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Consider three expressions (distributions):
 
Consider three expressions (distributions):
  
1. Likehood:
+
===1. Likehood===
  
 
<math>p(X; \theta)</math> (discrete)
 
<math>p(X; \theta)</math> (discrete)
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used for MLE: <math>\overset{\land}\theta_{ML} = f_x(X | \theta)</math>
 
used for MLE: <math>\overset{\land}\theta_{ML} = f_x(X | \theta)</math>
  
2. Prior:
+
===2. Prior===
  
 
<math>P(\theta)</math> (discrete)
 
<math>P(\theta)</math> (discrete)
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Indicates some prior knowledge as to what <math>\theta</math> should be.  Prior refers to before seeing observation.
 
Indicates some prior knowledge as to what <math>\theta</math> should be.  Prior refers to before seeing observation.
  
3. Posterior:
+
===3. Posterior===
  
 
<math>p(\theta | x)</math> (discrete)
 
<math>p(\theta | x)</math> (discrete)
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<math>\overset{\land}\theta_{\mbox{map}} = \overset{\mbox{argmax}}\theta f_{X | \theta}(X | \theta) F_\theta(\theta)</math>
 
<math>\overset{\land}\theta_{\mbox{map}} = \overset{\mbox{argmax}}\theta f_{X | \theta}(X | \theta) F_\theta(\theta)</math>
 +
 +
== Example 1 ==
 +
 +
<math>X \sim f_x(X) = \lambda e^{-\lambda X}</math>
 +
 +
but we don't know the parameter <math>\lambda</math>.  Let us assume, however, that <math>\lambda</math> is actually itself exponentially distributed, i.e.
 +
 +
<math>\lambda \sim f_\lambda(\lambda) = \Lambda e^{-\Lambda\lambda}</math>
 +
 +
where <math>\Lambda</math> is fixed and known.
 +
 +
Find <math>\overset{\land}\lambda_{\mbox{map}}</math>.
 +
 +
Solution:
 +
 +
<math>\overset{\land}\lambda_{\mbox{map}} = \overset{\mbox{argmax}}\lambda f_{\lambda | X}(\lambda | X)</math>
 +
 +
<math>\overset{\land}\lambda_{\mbox{map}} = \overset{\mbox{argmax}}\lambda f_x(\lambda)f_{x|\lambda}(x; \lambda)</math>
 +
 +
<math>\overset{\land}\lambda_{\mbox{map}} = \overset{\mbox{argmax}}\lambda \Lambda e^{-\lambda \Lambda}\lambda e^{-\lambda X}</math>
 +
 +
<math>\frac{d}{d\lambda} \lambda \Lambda e^{-\lambda(\Lambda + X)} = 0</math>

Revision as of 11:13, 17 November 2008

Given observation X used to estimate an unknown parameter $ \theta $ of distribution $ f_x(X) $ (i.e. $ f_x(X) = $ some function $ g(\theta) $

Consider three expressions (distributions):

1. Likehood

$ p(X; \theta) $ (discrete)

$ f_x(X; \theta) $ (continuous)

used for MLE: $ \overset{\land}\theta_{ML} = f_x(X | \theta) $

2. Prior

$ P(\theta) $ (discrete)

$ P_\theta(\theta) $ (continuous)

Indicates some prior knowledge as to what $ \theta $ should be. Prior refers to before seeing observation.

3. Posterior

$ p(\theta | x) $ (discrete)

$ f_x(\theta, x) $ (continuous)

"Posterior" refers to after seeing observations. Use Posterior to define maximum a-posterior i (map) estimate:

$ \overset{\land}\theta_{\mbox{MAP}} = \overset{\mbox{argmax}}\theta f_{\theta | X}(\theta | X) $

Using Bayes' Rule, we can expand the posterior $ f_{\theta | X}(\theta | X) $:

$ f_{\theta | X}(\theta | X) = \frac{f_{x|\theta}f_\theta(\theta)}{f_X(X)} $

$ \overset{\land}\theta_{\mbox{map}} = \overset{\mbox{argmax}}\theta f_{X | \theta}(X | \theta) F_\theta(\theta) $

Example 1

$ X \sim f_x(X) = \lambda e^{-\lambda X} $

but we don't know the parameter $ \lambda $. Let us assume, however, that $ \lambda $ is actually itself exponentially distributed, i.e.

$ \lambda \sim f_\lambda(\lambda) = \Lambda e^{-\Lambda\lambda} $

where $ \Lambda $ is fixed and known.

Find $ \overset{\land}\lambda_{\mbox{map}} $.

Solution:

$ \overset{\land}\lambda_{\mbox{map}} = \overset{\mbox{argmax}}\lambda f_{\lambda | X}(\lambda | X) $

$ \overset{\land}\lambda_{\mbox{map}} = \overset{\mbox{argmax}}\lambda f_x(\lambda)f_{x|\lambda}(x; \lambda) $

$ \overset{\land}\lambda_{\mbox{map}} = \overset{\mbox{argmax}}\lambda \Lambda e^{-\lambda \Lambda}\lambda e^{-\lambda X} $

$ \frac{d}{d\lambda} \lambda \Lambda e^{-\lambda(\Lambda + X)} = 0 $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett