Lecture 4: Probability Theory and Probabilistic Modeling
Introduction to Machine Learning Semester 1, 2022
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Last time… Concepts and KNN classification
• data, features, classes
• K Nearest Neighbors algorithm • Application to classification
Today… Probability
• basics / refresher
• distributions and parameterizations • why probability in ML?
Estimating confidence in different possible outcomes
Probability Theory
“The calculus of probability theory provides us with a formal framework for considering multiple possible outcomes and their likelihood. It defines a set of mutually exclusive and exhaustive possibilities, and associates each of them with a probability — a number between 0 and 1, so that the total probability of all possibilities is 1. This framework allows us to consider options that are unlikely, yet not impossible, without reducing our conclusions to content-free lists of every possibility.”
From Probabilistic Graphical Models: Principles and Techniques (2009; Koller and Friedman) http://pgm.stanford.edu/intro.pdf
(Very) Basics of Probability Theory
P(A=a): the probability that random variable A takes value a
0 <= P(A=a) <= 1
the fraction of times the event A=a is true in independent trials
P(True) = 1 P(False) = 0
(Very) Basics of Probability Theory
P(A=a): the probability that random variable A takes value a
0 <= P(A=a) <= 1
Given a deck of 52 cards
the fraction of times the event A=a is true in independent trials
P(True) = 1 P(False) = 0
• 13 ranks (ace, 2-10, jack, queen, king)
• of each of four suits (clubs, spades = black; hearts, diamonds = red)
• A is a random variable denoting the value of a randomly selected card.
https://commons.wikimedia.org/wiki/File:English pattern playing cards deck.svg
(Very) Basics of Probability Theory
P(A=a): the probability that random variable A takes value a
0 <= P(A=a) <= 1
Given a deck of 52 cards
the fraction of times the event A=a is true in independent trials
P(True) = 1 P(False) = 0
• 13 ranks (ace, 2-10, jack, queen, king)
• of each of four suits (clubs, spades = black; hearts, diamonds = red)
• A is a random variable denoting the value of a randomly selected card.
P(A = queen) = P(A = red) = P(A = heart) =
Joint Probability
P(A, B): joint probability of the probability of both A and two events A and B B occurring = P(A ∩ B)
P(A = ace,B = heart) = P(A = heart,B = red) =
Joint Probability
P(A, B): joint probability of the probability of both A and two events A and B B occurring = P(A ∩ B)
P(A = ace,B = heart) = P(A = heart,B = red) =
P(A, B) = P(A) ∗ P(B) iff A and B are independent
Conditional Probability
P(A|B): conditional probability the probability of A=a given
the observation B=b = P(A∩B) P(B)
P(A = ace|B = heart) = P(A = heart|B = red) =
1. P(A = x) probability that random variable A takes on value x
2. P(A) probability distribution over random variable A
3. P(x) I’ll often use this as a short-hand for 1. if clear from the context
What type of probability?
https://imgs.xkcd.com/comics/conditional risk.png
Rules of Probability I
• Independence: A and B are independent iff P(A ∩ B) = P(A)P(B) • Disjoint events: The probability of two disjoint events, such that
A ∩ B = ∅, is P(A or B) = P(A) + P(B)
• Product rule: P(A ∩ B) = P(A|B)P(B) = P(B|A)P(A) • Chain rule:
P(A1 ∩...∩An)=P(A1)P(A2|A1)P(A3|A2 ∩A1)...P(An|∩n−1 Ai) i=1
Rules of Probability I
• Independence: A and B are independent iff P(A ∩ B) = P(A)P(B) • Disjoint events: The probability of two disjoint events, such that
e.g., draw an ace or a king: A= A ∩ B = ∅, is P(A or B) = P(A) + P(B)
draw an ace; B=draw a king.
• Product rule: P(A ∩ B) = P(A|B)P(B) = P(B|A)P(A) • Chain rule:
P(A1 ∩...∩An)=P(A1)P(A2|A1)P(A3|A2 ∩A1)...P(An|∩n−1 Ai) i=1
Rules of Probability I
• Independence: A and B are independent iff P(A ∩ B) = P(A)P(B) • Disjoint events: The probability of two disjoint events, such that
e.g., draw an ace or a king: A= A ∩ B = ∅, is P(A or B) = P(A) + P(B)
draw an ace; B=draw a king. • Product rule: P(A ∩ B) = P(A|B)P(B) = P(B|A)P(A)
• Chain rule:
P(A1 ∩...∩An)=P(A1)P(A2|A1)P(A3|A2 ∩A1)...P(An|∩n−1 Ai)
again, we can choose the factorization, e.g., :
P(July, 5◦C, sick) = P(July) × P(5◦C|July) × P(sick|5◦C, July) = P(5◦C) × P(sick|5◦C) × P(July|5◦C, sick)
makes sense
Rules of Probability II
Bayes Rule
P(A|B) = P(A)P(B|A) (cf., P(A|B)= P(A∩B) ) P(B) P(B)
Basic rule of probability
• Bayes’ Rule allows us to compute P(A|B) given knowledge of the ‘inverse’ probability P(B|A).
