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You build a probabilistic model taking all background information into account. This gives the prior probability. All other information must be conditioned on.
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CPSC 322 Lecture 25
March 21, 2007 Textbook §9.2 – §9.
(^1) Recap
2 Marginal Independence
3 Conditional Independence
The conditional probability of formula h given evidence e is
P (h|e) = P (h ∧ e) P (e)
Chain rule:
P (f 1 ∧ f 2 ∧... ∧ fn) =
∏^ n
i=
P (fi|f 1 ∧ · · · ∧ fi− 1 )
Bayes’ theorem:
P (h|e) =
P (e|h) × P (h) P (e)
(^1) Recap
2 Marginal Independence
3 Conditional Independence
The probability that the Canucks will win the Stanley Cup is independent of whether light l 1 is lit. remember the diagnostic assistant domain—the picture will recur in a minute! Whether there is someone in a room is independent of whether a light l 2 is lit. Whether light l 1 is lit is not independent of the position of switch s 2.
(^1) Recap
2 Marginal Independence
3 Conditional Independence
Kevin separately phones two students, Alice and Bob. To each, he tells the same number, nk ∈ { 1 ,... , 10 }. Due to the noise in the phone, Alice and Bob each imperfectly (and independently) draw a conclusion about what number Kevin said. Let the numbers Alice and Bob think they heard be na and nb respectively. Are na and nb marginally independent?
Kevin separately phones two students, Alice and Bob. To each, he tells the same number, nk ∈ { 1 ,... , 10 }. Due to the noise in the phone, Alice and Bob each imperfectly (and independently) draw a conclusion about what number Kevin said. Let the numbers Alice and Bob think they heard be na and nb respectively. Are na and nb marginally independent? No: we’d expect (e.g.) P (na = 1|nb = 1) > P (na = 1).
Kevin separately phones two students, Alice and Bob. To each, he tells the same number, nk ∈ { 1 ,... , 10 }. Due to the noise in the phone, Alice and Bob each imperfectly (and independently) draw a conclusion about what number Kevin said. Let the numbers Alice and Bob think they heard be na and nb respectively. Are na and nb marginally independent? No: we’d expect (e.g.) P (na = 1|nb = 1) > P (na = 1). Why are na and nb conditionally independent given nk? Because if we know the number that Kevin actually said, the two variables are no longer correlated. e.g., P (na = 1|nb = 1, nk = 2) = P (na = 1|nk = 2)
light
two-wayswitch
switch
off on
power outlet
circuit breaker
outside power
l 1
l 2
w 1
w 0
w 2
w 4
w 3
w 6
w 5
p 2
p 1
cb 2
cb 1 s (^1)
s (^2) s (^3)
The probability that the Canucks will win the Stanley Cup is independent of whether light l 1 is lit given whether there is outside power. sometimes, when two random variables are marginally independent, they’re also conditionally independent given a third variable. But not always... Let C 1 be the proposition that coin 1 is heads; let C 2 be the proposition that coin 2 is heads; let B be the proposition that coin 1 and coin 2 are both either heads or tails. P (C 1 |C 2 ) = P (C 1 ): C 1 and C 2 are marginally independent. But P (C 1 |C 2 , B) 6 = P (C 1 |B): if I know both C 2 and B, I know C 1 exactly, but if I only know B I know nothing. Hence C 1 and C 2 are not conditionally independent given B.
