(New page: Example: Let <math>\mathbb{X}</math> and <math>\mathbb{Y}</math> be jointly distributed discrete random variables with ranges <math>X = \{0, 1, 2, 3, 4\}</math> and <math>Y = \{0, 1, 2\}</...)
 
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Example: Let <math>\mathbb{X}</math> and <math>\mathbb{Y}</math> be jointly distributed discrete random variables with ranges <math>X = \{0, 1, 2, 3, 4\}</math> and <math>Y = \{0, 1, 2\}</math> respectively.  Suppose that the conditional distributions <math>P_{\mathbb{X}|\mathbb{Y}}</math> are empirically estimated as follows:
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Example: Let <math>\mathbb{X}</math> and <math>\mathbb{Y}</math> be jointly distributed discrete random variables with ranges <math>X = \{0, 1, 2, 3, 4\}</math> and <math>Y = \{0, 1, 2\}</math> respectively.   
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Suppose that the conditional distributions <math>P_{\mathbb{X}|\mathbb{Y}}</math> are empirically estimated as follows:
  
  

Revision as of 09:37, 1 February 2010

Example: Let $ \mathbb{X} $ and $ \mathbb{Y} $ be jointly distributed discrete random variables with ranges $ X = \{0, 1, 2, 3, 4\} $ and $ Y = \{0, 1, 2\} $ respectively.

Suppose that the conditional distributions $ P_{\mathbb{X}|\mathbb{Y}} $ are empirically estimated as follows:


$ x $ 0 1 2 3 4
$ P_{\mathbb{X}|\mathbb{Y}}(x|y=0) $ .175 .635 .159 .000 .031


$ x $ 0 1 2 3 4
$ P_{\mathbb{X}|\mathbb{Y}}(x|y=1) $ .048 .000 .143 .238 .571


$ x $ 0 1 2 3 4
$ P_{\mathbb{X}|\mathbb{Y}}(x|y=2) $ .188 .562 .250 .000 .000


and the marginal $ P_{\mathbb{Y}} $ is empirically estimated as:


$ y $ 0 1 2
$ P_{\mathbb{Y}}(y) $ .63 .21 .16

Estimate the conditional distributions $ P_{\mathbb{Y}|\mathbb{X}} $



By definition $ P_{\mathbb{X}|\mathbb{Y}}(x|y) = \frac{P_{\mathbb{X},\mathbb{Y}}(x,y)}{P_{\mathbb{Y}}(y)} $, so the joint distribution $ P_{\mathbb{X},\mathbb{Y}}(x,y) $ can be computed.

$ P_{\mathbb{X},\mathbb{Y}}(0,0) = P_{\mathbb{X}|\mathbb{Y}}(0|0)P_{\mathbb{Y}}(0) = .175 \cdot .63 = .11 $

Computing the rest of the distribution similarly:

$ P_{\mathbb{X},\mathbb{Y}}(x,y) $
0 1 2 3 4
0 .11 .40 .10 .00 .02
1 .01 .00 .03 .05 .12
2 .03 .09 .04 .00 .00

The marginal distribution $ P_\mathbb{X} $ can be extracted from the joint distribution as:

$ P_\mathbb{X}(x) = \sum_{y\in Y} P_{\mathbb{X},\mathbb{Y}}(x,y) $

$ P_\mathbb{X}(0) = .11 + .01 + .03 = .15 $

Computing the rest of the distribution similarly:


$ x $ 0 1 2 3 4
$ P_{\mathbb{X}}(x) $ .15 .49 .17 .05 .14


Finally $ P_{\mathbb{Y}|\mathbb{X}} $ can be computed by definition.

$ P_{\mathbb{Y}|\mathbb{X}}(0|0) = \frac{P_{\mathbb{X},\mathbb{Y}}(0,0)}{P_{\mathbb{X}}(0)} = \frac{.11}{.15} = .733 $

Computing the rest similarly:


$ y $ 0 1 2
$ P_{\mathbb{Y}|\mathbb{X}}(y|x=0) $ .733 .067 .200


$ y $ 0 1 2
$ P_{\mathbb{Y}|\mathbb{X}}(y|x=1) $ .816 .000 .184


$ y $ 0 1 2
$ P_{\mathbb{Y}|\mathbb{X}}(y|x=2) $ .588 .176 .236


$ y $ 0 1 2
$ P_{\mathbb{Y}|\mathbb{X}}(y|x=3) $ .000 1.00 .000


$ y $ 0 1 2
$ P_{\mathbb{Y}|\mathbb{X}}(y|x=4) $ .143 .857 .000


Note from these $ P_{\mathbb{Y}|\mathbb{X}} $ distributions that for large $ x $ it is highly probable that $ y=1 $ and for small $ x $ it is highly probable that $ y=0 $.

--Jvaught 22:34, 29 January 2010 (UTC)

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

Dr. Paul Garrett