More philosophically,
• Bayes’ Rule allows us to update prior belief with empirical evidence
Rules of Probability II
Bayes Rule
P(A|B) = P(A)P(B|A) P(B)
(cf., P(A|B)= P(A∩B) ) P(B)
Posterior Probability P(A|B)
• the degree of belief having accounted for B.
Prior Probability P(A)
• the initial degree of belief in A.
• the probability of A occurring, given no additional knowledge about A Likelihood P(B|A)
• the support B provides for A
Normalizing constant (‘Evidence’) P(B) = A P(B|A)P(A)
Rules of Probability II
Bayes Rule
P(A|B) = P(A)P(B|A) (cf., P(A|B)= P(A∩B) ) P(B) P(B)
Estimate the probability of a student being smart given that (s)he achieved H1 score, P(smart|H1) from the following information:
P(Smart) = 0.3 P(H1|Smart) = 0.6 P(H1) = 0.2
(What if P(H1) = 0.4?)
prior rate of smart students empirically measured H1|smart emprirically measured
Gaussian Distribution
Models the probability of an event that follows a typical pattern, but with some random variation.
• A continuous probability distribution over real-valued values. • E.g.,
It has two parameters
• The mean μ: what is the most typical value
• The standard deviation σ: the spread around the mean (σ2 is the variance)
1 − 1 ( x−μ )2 •P(X=x|μ,σ)=σ√2πe 2 σ
https://en.wikipedia.org/wiki/Normal_distribution#/media/File:Normal_Distribution_PDF.svg
Binomial Distributions
• A binomial distribution results from a
series of independent trials with only two outcomes (aka Bernoulli trials)
e.g. multiple coin tosses (⟨H, T , H, H, ..., T ⟩)
• The probability P of an event with probability p occurring exactly m out of
n times is given by
P(m,n,p) = m pm(1−p)n−m
P(m,n,p) = n! pm (1−p)n−m m!(n − m)!
m successes n − m failures possible distributions
of m successes over n trials
What is the probability of getting times 2 heads out of 3 tosses of a fair coin?
Binomial Example: Coin Toss
Go through solution:
What is the probability of getting times 2 heads out of 3 tosses of a fair coin?
Binomial Example: Coin Toss [SOLUTION]
What is the probability of getting times 2 heads out of 3 tosses of a fair coin?
1. m = 2 successes (heads) when flipping coin n = 3 times; P(X = 2)
2. number of possible outcomes e from 3 coin flips: 2∗2∗2=23 =8 eachwithP(e)= 18
3. Choose2outof3: C(3,2)= 3! =3 2!1!
4. 3 possible outcomes, 81 for each: P(X = 2) = 83
1 3! 12 13−2
P 2,3,2 = 2!(3−2)! 2 2
P(m, n, p) = n! pm(1 − p)n−m m!(n−m)!
11 =3 4 2
Multinomial Distributions
• A multinomial distribution models the probability of counts of different events from a series of independent trials with more than two possible outcomes, e.g.,
- a fair 6-sided dice is rolled 5 times
- what is the probability of observing exactly 3 ’ones’ and 2 ’fives’? - what is the probability of observing 5 ’threes’?
• TheprobabilityofeventsX1,X2,...,Xn withprobabilitiesp=p1,p2,...,pn occurringexactlyx1,x2,...,xn times,respectively,isgivenby
P(X1 = x1,X2 = x2,...,Xn = xn;p) = (i xi)!px1 × px2 × ··· × pxn x1!...xn! 1 2 n
= (i xi)! pxi
x1 !...xn !
Categorical Distributions
• The categorical distribution models the probability of events resulting from a single trial with more than two possible outcomes, e.g.,
- we roll a fair-sided dice once
- what is the probability of observing a ’five’?
• TheprobabilityofeventsX1,X2,...,Xn withprobabilitiesp=p1,p2,...,pn occurringexactlyx1,x2,...,xn times,respectively,isgivenby
P(X1 = x1,X2 = x2,...,Xn = xn;p) = px1 ×px2 ×···×pxn 12n
Marginalization
We want to know the probability of an event A irrespective of the outcome of another event B. We can obtain it, by summing over all possible outcomes B of B.
