Download - Discrete Math CS 280
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Discrete MathCS 280
Prof. Bart [email protected]
Module Probability --- Part b)
Bayes’ RuleRandom Variables
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Bayes’ Theorem
How to assess the probability that a particular event will occur on the basis of partial evidence?
Examples:
What is the likelihood that people who test positive to a particular disease (e.g., HIV), actually have the disease?
What is the probability that an e-mail message is spam?
Key idea: one should factor in additional information regarding occurrence of events.
Assume that with respect to events F and E (“E” for “Evidence”):
We know P(F) – probability that event F occurs
(e.g. probability that email message is spam;
this is given by what fraction of email is spam)
We also know event E has occurred.
(e.g., email message contains words “sale” and “bargain”)
Therefore the probability conditional probability that F occurs given
that E occurs, P(F|E), is a more realistic estimate that F occurs than P(F).
How do we compute P(F|E)?
E.g., based on P(F), P(E|F), and P(E| ¬F)
Note: ¬F is also referred to as complement of F (FC or F).
BayesianBayesianInferenceInference
EvidenceEvidence
Original BeliefOriginal Belief(Prior Probability)(Prior Probability)
ModifiedModifiedBeliefBelief
HypothesisHypothesis TheoryTheory
P(F)
E
P(F|E)
Experiment: Pick one box at random (p = 0.5) and than a ball at random from that box.
Assume you got a red ball.
What’s the probability that it came form the left box?
Define:E – you choose a red ball. (therefore ¬ E – you choose the green ball)F – you choose the left box. (therefore ¬ F– you choose the right box)
We want to know P(F|E)
Box A Box B
What we know:
P(E|F) =
P(E|¬F) =
Given that the boxes are selected at random: P(F) = P(¬F)=1/2
P(F|E) = P(E∩F)/P(E) so we need to compute P(E∩F) and P(E).
P(F|E)?
We know P(E|F) = P(E∩F)/P(F). So, P(E∩F) = P(E|F) P(F) = 7/9 * 1/2 = 7/18.What about P(E)? Note that P(E) = P(EF) +P (E∩ ¬F). Why?Note also that P (E ∩ ¬F)= P(¬F) P(E|¬F) = 1/2 * 3/7 = 3/14So, P(E) = P(EF) +P (E∩ ¬F) = 7/18 + 3/14 = 38/63
And therefore P(F|E) = P(E∩F)/P(E) = (7/18) / (38/63) = 49/76 0.645
7/93/7
E – red colorF – left box
BayesianBayesianInferenceInference
Concrete (new) EvidenceConcrete (new) EvidenceRed ball picked (E)Red ball picked (E)
Original BeliefOriginal Belief there is a 0.5 that you will pick left box there is a 0.5 that you will pick left box
(P(F)).(P(F)).
Modified BeliefModified BeliefIncreased to 0.65 Probability Increased to 0.65 Probability
(P(F|E))(P(F|E))
( | ) ( )( | )
( | ) ( ) ( | ) ( )C CP E F P F
P F EP E F P F P E F P F
Theorem: Bayes’ Theorem
Suppose that E and F are events from a sample space S such that P(E) ≠0 and P(F)≠0. Then
Proof:
, ( ) ( | ) ( )
exp ( )
( ) ( ) ( )
( ) ( ) ( ) (
( | ) ( )( | )
(
( | ) ( ) / ( )
( | ) (
| ) ( ) ( | ) ( )
( |
)
) / (
)
)
( |
C C
C C C
So P E F P E F P F therefore
We only need an ression for P E
E E S E F F E F E F
So
P E P E F
P F E P
P
P E F P FP
E F
E F P E
P E F P
F EP
P E F P F P E F P F
So
P EP F E
F P
E
E F
( | ) ( )
( | ) ( ) ( | )
) ( )
( )( ) C C
P E F P F
P E F
F
P F
P
P E F P
F
P E F
Why?
