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Revision as of 08:40, 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:
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)