Example of Turning Conditional Distributions Around
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}} $
Solution 1
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)
Solution 2
Or, equivalently, we can use Bayes' Rule explicity.
Bayes' Rule is:
$ P_{\mathbb{Y}|\mathbb{X}}(y|x) = \frac{P_{\mathbb{X}|\mathbb{Y}}(x|y)P_{\mathbb{Y}}(y)}{P_{\mathbb{X}}(x)} $
$ P_{\mathbb{X}}(x) $ can be computed using:
$ P_\mathbb{X}(x) = \sum_{y\in Y} P_{\mathbb{X}|\mathbb{Y}}(x|y)P_{\mathbb{Y}}(y) $
Thus, calculation of $ P_{\mathbb{Y}|\mathbb{X}}(Y=0|X=0) $ would proceed as follows:
$ P_{\mathbb{Y}|\mathbb{X}}(Y=0|X=0) = \frac{P_{\mathbb{X}|\mathbb{Y}}(X=0|Y=0)P_{\mathbb{Y}}(Y=0)}{P_{\mathbb{X}}(X=0)} = \frac{P_{\mathbb{X}|\mathbb{Y}}(X=0|Y=0)P_{\mathbb{Y}}(Y=0)}{\sum_{y\in Y} P_{\mathbb{X}|\mathbb{Y}}(X=0|Y=y)P_{\mathbb{Y}}(Y=y)} $
$ = \frac{0.175 \cdot 0.63}{0.175 \cdot 0.63 + 0.048 \cdot 0.21 + 0.188 \cdot 0.16} = \frac{.11025}{.15041} = 0.732996476 \approx 0.733 $
The rest of the conditional distribution can be computed similarly using Bayes' Rule and will result in the same answer as in Solution 1.
- This solution is nearly identical to Solution 1, differing only in that this solution does not contruct the joint probability mass function.
--Pritchey 14:38, 1 February 2010 (UTC)