( | ) ( )( | )
( )
P E F P FP F E
P E
Example:
Suppose that 1 person in 100,000 has a particular rare disease. There is an
accurate test for the disease that is correct in 99% of the time when given
to someone with the disease; it is correct in 99.5% of the time when given
to someone without the disease.
Find:
a) Probability that someone who tests positive has the disease.
b) Probability that someone who tests negative does not have the disease.
Solution:
a)
Always start by defining the events!
F – the person has the diseaseE – the person tests positive to the diseaseP(F|E) – probability of having the disease given positive test
P(F)=1/100,000 = 0.00001; P(FC) = 0.99999P(E|F) = 0.99; P(EC|F) = 0.01P(E|FC) = 0.005
002.0)99999.0)(005.0()00001.0)(99.0(
)00001.0)(99.0(
)()|()()|(
)()|()|(
CC FPFEPFPFEP
FPFEPEFP
Only 0.2% of people who test positive actually have the disease!!!
Note: These arethe probabilitiesmost easily measured!
b) F – the person has the disease
E – the person tests positive to the disease
P(FC|EC) – probability of not having the disease given negative test
P(F)=1/100,000=0.00001; P(FC)=0.99999
P(E|F)=0.99; P(EC|F)=0.01
P(E|FC)=0.005
9999999.0)00001.0)(01.0()99999.0)(995.0(
)999991.0)(995.0(
)()|()()|(
)()|()|(
FPFEPFPFEP
FPFEPEFP
CCCC
CCCCC
That’s… pretty good!
MarblesMarbles
TOYS R US sells two kinds of bags of marbles:(1) Bags of all black marbles, and(2) Bags of mixed marbles in which 20% of the marbles are black.
The bags are opaque and wrapped in plastic, and I have no idea which bag is more common. I buy a bag and figure there is a 50:50 chance that the bag I purchased contains all black marbles. A guess!
I pull a marble out of the bag and see that it is black. How should this new evidence affect the 50:50 assessment I assigned to the probability of my having purchased an all black bag of marbles? (as previous example)
F – bag of all black marbles; FC – bag with 20% black marblesE – black marble
Prior BeliefThere is a 1/2 chance that I have an all-black bag of marbles … a guess (P(F))
MarblesMarbles
1 0.5( | ) 0.83
(1 0.5) (0.2 0.5)P F E
0.5 chance of all-black (100%) marble bag.
0.5 chance of 0.2 black marble bag.
Posterior BeliefProbability that my
bag of marbles is all black = 0.833 P(F|E).
( | ) ( )( | )
( | ) ( ) ( | ) ( )C C
P E F P FP F E
P E F P F P E F P F
MarblesMarbles
96.0)2.017(.)183.0(
183.0)|(
EFP
Prior Belief0.83
0.83 chance of all-black (1) marble bag.
0.17 chance of 0.2 black marble bag.
New Belief0.96
I put the marble back, shake the bag, and draw another marble. It is black? What happens now that my new prior probability is 0.83?
Remember, I don’t know which type of marble bag is most popular … Wal-Mart may have 100 bags of mixed marbles on the shelf for every bag of all black marbles.
Bayes’ Theorem doesn’t tell me the probability of my marble bag being all black – it only tells me how I should revise my initial best guess based on the newly obtained information.
Warning: Correct but slightly informal! Instead of changing prior, we could consider new experiment and evidence drawing two marbles.
( | ) ( )( | )
( | ) ( ) ( | ) ( )C C
P E F P FP F E
P E F P F P E F P F
BayesianBayesianInferenceInference
Concrete EvidenceConcrete Evidence1st Black Marble1st Black Marble
Original BeliefOriginal BeliefI shrug my shoulders and guess is that I shrug my shoulders and guess is that
there is a 0.5 chance that my bag contains there is a 0.5 chance that my bag contains all black marbles.all black marbles.
