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Dec 29, 2016 at 22:16 comment added Puzzled OK, got it. $P(Y'\le 4^n)<P(X'\le 4^n)$, $P(Y'\le 2\times 4^n)>P(X'\le 2\times 4^n)$, $P(Y'\le 3\times 4^n) = P(X'\le 3\times 4^n)$. For this counterexample to hold we need to have unbounded random variable. I will add a word bounded in the original post.
Dec 29, 2016 at 15:32 history answered Douglas Zare CC BY-SA 3.0