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VincentYu comments on Solving the two envelopes problem - Less Wrong
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This does not follow; a counterexample:
Suppose X and Y are independent random variables, with X taking the values {2,100}, and Y the values {1,150}, each with equal probability (i.e., each of X and Y follows the Bernoulli distribution with p = 0.5). Then we have
E(X/Y) = (2/1 + 100/1 + 2/150 + 100/150) / 4 = 25.67 > 1,
but
E(X) = 51 < 75.5 = E(Y).
Keep in mind that the equation E(1/Y) = 1/E(Y) does not hold in general, because taking the inverse is not a linear transformation. To evaluate the expectation after a nonlinear transformation, one requires not just the orignal expectation, but also the pdf of the distribution. (I can't be sure this is what you did, but misapplying this gives: E(X/Y) = E(X)E(1/Y) = E(X)/E(Y). The first equality holds if X and Y are independent or by definition if X and 1/Y are uncorrelated, but the second equality does not hold in general.)
Thanks, good point. Hopefully I don't turn out to be falsely complacent about this conclusion too :)
Addressed by making a few edits to the "Solution" section. Thank you!
An even easier example: X is always -4 and Y is always -2.
Since amounts of money are positive, this is a counterexample to a mistake not in the original article.
Thanks, this is indeed a simpler counterexample. Note that it is a counterexample to a different error in the original statement. In particular, the error you've caught here is that from
X/Y > 1,
one cannot conclude that
X > Y.