Modified BeliefModified BeliefIncreased to 0.83 ProbabilityIncreased to 0.83 Probability
BayesianBayesianInferenceInference
Concrete EvidenceConcrete Evidence2nd Black Marble2nd Black Marble
Modified BeliefModified BeliefIncreased to 0.96 Increased to 0.96
Generalized Bayes’ Theorem
Suppose that E is an event from a sample space S and F1, F2 ,…, Fn are
mutually exclusive events such that
Asume that P(E) ≠ 0 and P(Fi) ≠ 0 for i=1, 2,…, n. Then
SFini 1
)
1
( | ( )( | )
( | ) ( )
j jj n
i ii
P E F P FP F E
P E F P F
P(E)
( | ) ( )( | )
( | ) ( ) ( | ) ( )C C
P E F P FP F E
P E F P F P E F P F
Compare:
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Bayesian Spam Filters
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Applying Bayes’ Theorem:SPAM or HAM?
Let our sample space or universe be the set of emails. (So, we’re sampling fromLet our sample space or universe be the set of emails. (So, we’re sampling from
the space of possible emails.)the space of possible emails.)
Let S be the event a message is spam; hence is the event a message is not spamLet S be the event a message is spam; hence is the event a message is not spam
Let E be the event a message contains a word Let E be the event a message contains a word ww. .
Since we have no idea of likelihood of SPAM, we assume P(S)=P(SC)=1/2.
Can we do better?
How do we get and ? ( | )p E S ( | )p E S
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Estimations
Note these are estimates based on frequencies in samples.
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Estimation Continued
Note P(S) = P(SC) = ½ divides out.
So,
becomes
So, a quite straightforward formula for our first Bayesian spam filter!
So, what do we want forp(w) and q(w) ??
( | ) ( )p E S p w
( | ) ( )p E S q w
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Spam based on single words?
Probabilities based on single words: Bad IdeaProbabilities based on single words: Bad Idea
– False positives AND false negatives a plentyFalse positives AND false negatives a plenty
Calculate based on Calculate based on n n words, assuming each event Ewords, assuming each event Eii|S (E|S (Eii|S|SCC) is ) is
independent;independent;
P(S) = P(SP(S) = P(SCC).).
Derivation see Sect. 6.3.
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Final Approximation
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Compare to single word:
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How do we use this?
User must train the filter based on messages in his/her inbox to estimate User must train the filter based on messages in his/her inbox to estimate probabilities.probabilities.
The program or user must define a threshold probability The program or user must define a threshold probability rr: :
If , the message is considered spam.If , the message is considered spam.
Gmail: Train on all users! (note: report spam button)Gmail: Train on all users! (note: report spam button)
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Example
Suppose the filter has the following dataSuppose the filter has the following dataThreshold Probability: .9Threshold Probability: .9““Nigeria” occurs in 250 of 2000 spam messages Nigeria” occurs in 250 of 2000 spam messages ““Nigeria” occurs in only 5 of 1000 non-spam messagesNigeria” occurs in only 5 of 1000 non-spam messagesLet’s try to estimate the probability, using the process we just definedLet’s try to estimate the probability, using the process we just defined
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Example Cont.
Step 1: Find the probability that the message has the Step 1: Find the probability that the message has the word “Nigeria” in it and is spam. word “Nigeria” in it and is spam. – p(p(NigeriaNigeria) = 250 / 2000 = 0.125) = 250 / 2000 = 0.125
Step 2: Find the probability that the message has the Step 2: Find the probability that the message has the word “Nigeria” in it and is not spam.word “Nigeria” in it and is not spam.– q(q(NigeriaNigeria) = 5 / 1000 = 0.005) = 5 / 1000 = 0.005
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Since we are assuming that it is equally likely that an Since we are assuming that it is equally likely that an incoming message is or is not spam, we can incoming message is or is not spam, we can estimate the probability with this equation:estimate the probability with this equation:
r(Nigeria) = r(Nigeria) = p(Nigeria) p(Nigeria)
p(Nigeria) + q(Nigeria)p(Nigeria) + q(Nigeria)
Example Cont.