• Take an event B. The set of all possible individual outcomes of B, B is the partition of the outcome space
• E.g., B ={head, tail} for a coin flip; B ={king, heart, diamond, spades} for card suits
• We can marginalize over the set of outcomes of B as follows P(A) = P(A,B = b)
or equivalently (remember the product rule?)
P(A) = P(A|B = b)P(B = b) b∈B
and even for conditional probabilities
P(A|C) = P(A|C, B = b)P(B = b|C) b∈B
Marginalization
We want to know the probability of success of movies of a specific genre
(A = {comedy , thriller , romance}). But we only have data on movie success probabilities in a specific market, namely (B = {EU, NA, AUS}).
P(A) = P(A,B = b) b∈B
A B P(A,B)
romance romance romance thriller thriller thriller comedy comedy comedy
EU 0.05 NA 0.1 AUS 0.3 EU 0.1 NA 0.2 AUS 0.1 EU 0.1 NA 0.025 AUS 0.025
Marginalization
We want to know the probability of success of movies of a specific genre
(A = {comedy , thriller , romance}). But we only have data on movie success probabilities in a specific market, namely (B = {EU, NA, AUS}).
P(A) = P(A,B = b) b∈B
P(A,B) A P(A)
romance romance romance thriller thriller thriller comedy comedy comedy
EU 0.05 romance
NA 0.1 AUS 0.3 EU 0.1 NA 0.2 AUS 0.1 EU 0.1 NA 0.025 AUS 0.025
Marginalization
We want to know the probability of success of movies of a specific genre
(A = {comedy , thriller , romance}). But we only have data on movie success probabilities in a specific market, namely (B = {EU, NA, AUS}).
P(A) = P(A,B = b) b∈B
P(A,B) A P(A)
romance romance romance thriller thriller thriller comedy comedy comedy
EU 0.05 NA 0.1 AUS 0.3 EU 0.1 NA 0.2 AUS 0.1 EU 0.1 NA 0.025 AUS 0.025
romance 0.45
Marginalization
We want to know the probability of success of movies of a specific genre
(A = {comedy , thriller , romance}). But we only have data on movie success probabilities in a specific market, namely (B = {EU, NA, AUS}).
P(A) = P(A,B = b) b∈B
P(A,B) A P(A)
romance romance romance thriller thriller thriller comedy comedy comedy
EU 0.05 NA 0.1 AUS 0.3 EU 0.1 NA 0.2 AUS 0.1 EU 0.1 NA 0.025 AUS 0.025
romance 0.45
Marginalization
We want to know the probability of success of movies of a specific genre
(A = {comedy , thriller , romance}). But we only have data on movie success probabilities in a specific market, namely (B = {EU, NA, AUS}).
P(A) = P(A,B = b) b∈B
P(A,B) A P(A)
romance romance romance thriller thriller thriller comedy comedy comedy
EU 0.05 NA 0.1 AUS 0.3 EU 0.1 NA 0.2 AUS 0.1 EU 0.1 NA 0.025 AUS 0.025
romance 0.45
Marginalization
We want to know the probability of success of movies of a specific genre
(A = {comedy , thriller , romance}). But we only have data on movie success probabilities in a specific market, namely (B = {EU, NA, AUS}).
P(A) = P(A,B = b) b∈B
P(A,B) A P(A)
romance romance romance thriller thriller thriller comedy comedy comedy
EU 0.05 NA 0.1 AUS 0.3 EU 0.1 NA 0.2 AUS 0.1 EU 0.1 NA 0.025 AUS 0.025
romance 0.45
Marginalization
We want to know the probability of success of movies of a specific genre
(A = {comedy , thriller , romance}). But we only have data on movie success probabilities in a specific market, namely (B = {EU, NA, AUS}).
P(A) = P(A,B = b) b∈B
P(A,B) A P(A)
romance romance romance thriller thriller thriller comedy comedy comedy
EU 0.05 NA 0.1 AUS 0.3 EU 0.1 NA 0.2 AUS 0.1 EU 0.1 NA 0.025 AUS 0.025
romance 0.45
thriller 0.4
comedy 0.15
Please go to
for a quick quiz on probabilities!
https://pollev.com/iml2022
Probability and Machine Learning
We probably all agree that probabilities are useful for thinking about card games or coin flips
... but why should we care in machine learning? Consider typical classification problems
• document → {spam, no spam}
• hand-written digit → {0,1,2,3,4,5,6,7,8,9}
• purchase history → recommend {book a, book b, book c, ...}
Probability and Machine Learning
We probably all agree that probabilities are useful for thinking about card games or coin flips
... but why should we care in machine learning? Consider typical classification problems
• document → {spam, no spam}
• hand-written digit → {0,1,2,3,4,5,6,7,8,9}
• purchase history → recommend {book a, book b, book c, ...}
• uncertainty, due to few observations, noisy data, ...
