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1: Introduction

Notation and terminology . . .

CSE454 2002 : This document is online at   http://www.csse.monash.edu.au/~lloyd/tilde/CSC4/CSE454/   and contains hyper-links to other resources - Lloyd Allison ©.

 
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NB. The term data space is often used, in machine learning.

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Inference

People often distinguish between

e.g.
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Bayes

If B1, B2, ..., Bk is a partition of a set B (of causes) then


             P(A|Bi) P(Bi)
P(Bi|A) = -------------------     i=1, 2, ..., k
             k
          SUM   P(A|Bj) P(Bj)
           j=1

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. . . applied to data D and hypotheses Hi:
   k
SUM P(D|Hj) P(Hj) = P(D)
 j=1

P(Hi|D) = P(D|Hi) P(Hi) / P(D)   posterior

P(Hi|D)    P(D|Hi) P(Hi)
------- = --------------        posterior odds-ratio
P(Hj|D)    P(D|Hj) P(Hj)

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NB. Can ignore P(Hi) in posterior odds-ratio if, and only if, P(Hi)=P(Hj). Maximum likelihood may can cause problems when we have inequality.

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Example

C1, a fair coin, P(H) = P(T) = 0.5.

C2, a biased coin, P(H) = 2/3, P(T) = 1/3.

One of the coins is thrown 4 times, giving H, T, T, H.

Which coin was thrown? H1 : was C1.   H2 : was C2.


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Prior, P(C1) = P(C2) = 0.5.

Likelihood, P(HTTH | C1) = 1/16

and   P(HTTH | C2) = 4/9 . 1/9 = 4/81.

Posterior odds-ratio, P(C1|HTTH)/P(C2|HTTH) = (1/16 . 1/2) / (4/81 . 1/2) = 81/64.


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Now, P(C1|HTTH) + P(C2|HTTH) = 1

and if x/(1-x) = 81/64, then 64.x = 81 - 81.x, x = 81/145

P(C1|HTTH) = 81/145.

This case is simple because the model space is discrete, in fact finite (2).


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e.g. prediction

Know P(C1) = 81/145,   P(C2) = 64/145.

The more likely coin is C1.

If we assumed the coin really was C1, would predict P(H) = 0.5 in future.

But the coin might be C2.

Should predict   P(H) = 81/145 . 1/2 + 64/145 . 2/3 = (243 + 256) / (145 . 6) = 499 / 870 = 0.57

i.e. use a weighted average of the hypotheses.


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Conclusion

We have looked at simple examples!

© L. Allison, School of Computer Science and Software Engineering, Monash University, Australia 3168.
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