Here we demonstrate that, when random variables X and Y from two different distributions are compared, P(X > Y) and P(Y > X) need bear no particular relationship to the relative locations of the medians of the two distributions. This happens because the location of the median gives you only one reference point on the distribution. The behaviour of the tails will help determine the probabilities P(X > Y) and P(Y > X).
Here X and Y have identical medians, but P(X > Y) != P (Y > X).
| Value | Prob of X | Prob of Y |
|---|---|---|
| 2 | 0.49 | 0.0 |
| 1 | 0.0 | 0.49 |
| 0 | 0.02 | 0.02 |
| -1 | 0.49 | 0.0 |
| -2 | 0.0 | 0.49 |
P(X > Y) = 0.49 + 0.02 * 0.49 + 0.49 * 0.49 = 0.7399
P(Y > X) = 0.49 * 0.51 + 0.02 * 0.49 = 0.2597
Here the median of X is 1 and the median of Y is -1. Despite this, p(X > Y) = P(Y > X).
| Value | Prob of X | Prob of Y |
|---|---|---|
| 3 | 0.175 | 0.225 |
| 2 | 0.225 | 0.175 |
| 1 | 0.2 | 0.0 |
| 0 | 0.0 | 0.0 |
| -1 | 0.0 | 0.2 |
| -2 | 0.175 | 0.225 |
| -3 | 0.225 | 0.175 |
P(X > Y) = 0.175 * 0.775 + 0.225 * 0.6 + 0.2 * 0.6 + 0.175 * 0.175 = 0.42125
P(Y > X) = 0.225 * 0.825 + 0.175 * 0.6 + 0.2 * 0.4 + 0.225 * 0.225 = 0.42125