PSM1 recap and outlook

PSM2 UCL

Bennett Kleinberg

8 Jan 2019

Welcome

Probability, Statistics & Modeling II

Lecture 1

Quick recap 1

Predicting crimes

Behind the problem:

What is the claim?

Formalising the problem

chance_day_1 = 0.5
chance_day_2 = 0.5
chance_day_3 = 0.5
#...

Solving the problem

Probability for correct prediction?

P(prediction == 1) = p_correct = 0.5

… on 10 consecutive days?

p_correct * p_correct * p_correct ...

p_correct = 0.5

# for d = 10 days

d = 10

#Formal:
p_correct ^ d

## [1] 0.0009765625

Equivalent to: 1/2^10 = 1/1024

MARGINAL Probability:

P(EVENT)

Even very, very, rare events happen…

… but most of the time they don’t.

You need probability theory to tell the lucky from the likely.

(and proper statistics notations)

Quick recap 2

About Maria

Maria is 26 years old, single, outspoken, and very bright. She majored in law. As a student, she was deeply concerned with issues of discrimination and miscarriage of justice, and also participated in animal-rights demonstrations.

Adapted from Tversky & Kahneman (1983)

Which is more probable?

A: Maria works in a law firm
B: Maria works in a law firm and does pro bono work for disadvantaged defendants

Formalising the problem

Two events:

P(A) #prob of answer A
P(B) #prob of answer B

… BUT:

There’s something special with P(B)

P(B) = P(A) + "something else"

P(B) contains two ‘events’: P(A) and ‘pro bono work’

Let 'pro bono work' be P(C)

P(B) = P(A) and P(C)

Solving the problem

Joint probability

P(B) = P(A and C)

Let’s try:

Prob_A = 0.4
Prob_C = 0.3

Formula: P(A and B) = P(A)*P(C)

(Prob_A_and_C = Prob_A * Prob_C)

## [1] 0.12

By definition: P(X) > P(X and Y)

Therefore:

P(‘M is a lawyer’) > P(‘M is a lawyer’ and ‘pro-bono work’)

JOINT Probability:

P(EVENT_A AND EVENT_B) = P(EVENT_A)*P(EVENT_B)

Probability of two independent events is always smaller than the probability of each single events.

Quick recap 3

Screening terrorists

Your turn

What are the chances that this man is a terrorist?

Formalising the problem

CONDITIONAL Probability:

Probability of TERRORIST given that there is an ALARM

Looking for: P(terrorist GIVEN alarm)

Formal: P(terrorist|alarm)

Solving the problem (method 1)

	Terrorist	Passenger
Terrorist	950	50	1,000
Passenger	4,950	94,050	99,000
	5,900	94,100	100,000

P(terrorist|alarm) = 950/5900 = 16.10%

Solving the problem (method 2)

Bayes’ rule

Setting the stage:

P(T) -> probability of terrorist
P(A) -> probability of alarm

We want:

P(T|A)

We know:

accuracy = P(A|T) = 0.95
baserate = P(T) = 0.01

Bayes’ rule (cont’d)

accuracy = 0.95 #P(A|T)
baserate = 0.01 #P(T)

Bayes’ rule: P(T|A) = ( P(A|T) * P(T) ) / P(A)

P(A) –> probability of any alarm???

P(A) = P(A|T) * P(T) + P(A|notT) * P(notT)

(Prob_notT = 1 - baserate) #P(notT) = 1 - P(T)

## [1] 0.99

(Prob_A_given_notT = 1 - accuracy) #P(A|notT) = 1 = P(A|T)

## [1] 0.05

Bayes’ rule (cont’d)

Putting it together:

#Bayes' rule:
Prob_A = accuracy * baserate + Prob_A_given_notT * Prob_notT #P(A) = P(A|T) * P(T) + P(A|notT) * P(notT)
Prob_A

## [1] 0.059

Prob_T_given_A = (accuracy * baserate) / Prob_A #P(T|A) = ( P(A|T) * P(T) ) / P(A)
Prob_T_given_A

## [1] 0.1610169

! Revise this rule here

CONDITIONAL Probability:

P(EVENT_A GIVEN EVENT_B) = P(EVENT_A|EVENT_B)

Probability of one event given that another event is true.

BEWARE OF THE BASERATE FALLACY

Quick recap 4

Solving gang crime

The context

Problem: gang crime in London

Mayor proposes two programmes:

A: zero-tolerance
B: work-and-integration

100 gang-members in two areas.

Outcome measure: number of gang members who disengaged

Results

	Programme A	Programme B
Camden	63/90	8/10
Lambeth	4/10	45/90

Mayor has GBP 5m to invest in one programme.

Your decision?

Solving the problem

	Programme A	Programme B
Camden	63/90 = 70%	8/10 = 80%
Lambeth	4/10 = 40%	45/90= 50%
	67/100 = 67%	53/100 = 53%

[a] phenomenon wherein an association or a trend observed in the data at the level of the entire population disappears or even reverses when data is disaggregated by its underlying subgroups Alipourfard et al., 2018