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= = 0.1250.125
0.1300.130
= 0.962= 0.962
Since Since r(Nigeria)r(Nigeria) is greater than the threshold of 0.9, we can reject is greater than the threshold of 0.9, we can reject
this message as spam.this message as spam.
Example Cont.
0.125____0.125____0.125 + 0.0050.125 + 0.005
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Multiple Words
2000 Spam messages; 1000 real messages2000 Spam messages; 1000 real messages
““Nigeria” appears in 400 spam messagesNigeria” appears in 400 spam messages
““Nigeria” appears in 60 real messagesNigeria” appears in 60 real messages
““bank” appears in 200 spam and 25 real messagesbank” appears in 200 spam and 25 real messages
Threshold Probability: .9Threshold Probability: .9
Let’s calculate the probability that message with “Nigeria” and “bank” is Let’s calculate the probability that message with “Nigeria” and “bank” is spam.spam.
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Example Cont.
Step 1: Find the probability that the message has the word “Nigeria” in it Step 1: Find the probability that the message has the word “Nigeria” in it and is spam.and is spam.– p(Nigeria) = 400 / 2000 = 0.2p(Nigeria) = 400 / 2000 = 0.2
Step 2: Find the probability that the message has the word “Nigeria” and is Step 2: Find the probability that the message has the word “Nigeria” and is not spam.not spam.– q(Nigeria) = 60 / 1000 = 0.06q(Nigeria) = 60 / 1000 = 0.06
Step 3: Find the probability that the message contains the word “bank” and Step 3: Find the probability that the message contains the word “bank” and is spam.is spam.– p(bank) = 200 / 2000 = 0.1p(bank) = 200 / 2000 = 0.1
Step 4: Find the probability that the message contains the word “bank” and Step 4: Find the probability that the message contains the word “bank” and is not spam.is not spam.– q(bank) = 25 / 1000 = 0.025q(bank) = 25 / 1000 = 0.025
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Example Cont
Using our approximation, we have:Using our approximation, we have:
r(Nigeria,bank) = r(Nigeria,bank) = p(Nigeria) * p(bank) p(Nigeria) * p(bank) p(Nigeria) * p(bank) + q(Nigeria) * q(bank)p(Nigeria) * p(bank) + q(Nigeria) * q(bank)
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Example Cont.
Using our approximation, we have:Using our approximation, we have:
r(Nigeria,bank) = r(Nigeria,bank) = p(Nigeria) * p(bank) p(Nigeria) * p(bank) p(Nigeria) * p(bank) + q(Nigeria) * q(bank)p(Nigeria) * p(bank) + q(Nigeria) * q(bank)
r(Nigeria,bank)r(Nigeria,bank) = = (0.2)(0.1) (0.2)(0.1)
(0.2)(0.1) + (0.6)(0.025)(0.2)(0.1) + (0.6)(0.025)
= = 0.9300.930
This message will be rejected however since we set the threshold probability at 0.9.This message will be rejected however since we set the threshold probability at 0.9.
Concludes Bayes Reasoning
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Probability Paradox I
Magic Dice: Or How to Win Every Time!
a) You select any one of the four dice (A, B, C, or D). b) I’ll select another.Both dice are thrown, highest number wins throw. Do series of 10 throws. The person with the most highest throws wins the series. (I.e. die “more likely to get a higher number” wins.)
Claim: In a game of 'The Best of Ten Throws’, I will almost certainly win --- no matter which die you pick!!
Why is this strange? Say, you pick die A. Let’s assume, die B is better. So, I pick B. But, then, next game & next person picks B. Let’s assume C is better. I’ll select C. Next person, will pick C. I’ll pick D. Next person, will pick D… Hmm… I’ll pick A and will win!!A < B < C < D …. < A !! Failure of transitivity!
But, could such a set ofdice exist?
Surprisingly, yes!
