Some of the basic probability and statistics concepts. joint probabilities, combinatories; conditional probabilities and independence and their examples; Bayes' rule.

Portal to all the other notes

• Lecture 01 - 2017.09.06
• Lecture 02 - 2017.09.13 -> This post
• Lecture 03 - 2017.09.20
• Lecture 04 - 2017.09.27
• Lecture 05 - 2017.10.04
• Lecture 06 - 2017.10.11
• Lecture 07 - 2017.10.18
• Lecture 08 - 2017.10.25
• Lecture 09 - 2017.11.01
• Lecture 10 - 2017.11.08
• Lecture 11 - 2017.11.15
• Lecture 12 - 2017.11.20
• Lecture 13 - 2017.11.29
• Lecture 14 - 2017.12.06

Lecture 02 - Sept 13 2017

Recall: If A & B are 2 disjoint events, then P[A∪B]=P[A]+P[B], P[A∩B]=P[A]⋅P[B].

If A∪B≠∅, then P[A∩B]=P[A]⋅P[B]−P[A∩B].

General formula (“inclusion-exclusion formula”)

Example 1

Weather data for 2 consecutive days in some month in some location. We aggregate this data and find: ‘R1 is rain on 1st day’, ‘R2 is rain on 2nd day’. P[R1]=0.6,P[R2]=0.5,P[R1∩R2C]=0.2. Find P[R1∩R2] and P[R1∪R2].

Answer:

P[(R1∩R2C)∪(R1∩R2)]=P[R1].

P[R1∩R2C]+P[R1∩R2]=P[R1]→P[R1∩R2]=0.4

P[R1∪R2]=P[R1]+P[R2]−P[R1∩R2]=0.6+0.5−0.4=0.7

Example to introduce the use of some combinatories

Form a jury with 6 people. We choose from a group with 8 men & 7 women. Find the prob that there are exactly 2 women in the selected jury.

Need an assumption about the relative prob. of people being picked: 6/15. Every one has equal chance of being picked.

P=({# of ways to pick exactly 2 women})/({# of ways to pick 6 out of 15})

(C27×C48)/C615

(tree idea)

C615=5005

numerator=21×70=1470

P=1470/5005=0.2937

After the break.

Chapter 1.3 Conditional prob. & independence

Definition: A & B are independence events, if P[A∩B]=P[A]×P[B].

The definition comes from the definition of conditional prob.

Definition: Let A & B be two events. The conditional prob of B given A denoted by P[B|A], is P[A∩B]/P[A].

Note: if P[B|A]=P[B] (given information of A does not affect/influence the chance of B), then the definition is proven.

Example 2

P[disease]=0.006. Test: positive vs negative. P[positive|dissease]=0.98 (sensitivity). P[pos|nodisease]=0.01.

Q1: Find prob P[pos].

P[pos]=P[pos∩disease]+P[neg∩nodisease]

P[pos|disease]=P[pos∩disease]/P[disease]

P[pos∩disease]=0.98×0.006

similarly, P[neg∩no disease]=(1−0.99)×(1−0.006)

P[pos]=0.98×0.006+(1−0.99)×(1−0.006)=1.582%

Q2: Find P[dis|pos].

P[dis|pos]=P[dis∩pos]/P[pos]=0.00588/0.01582=0.3717

The law of total probabilities:

Let A1,A2,…,An be a partition of Ω. (Ai are all mutually incompatible, and ∑Ai=Ω)

then for any B∈Ω, P[B]=∑iP[B|Ai]P[Ai].

Try to prove it at home.

Bayes’ rule

denominator: from law of total prob when n=2, A1=A,A2=AC

General Bayes’ Rule:

For every fixed i:

Role of different pieces of information in the formula:

• P[A|B]: B has happend, the prob of A.
• P[A] and P[AC]: “The prori” a priori info about the events you are interested in, regard of the observation.
• P[B|A] and P[B|AC]: “Likelihood” likelihoods that eventsthat you observe B actually happen given the events of interest to you.

At home, relate disease example problem to Bayes’ rule.