdf object for the sum of three dice:
cdf = Cdf(thrice) cdf.Print()
3 0.00462962962963 4 0.0185185185185 5 0.0462962962963 6 0.0925925925926 7 0.162037037037 8 0.259259259259 9 0.375 10 0.5 11 0.625 12 0.740740740741 13 0.837962962963 14 0.907407407407 15 0.953703703704 16 0.981481481481 17 0.99537037037 18 1.0
Because we have to sort the values, the time to compute a
Cdf is $O(n \log n)$.
Here's what the CDF looks like:
The range of the CDF is always from 0 to 1.
Now we can compute $CDF(x)$ by searching the
xs to find the right location, or index, and then looking up the corresponding probability. Because the
xs are sorted, we can use bisection search, which is $O(\log n)$.
Probs, which takes an array of values and returns the corresponding probabilities:
def Probs(self, xs): """Gets probabilities for a sequence of values. xs: any sequence that can be converted to NumPy array returns: NumPy array of cumulative probabilities """ xs = np.asarray(xs) index = np.searchsorted(self.xs, xs, side='right') ps = self.ps[index-1] ps[xs < self.xs] = 0 return ps
The details here are a little tricky because we have to deal with some "off by one" problems, and if any of the values are less than the smallest value in the
Cdf, we have to handle that as a special case. But the basic idea is simple, and the implementation is efficient.
Now we can look up probabilities for a sequence of values:
cdf.Probs((2, 10, 18))
array([ 0. , 0.5, 1. ])
Cdf also provides
__getitem__, so we can use brackets to look up a single value:
Exercise: If you roll three dice, what is the probability of getting 15 or more?
# Solution 1 - cdf
You might wonder why I represent a
Cdf with two lists rather than a dictionary. After all, a dictionary lookup is constant time and bisection search is logarithmic. The reason is that we often want to use a
Cdf to do a reverse lookup; that is, given a probability, we would like to find the corresponding value. With two sorted lists, a reverse lookup has the same performance as a forward loopup, $O(\log n)$.
Here's the implementation:
def Values(self, ps=None): """Returns InverseCDF(p), the value that corresponds to probability p. If ps is not provided, returns all values. Args: ps: NumPy array of numbers in the range [0, 1] Returns: NumPy array of values """ if ps is None: return self.xs ps = np.asarray(ps) if np.any(ps < 0) or np.any(ps > 1): raise ValueError('Probability p must be in range [0, 1]') index = np.searchsorted(self.ps, ps, side='left') return self.xs[index]
And here's an example that finds the 10th, 50th, and 90th percentiles:
cdf.Values((0.1, 0.5, 0.9))
array([ 7, 10, 14])
Cdf representation is also good at generating random samples, by choosing a probability uniformly from 0 to 1 and finding the corresponding value. Here's the method
def Sample(self, n): """Generates a random sample from this distribution. n: int length of the sample returns: NumPy array """ ps = np.random.random(n) return self.ValueArray(ps)
The result is a NumPy array with the given
shape. The time to generate each random choice is $O(\log n)$
Here are some examples that use it.
array([10, 9, 6, 13, 14, 9])
array([[10, 7], [12, 14]])
Exercise: Write a function that takes a
Cdf object and returns the interquartile range (IQR), which is the difference between the 75th and 25th percentiles.
# Solution def iqr(cdf): values = cdf.Values((0.25, 0.75)) return np.diff(values) iqr(cdf)
Cdf representation is particularly good for finding the distribution of a maximum. For example, in Dungeons and Dragons, players create characters with random properties like strength and intelligence. The properties are generated by rolling three dice and adding them, so the CDF for each property is the
Cdf we used in this example. Each character has 6 properties, so we might wonder what the distribution is for the best of the six.
Here's the method that computes it:
def Max(self, k): """Computes the CDF of the maximum of k selections from this dist. k: int returns: new Cdf """ cdf = self.Copy() cdf.ps **= k return cdf
To get the distribution of the maximum, we make a new
Cdf with the same values as the original, and with the
ps raised to the
kth power. Simple, right?
To see how it works, suppose you generate six properties and your best is only a 10. That's unlucky, but you might wonder how unlucky. So, what is the chance of rolling 3 dice six times, and never getting anything better than 10?
Well, that means that all six values were 10 or less. The probability that each of them is 10 or less is $CDF(10)$, because that's what the CDF means. So the probability that all 6 are 10 or less is $CDF(10)^6$.
Now we can generalize that by replacing $10$ with any value of $x$ and $6$ with any integer $k$. The result is $CDF(x)^k$, which is the probability that all $k$ rolls are $x$ or less, and that is the CDF of the maximum.
Here's how we use
best = cdf.Max(6) thinkplot.Cdf(best);
So the chance of generating a character whose best property is 10 is less than 2%.
Exercise: Write a function that takes a CDF and returns the CDF of the minimum of
Hint: If the minimum is less than $x$, that means all
k values must be less than $x$.
# Solution def Min(cdf, k): return Cdf(cdf.xs, 1 - (1-cdf.ps)**k) worst = Min(cdf, 6) thinkplot.Cdf(worst);
At this point we've answered all the questions on the list, but I want to come back to addition, because the algorithm we used with the
Pmf representation is not as efficient as it could be. It enumerates all pairs of outcomes, so if there are $n$ values in each
Pmf, the run time is $O(n^2)$. We can do better.
The key is the characteristic function, which is the Fourier transform (FT) of the PMF. If you are familiar with the Fourier transform and the Convolution Theorem, keep reading. Otherwise, skip the rest of this cell and get to the code, which is much simpler than the explanation.
If you are familiar with the FT in the context of spectral analysis of signals, you might wonder why we would possibly want to compute the FT of a PMF. The reason is the Convolution Theorem.
It turns out that the algorithm we used to "add" two
Pmf objects is a form of convolution. To see how that works, suppose we are computing the distribution of $Z = X+Y$. To make things concrete, let's compute the probability that the sum, $Z$ is 5. To do that, we can enumerate all possible values of $X$ like this:
$Prob(Z=5) = \sum_x Prob(X=x) \cdot Prob(Y=5-x)$
Now we can write each of those probabilities in terms of the PMF of $X$, $Y$, and $Z$:
$PMF_Z(5) = \sum_x PMF_X(x) \cdot PMF_Y(5-x)$
And now we can generalize by replacing 5 with any value of $z$:
$PMF_Z(z) = \sum_x PMF_X(x) \cdot PMF_Y(z-x)$
You might recognize that computation as convolution, denoted with the operator $\ast$.
$PMF_Z = PMF_X \ast PMF_Y$
Now, according to the Convolution Theorem:
$FT(PMF_X \ast Y) = FT(PMF_X) \cdot FT(PMF_Y)$
Or, taking the inverse FT of both sides:
$PMF_X \ast PMF_Y = IFT(FT(PMF_X) \cdot FT(PMF_Y))$
In words, to compute the convolution of $PMF_X$ and $PMF_Y$, we can compute the FT of $PMF_X$ and $PMF_Y$ and multiply them together, then compute the inverse FT of the result.
Let's see how that works. Here's a class that represents a characteristic function.
import matplotlib.pyplot as plt class CharFunc: def __init__(self, hs): """Initializes the CF. hs: NumPy array of complex """ self.hs = hs def __mul__(self, other): """Computes the elementwise product of two CFs.""" return CharFunc(self.hs * other.hs) def make_pmf(self, thresh=1e-11): """Converts a CF to a PMF. Values with probabilities below `thresh` are dropped. """ ps = ifft(self.hs) d = dict((i, p) for i, p in enumerate(ps.real) if p > thresh) return Pmf(d) def plot_cf(self, **options): """Plots the real and imaginary parts of the CF.""" n = len(self.hs) xs = np.arange(-n//2, n//2) hs = np.roll(self.hs, len(self.hs) // 2) plt.plot(xs, hs.real, label='real', **options) plt.plot(xs, hs.imag, label='imag', **options) plt.legend()
hs, is the Fourier transform of the
Pmf, represented as a NumPy array of complex numbers.
The following function takes a dictionary that maps from outcomes to their probabilities, and computes the FT of the PDF:
def compute_fft(d, n=256): """Computes the FFT of a PMF of integers. Values must be integers less than `n`. """ xs, freqs = zip(*d.items()) ps = np.zeros(256) ps[xs,] = freqs hs = fft(ps) return hs
fft computes the Fast Fourier Transform (FFT), which is called "fast" because the run time is $O(n \log n)$.
Here's what the characteristic function looks like for the sum of three dice (plotting the real and imaginary parts of
hs = compute_fft(thrice.d) cf = CharFunc(hs) cf.plot_cf()
The characteristic function contains all of the information from the
Pmf, but it is encoded in a form that is hard to interpret. However, if we are given a characteristic function, we can find the corresponding
make_pmf, which uses the inverse FFT to get back to the
Pmf representation. Here's the code:
def make_pmf(self, thresh=1e-11): """Converts a CF to a PMF. Values with probabilities below `thresh` are dropped. """ ps = ifft(self.hs) d = dict((i, p) for i, p in enumerate(ps.real) if p > thresh) return Pmf(d)
And here's an example:
Now we can use the characteristic function to compute a convolution.
