# 留学生辅导 Inference by enumeration – cscodehelp代写

Inference by enumeration

We saw that with our joint distribution table

we can calculate any probability Obvious problems:

1. Worst-casetimecomplexityOpdnqwheredisthelargestarity 2. SpacecomplexityOpdnqtostorethejointdistribution

3. HowtofindthenumbersforOpdnqentries???

These problems effectively stopped the use of probability in AI until the mid 80s

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Computational efficiency

To get efficient probabilistic computations, we need two things. 1. (Conditional)independence.

2. Bayesrule.

Will cover these now, in order.

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Independence

A and B are independent iff Why is this interesting?

PpAqPpBq

PpA, Bq “ Can help with the size of the problem.

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Independence

PpT oothache, Catch, Cavity, W eatherq

“ PpT oothache, Catch, Cavityq d PpW eatherq

If you store all values naively, this requires 2 ̈ 2 ̈ 2 ̈ 4 “ 32 entries.

You can do it in 31 by leaving one entry empty. This is possible since you know

that the probabilities add up to 1.

Using the independence, you can even reduce the 31 values to 10:

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Independence

PpT oothache, Catch, Cavity, W eatherq

“ PpT oothache, Catch, Cavityq d PpW eatherq

If you store all values naively, this requires 2 ̈ 2 ̈ 2 ̈ 4 “ 32 entries.

You can do it in 31 by leaving one entry empty. This is possible since you know

that the probabilities add up to 1.

Using the independence, you can even reduce the 31 values to 10:

You store 7 for the 3 dependent variables and 3 for weather which is independent (note that we’re using the probabilities adding up to 1 trick twice here)

For n independent biased coins, 2n Ñ n

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Independence

Absolute independence powerful but rare

Dentistry is a large field with hundreds of variables, none of which are independent. What to do?

Conditional independence

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Conditional independence

PpT oothache, Cavity, Catchq has:

Three binary variables.

Thus 23 entries in the joint probability table. But these sum to 1.

So 23 ́ 1 independent entries

That’s 7 independent entries

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Conditional independence

But, wait! There’s more!

If I have a cavity, the probability that the probe catches in it doesn’t depend on whether I have a toothache:

P pcatch|toothache, cavityq “ P pcatch|cavityq (1) The same independence holds if I haven’t got a cavity:

P pcatch|toothache, cavityq “ P pcatch| cavityq (2) Catch is conditionally independent of Toothache given Cavity

We write

PpCatch|T oothache, Cavityq “ PpCatch|Cavityq Catch K T oothache|Cavity

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Conditional independence PIG

INDEP: A)

Equivalent statements:

1*7

–

PLA 8) =P(A)

,

pld-iolid-pftld.NO/ c)

, MANIC)=PCt18 c) PCB/C

#-) .co.jp#.)a)–PlT1ca7-PGaHIc-isCI

PIT/Cat

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) =PCAÑo

( PIA 8) )) ,

PpT oothache|Catch, Cavityq “ PpT oothache|Cavityq ¥⇒PpToothache,Catch|Cavityq “ PpToothache|CavityqdPpCatch|Cavityq

Conditional independence

Write out full joint distribution using chain rule:

PpT oothache, Catch, Cavityq

“ PpT oothache|Catch, Cavityq d PpCatch, Cavityq

“ PpT oothache|Catch, Cavityq d PpCatch|CavityqPpCavityq “ PpT oothache|Cavityq d PpCatch|Cavityq d PpCavityq

2 + 2 + 1 = 5 independent numbers Equations 1 and 2 remove 2.

Icar PCT )=✗

PATI can )=t✗ 11

Car a Car T 0 0

it☒☒

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Conditional independence

In most cases, the use of conditional independence reduces the size of the representation of the joint distribution from exponential in n to (close to) linear in n.

Conditional independence is our most basic and robust form of knowledge about uncertain environments.

Can often make conditional independence statements when little else is known.

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