• model features as following certain probability distributions
• soft predictions (“we are 60% confident that Bob will like given his purchase history”)
Probabilistic Models
“All models are wrong, but some are useful.” ( , Statistician)
Probabilistic Models
• allow to reason about random events in a principled way.
• allow to formalize hypotheses as different types of probability
distributions, and use the laws of probability to derive predictions
Example: Spam classification
• An email is a random event with two possible outcomes: spam, not spam
• The probability of observing a spam email P(spam) = θ, and trivially
P(not spam) = 1 − θ.
• We might care about a random variable X as the number of spam emails
in an inbox of 100 emails. X is distributed according to the binomial distribution, and depends on the parameters θ and N = 100
X ∼ Binomial(θ,N = 100)
Learning Probabilistic Models I
X is distributed according to the binomial distribution, and depends on the parameters θ and N = 100
X ∼ Binomial(θ,N = 100)
• In order to make predictions of X we need to know the parameters θ. How do we learn them?
• Typically, θ is unknown, but if we have data available we can estimate θ
• One common choice is to pick θ that maximizes the probability of the
observed data
θˆ= argmaxP(X;θ,N) θ
That is the maximum likelihood estimate (MLE) of θ.
• Once we have estimated θ we can use it to predict values for unseen data
Learning Probabilistic Models II
The maximum likelihood principle
θˆ= argmaxP(X;θ,N) (1)
• Consider a data set consisting of 100 emails, 20 of which are spam.
• Following from the binomial distribution
n L(θ)=P(X;θ,N)= m θx(1−θ)N−x
the likelihood of the data1 is ∝ θ20(1 − θ)100−20
• What do you think would be a good value for θ = p(spam = 1)? Why?
• Next lecture, we will see how to derive this value in a principled way
1∝ means ‘proportional to’. n can be ignored because it is independent of θ. m
Learning Probabilistic Models III
Maximum likelihood is only one choice of estimator among many
• Consider a data set of one inbox with no spam email. MLE: θ = 1, and hence P(not spam) = θ = 1 and P(spam) = 1 − θ = 0.
→ “spam emails don’t exist”
• We could modify this estimate with our prior belief. E.g., we might believe that about 80 of 100 emails are not spam. We ‘nudge’ θ from θ = 1 towards θ = 0.80
• We can combine our prior belief with the estimate from the data to arrive at a posterior probability distribution over θ: P(θ).
P(θ|x) = P(θ)P(x|θ) ∝ P(θ)P(x|θ) P(x)
• The maximum a posteriori estimate is then
θˆ= argmaxP(θ)P(x|θ) θ
(looks familiar)?
Probability underlies many modern knowledge technologies
• estimate the (conditional, joint) probability of observations • Bayes rule
• Expectations and marginalization
• Probabilistic models
• Maximum likelihood estimation (taster) • Maximum aposteriori estimation (taster)
Next Lecture(s):
• Optimization
• Naive Bayes Classification
References
Chris Bishop. Pattern Recognition and Machine Learning. Chapters: 1.2 (intro), 1.2.3, 2 (intro), 2.1 (up to 2.1.1), 2.2 (up to 2.2.1)
Optional / If time permits: Expectations
The expectation of a function (like a probability distribution) is the weighted average of all possible outcomes, weighted by their respective probability.
• For functions with discrete outputs
E[f(x)] = f(x)P(x) x∈X
• For functions with continuous outputs
f(x)P(x)dx
Optional / If time permits: Expectations
The expectation of a function (like a probability distribution) is the weighted average of all possible outcomes, weighted by their respective probability.
• On sunny days Bob watches 1 movie
• On rainy days Bob watches 3 movies
• Bob lives in Melbourne, it rains on 70% of all days
• What is the expected number of movies Bob watches per day?
Optional / If time permits: Expectations
The expectation of a function (like a probability distribution) is the weighted average of all possible outcomes, weighted by their respective probability.
• On sunny days Bob watches 1 movie
• On rainy days Bob watches 3 movies
• Bob lives in Melbourne, it rains on 70% of all days
• What is the expected number of movies Bob watches per day?
1∗0.3+3∗0.7 = 2.4
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