__mul__, which multiplies the
hs elementwise and returns a new
def __mul__(self, other): """Computes the elementwise product of two CFs.""" return CharFunc(self.hs * other.hs)
And here's how we can use it to compute the distribution of the sum of 6 dice.
sixth = (cf * cf).make_pmf() thinkplot.Pdf(sixth)
Here are the probabilities, mean, and variance.
6 2.14334705076e-05 7 0.000128600823045 8 0.000450102880659 9 0.00120027434842 10 0.00270061728395 11 0.0054012345679 12 0.00977366255144 13 0.0162037037037 14 0.0248842592593 15 0.0357081618656 16 0.0481610082305 17 0.0612139917695 18 0.0735382373114 19 0.0837191358025 20 0.0904706790123 21 0.0928497942387 22 0.0904706790123 23 0.0837191358025 24 0.0735382373114 25 0.0612139917695 26 0.0481610082305 27 0.0357081618656 28 0.0248842592593 29 0.0162037037037 30 0.00977366255144 31 0.0054012345679 32 0.00270061728395 33 0.00120027434842 34 0.000450102880658 35 0.000128600823045 36 2.14334705075e-05
This might seem like a roundabout way to compute a convolution, but it is efficient. The time to Compute the
CharFunc objects is $O(n \log n)$. Multiplying them together is $O(n)$. And converting back to a
Pmf is $O(n \log n)$.
So the whole process is $O(n \log n)$, which is better than
Pmf.__add__, which is $O(n^2)$.
Exercise: Plot the magnitude of
np.abs. What does that shape look like?
Hint: it might be clearer if you us
np.roll to put the peak of the CF in the middle.
#Solution n = len(cf.hs) mags = np.abs(cf.hs) plt.plot(np.roll(mags, n//2)) None # The result approximates a Gaussian curve because # the PMF is approximately Gaussian and the FT of a # Gaussian is also Gaussian
Finally, let's back to the question we started with: what is a distribution?
I've said that
CharFunc are different ways to represent the same information. For the questions we want to answer, some representations are better than others. But how should we represent the distribution itself?
One option is to treat each representation as a mixin; that is, a class that provides a set of capabilities. A distribution inherits all of the capabilities from all of the representations. Here's a class that shows what I mean:
class Dist(Pmf, Cdf, CharFunc): def __init__(self, d): """Initializes the Dist. Calls all three __init__ methods. """ Pmf.__init__(self, d) Cdf.__init__(self, d) CharFunc.__init__(self, compute_fft(d)) def __add__(self, other): """Computes the distribution of the sum using Pmf.__add__. """ pmf = Pmf.__add__(self, other) return Dist(pmf.d) def __mul__(self, other): """Computes the distribution of the sum using CharFunc.__mul__. """ pmf = CharFunc.__mul__(self, other).make_pmf() return Dist(pmf.d)
When you create a
Dist, you provide a dictionary of values and probabilities.
Dist.__init__ calls the other three
__init__ methods to create the
CharFunc representations. The result is an object that has all the attributes and methods of the three representations.
As an example, I'll create a
Dist that represents the sum of six dice:
dist = Dist(sixth.d) thinkplot.Pdf(dist)
Pmf, so we can look up the probability of a value.
We also get mean and variance from
But we can also use methods from
dist.ValueArray((0.25, 0.5, 0.75))
array([18, 21, 24])
dist.Probs((18, 21, 24))
[0.073538, 0.092850, 0.073538]
array([22, 19, 20, 21, 26, 26, 23, 16, 20, 19])
Pmf.__add__, which performs convolution the slow way:
twelfth = dist + dist thinkplot.Pdf(twelfth)
CharFunc.__mul__, which performs convolution the fast way.
twelfth_fft = dist * dist thinkplot.Pdf(twelfth_fft)
Either way, we get the answer, which is 42.
Abstractly, a distribution is an entity that can answer questions about the outcomes of random variables and their probabilities. There are many ways to represent a distribution; each representation is equivalent to the others in the sense that they contain the same information, and you can convert from any of them to the others.
Some representations make it easy and efficient to answer some questions, but none of the representations is best for all of the questions.
In my implementation, a
Dist object has the attributes and methods of all representations. From a software engineering point of view, that might not be the best design, but it is meant to illustrate what it means to be a distribution.
In short, if you give me any representation of a distribution, you have told me everything I need to know to answer questions about the possible outcomes and their probabilities. Converting from one representation to another is mostly a matter of convenience and computational efficiency.
Conversely, if you are trying to find the distribution of a random variable, you can do it by computing whichever representation is easiest to figure out.