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Dec 5, 2016 at 21:34 vote accept T. Amdeberhan
Dec 5, 2016 at 21:17 comment added Fedor Petrov Is not my counterexample to maj version correct? For a permutation $(k+2,\dots,n,1,\dots,k+1)$ we get $x$ in too small power, no?
Dec 5, 2016 at 18:45 comment added T. Amdeberhan Sure. We'll check then.
Dec 5, 2016 at 16:56 comment added Fedor Petrov Now I start to doubt that it is true for maj, please see my answer.
Dec 5, 2016 at 15:00 comment added T. Amdeberhan Thanks, Fedor. In that case, we need more work for the "maj" problem.
Dec 5, 2016 at 14:58 comment added Fedor Petrov For $k=1$ even the common distribution of $(maj,\pi_n)$ is the same as a common distribution of $(inv,\pi_n)$. For $k=2$ the common distribution of $(maj,\pi_{n-1}+\pi_n)$ is not already the same as a common distribution of $(inv,\pi_{n-1}+\pi_n)$, which may be seen from the case $\pi_{n-1}+\pi_n=2n-1$.
Dec 3, 2016 at 19:33 history edited T. Amdeberhan CC BY-SA 3.0
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Dec 3, 2016 at 18:15 history edited T. Amdeberhan CC BY-SA 3.0
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Dec 3, 2016 at 17:01 history edited T. Amdeberhan CC BY-SA 3.0
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Dec 3, 2016 at 8:52 answer added Fedor Petrov timeline score: 5
Dec 3, 2016 at 8:38 comment added Fedor Petrov Why this replacement does not affect $Q$?
Dec 3, 2016 at 5:53 history edited T. Amdeberhan CC BY-SA 3.0
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Dec 3, 2016 at 5:48 history asked T. Amdeberhan CC BY-SA 3.0