Transcript lecture14

Today’s Topics
• Graded HW1 in Moodle (Testbeds used for grading are linked to class home page)
• HW2 due (but can still use 5 late days) at 11:55pm tonight
• HW3 out tomorrow but not due to THURSDAY Nov 5 (no later than 11/12)
• Midterm next THURSDAY (10/22); Reviews 10/20 (me) & 10/21 (Dmitry)
• Basics of Probability
– We’re covering probabilistic reasoning next so that the final
programming HW can be undertaken right after the midterm
– Random Events and Random Variables
– Probability Distributions
– Conditional Probabilities
– Some Rules of Probability
Do Problem 1, HW3
before the midterm
(Lecture 14 is end of
midterm material)
– Independence of Random Events
• Joint Probability Distributions (our focus)
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CS 540 - Fall 2015 (Shavlik©), Lecture 14, Week 6
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Random Events
• Prob(event) or P(event)
– The probability event will happen
– Eg, prob(fair coin comes up heads) = 0.5
– A random variable represent a random event
• Probability Distributions
– Assume event has k possible disjoint outcomes
– Assign prob to each outcome; ∑ p(outcomei) = 1
– In cs540, we’ll ignore real-valued outcomes
(and probability density functions),
so we’ll SUM rather than INTEGRATE
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Sample Probability Distribution
Values of COLOR assumed disjoint and complete
(ie, all objects have one and only one color)
Probability
1.0
RED
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GREEN
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BLUE
Color
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What if Values of a Random Variable
are NOT Disjoint and Complete?
Assume some var has values x, y, and z,
but these are not disjoint nor complete
Create new values
a = x  y  z
b = x  y  z
c = x  y  z
g = x  y  z
h = x  y  z
d = x  y  z
e = x  y  z
f = x  y  z
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Conditional Prob’s
Prob(A | B) - probability event A
happens given event B occurred
Eg,
prob(you take bus to work) = 0.1
prob(you take bus to work | it is raining) = 0.9
NOTE:
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P( A  B | C  B) is shorthand for
P( (A  B) | (C  B) )
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Some Rules of Prob
1) 0  prob(A)  1
2) prob(false)  0 and prob(true)  1
3) prob(A  B) = prob(A) + prob(B) - prob(A  B)
Venn
4) prob(A  B) = prob(A | B) ₓ prob(B)
= prob(B | A) ₓ prob(A)
5) prob( A)
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A
B
= 1 – prob(A)
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Prob(A | B) = Prob(A  B)
Prob(B)
Given we’re in B,
what fraction is
also in A?
Prob(A) can be small, say 0.10 and Prob(B) can be smaller,
say 0.05, yet Prob(A | B) can be large, say, 0.75
A
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B
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Notational Complexity
• We really should state both variables and their values
P(A = valueA | B = valueB)
Eg
P(color = yellow | size = big)
• When we mean ‘whatever value A and B have’ we should say
something like
P(A = ? | B = ?)
• However, this gets complicated and so for clarity for
P(A = ? | B = ?), we’ll often just say
P(A | B)
• In a SPECIFIC calculation with Boolean-valued vars,
P(A | B) means P(A=true | B=true)
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Useful Equation for
Manipulating Conditional Probs
P(A  B | C ) = P(A | B  C)  P(B | C)
Derivation
1)
P(A | B  C)  P(A  B  C) / P(B  C)
2)
P(B | C)
3)
P(A | B  C)  P(B | C) = P(A  B  C) / P(C)
4)
P(A  B  C) / P(C)  P(A  B | C)
5)
P(A  B | C ) = P(A | B  C)  P(B | C)
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// Using P(  )  P( | )  P()
 P(B  C) / P(C)
CS 540 - Fall 2015 (Shavlik©), Lecture 14, Week 6
// Combine 1 and 2
// Combine 3 and 4
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Independence
If A and B are independent events, then
prob(A | B) = prob(A)
// Knowing B happened tells us
// nothing about A
and
prob(B | A) = prob(B)
// Ditto
So if independence holds (or is assumed)
prob(A  B) = prob(A) ₓ prob(B)
since prob(A  B) = prob(A | B) ₓ prob(B)
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I have two coins, if I flip them,
what is the prob one comes
up heads and the other tails?
I never promised you independent coins!
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Conditional Independence
• If A and B are independent events given C, then
prob(A  B | C) = prob(A | C)  prob(B | C)
We say “A and B are conditionally independent given C”
• A variant: “A is independent of B given C”
prob(A | B  C) = prob(A | C)
// Assuming we know the value of C,
// knowing B happened tells us nothing more about A
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Joint Prob Distributions
(probs for compound events)
• Assume we have three Boolean-valued
random variables; so 23 = 8 possible combo’s
Combination
Prob of this Combination
A=false, B=false, C=false
0.50
A=false, B=false, C=true
0.20
A=false, B=true, C=false
0.15
A=false, B=true, C=true
0.01
A=true, B=false, C=false
0.02
A=true, B=false, C=true
0.03
A=true, B=true, C=false
0.04
A=true, B=true, C=true
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? [hint: probs sum to 1]
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Visually – There are 8 Distinct Regions
B
A
C
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Calculating Prob of
Any Expression
Given any (arbitrarily complex) logical
expression, we can calculate its probability by
summing the probs in the cells of the
joint prob table where the expression is true
This is our key calculation! The bulk of this topic
in cs540 will address making this calculation
feasible because these tables get too big
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Combination
Examples
1) P(A)
=?
