www.lesswrong.com/posts/dKpC6wHFqDrGZwnah/ais-can-now-often-do-massive-easy-to-v...

This sentence states a general mathematical rule, but that rule does not hold.

Two key points:

1. Medians are not linear/additive. Unlike expectations, there is no general identity such as median(X+Y) = median(X) + median(Y).
2. Right-skewness does not fix this. Even if each summand is right-skewed, the median of the sum need not exceed the sum of the medians.

A concrete counterexample shows the claim is false:

• Let X take values 0 with probability 0.49, 1 with probability 0.50, and 100 with probability 0.01.
• This distribution is right-skewed: its mean is 1.5, which is greater than its median 1.
• Now take independent X and Y with this same distribution.
• median(X) = 1 and median(Y) = 1, so the sum of the medians is 2.
• But P(X+Y <= 1) = P(0,0)+P(0,1)+P(1,0) = 0.49^2 + 2(0.49)(0.50) = 0.7301, while P(X+Y <= 0) = 0.49^2 = 0.2401.
• So the median of X+Y is 1, not greater than 2.

That directly contradicts the article's statement.

What is true is weaker: for skewed distributions, the median of a sum/average can differ from the sum/average of medians, and for non-symmetric distributions the median of (X+Y)/2 is generally a different quantity (the pseudomedian), not the ordinary median. So the inequality used here is not a valid general justification for median(B) - median(A) > median(B-A).