Magic Dice
Non-transitive dice: http://www.sciencenews.org/20020420/mathtrek.asp
D
C
B
A
Prob(D wins over C) = 2/32/6 + (4/6)* (1/2) = 4/6
Prob(C wins over B) = 2/3since3/6 + (3/6)* (2/6) = 4/6Prob(B wins over A) = 4/6 = 2/3
(i.e. Prob(A wins over B) = 1/3)
Prob(A wins over D) = 4/6 = 2/3
A < B < C < D …. < A !!
However: transitivity inexpected value of dice throw
E[B] < E[A] = E[C] < E[D] 16/6 < 18/6 = 18/6 < 20/6
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Random Variables and Distributions
Random Variables
For a given sample space S, a random variable (r.v.) is any real valued
function on S, i.e., a random variable is a function that assigns a real
number to each possible outcome
Suppose our experiment is a roll of 2 dice. S is set of pairs.
Example random variables:
S0 2-2
X = sum of two dice. X((2,3)) = 5
Y = difference between two dice. Y((2,3)) = 1
Z = max of two dice. Z((2,3)) = 3
Sample spaceNumbers
Random variable
Suppose a coin is flipped three times. Let X(t) be the random variable that equals the number of heads that appear when t is the outcome.
X(HHH) = 3
X(HHT) = X(HTH)=X(THH)=2
X(TTH)=X(THT)=X(HTT)=1
X(TTT)=0
Note: we generally drop the argument! We’ll just say the “random variable X”.
And write e.g. P(X = 2) for “the probability that the random variable X(t) takes on the value 2”.
Or P(X=x) for “the probability that the random variable X(t) takes on the value x.”
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Distribution of Random Variable
Definition:
The distribution of a random variable X on a sample space S is the set of
pairs (r, p(X=r)) for all r X(S), where p(X=r) is the probability that X
takes the value r.
A distribution is usually described specifying p(X=r) for each r X(S).
A probability distribution on a r.v. X is just an allocation of
the total probability mass, 1, over the possible values of X.
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The Birthday Paradox
Birthdays
How many people have to be in a room to assure that the probability that at least two of them have the same birthday is greater than 1/2?
a) 23b) 183c) 365d) 730
Let pn be the probability that no people share a birthday among n people in a room.
We want the smallest n so that 1 - pn > 1/2.
Then 1 - pn is the probability that 2 or more share a birthday.
Hmm. Why does such an n exist? Upper-bound?
For L optionsanswer is inthe orderof sqrt(L) ?
Informally, why??
A: 23
Birthdays
Assumption:
Birthdays of the people are independent.
Each birthday is equally likely and that there are 366 days/year
Let pn be the probability that no-one shares a birthday among n people in a room.
Assume that people come in certain order; the probability that the second person has a birthday. Different than the first is 365/366; the probability that the third person has a different birthday. Form the two previous ones is 364/366.. For the jth person we have (366-(j-1))/366.
What is pn? (“brute force” is fine)
After several tries, when n=22 1= pn = 0.475.
n=23 1-pn = 0.506
So,366
367
366
363
366
364
366
365 npn
366
367
366
363
366
364
366
36511
npn
Relevant to “hashing”. Why?
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From Birthday Problem to Hashing Functions
Probability of a Collision in Hashing Functions
A hashing function h(k) is a mapping of the keys (or records, e.g., SSN, around 300x 106 in the US) to a much smaller storage location. A good hashing fucntio yields few collisions. What is the probability that no two keys are mapped to the same location by a hashing function?
Assume m is the number available storage locations, so the probability of mapping a key to a location is 1/m.Assuming the keys are k1, k2, kn, the probability of mapping the jth record to a free location is after the first (j-1) records is (m-(j-1))/m.
m
nm
m
m
m
mp
m
nm
m
m
m
mp
n
n
12111
121
Given a certain m, find the smallest nSuch that the probability of a collision is greater than a particular threshold p.
It can be shown that for p>1/2,
n 1.177 m
m = 10,000, gives n = 117. Not that many!