2) P(A  C) = ?
3) P(A 
4) P(C)
A=false, B=false, C=false
0.50
A=false, B=false, C=true
0.20
A=false, B=true, C=false
0.15
A=false, B=true, C=true
0.01
A=true, B=false, C=false
0.02
A=true, B=false, C=true
0.03
A=true, B=true, C=false
0.04
A=true, B=true, C=true
0.05
B) = ?
=?
5) P(B  (A  C)) = ?
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Prob of this Combo
Answers
1) 0.14
2) 0.06
3) 0.30
4) 0.71
5) 0.28
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Here, vars = true
unless NOT sign
present
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Then vs. Now
• In 1980s: from where do the probs come?
• Today: we have tons of data!
• We can (conceptually) fill a huge joint prob
table simply by counting, then can answer
any possible question about the ‘random
variables’ in our table!
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Evidence, Hidden, and Query Vars
• In general, we ask about some QUERY
variables CONDITIONED on some
EVIDENCE variables, but don’t mention
some HIDDEN variables
Prob(A   C  E |  S  P Y ) ?
Evidence vars: S, P and Y
Hidden vars: B, D, F, G, …, N, O, Q, R, T, U, … X, & Z
Query
vars: A, C, and E
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More Experience with
Joint Prob Tables
• Recall that a Joint Probability Table
specifies the probability of each
(discrete) “complete world state”
• A “complete world state” specifies the value
of each random variable used to represent the
world we’re modeling
• If N variables each with M possible values,
there are M N different states of the world
one Petabyte = 1015 bytes
if N = 50 and M = 2, then M N  1015
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Marginalizing (“Summing Out”)
A method to answer questions involving
partial world states
P(Y) =

P(Y, Z)
// Eq 13.6
Z  all possible
‘conjunctive’
settings for vars
not set by Y
This comma
means AND
Ex: Assume we have vars A, B, C, and D
And Y = A ˄ ¬C
Then Z  { B ˄ D, B ˄ ¬D, ¬B ˄ D, ¬B ˄ ¬D }
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CS 540 - Fall 2015 (Shavlik©), Lecture 15, Week 6
We did this earlier
when we added
up all cells where
Y was true
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Conditional Forms
Called ‘conditioning’
P(Y) =  P(Y | Z)  P (Z) // Eq 13.8
- uses P(A ˄ B) = P(A | B)  P(B)
- a bit like a weighted sum
P(Y | X)  P(Y ˄ X) / P (X)
=  P(Y ˄ X ˄ Z1) /  P(X ˄ Z2)
Sum over all vars
not set by X
Sum over all vars
not set by Y or X
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A ‘Weighted Sum’ Example
P(takeBus)
= P(takeBus | weather=sunny) x P(w=sunny)
+ P(takeBus | weather=cloudy) x P(w=cloudy)
+ P(takeBus | weather=rainy) x P(w=rainy)
+ P(takeBus | weather=snowy) x P(w=snowy)
Assumed disjoint and complete
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Some Worked Examples
Assume we have a joint prob table for A, …, E
Express the following using table entries
Here, vars = true unless
(ie, full world states)
Basic idea
P(B ˄ D ˄ E) =
1) create probs that only involve
AND and NOT
P(B | D ˄ E) =
2) “AND in” the remaining vars in all
possible (conjunctive) ways
P(A ˅ C)
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NOT sign present
3) look up fully specified ‘world states’
=
4) do the arithmetic
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Solutions for Prev Slide
P(B ˄ D ˄ E) = P(B ˄ D ˄ E ˄ A ˄ C) + P(B ˄ D ˄ E ˄ A ˄ ¬ C) +
P(B ˄ D ˄ E ˄ ¬ A ˄ C) + P(B ˄ D ˄ E ˄ ¬ A ˄ ¬ C)
P(B | D ˄ E) = P(B ˄ D ˄ E) / P(D ˄ E) = …
- numerator done in first question
- denominator involves EIGHT terms
P(A ˅ C)
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= P(A) + P(C) - P(A ˄ C) = …
- first TWO terms each involve summing SIXTEEN terms
- last term involves EIGHT
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Next
• What if too few examples to sufficiently
populate all the cells in the joint table?
• What if joint prob table too large for memory?
• Bayesian Networks (Bayes Nets for short)
provide a popular/successful answer
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