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Let $(M_{k,n})_{\substack{0\leqslant k\leqslant K\\ 0\leqslant n \leqslant N}}$ a matrix of nonnegative random variables and let $C$ be a constant. Assume that for all $0\leqslant n\leqslant N$, $\mathbb{E}\big[\max_{0\leqslant k\leqslant K} M_{k,n} \big] \leqslant C$ and for all $0\leqslant k\leqslant K$, $\mathbb{E}\big[\max_{0\leqslant n\leqslant N} M_{k,n} \big] \leqslant C$.

I'm trying to prove (or disprove ) this $\mathbb{E}\big[\max_{\substack{0\leqslant n\leqslant N\\ 0\leqslant k\leqslant K}} M_{k,n} \big] \leqslant C$ ?

Any hints would be appreciated.

20Xblog8x12
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1 Answers1

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It cannot be hold. Let $K=1$, and $M_{0,n}, M_{1,n}\sim \mathcal U (0,1), n=0,\dots,N$, for which we have $C=\frac{N}{N+1}$. However, $$\mathbb{E}\left[\max_{\substack{0\leqslant n\leqslant N\\ 0\leqslant k\leqslant K}} M_{k,n} \right] =\frac {2N}{2N+1}>C$$

considering that the distribution of the maximum of $m$ iid random variables with uniform distribution on $(0,1)$ follows $\text{Beta}(m,1).$

Amir
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  • The OP probably meant to specify that $K>0$ – FD_bfa Apr 27 '24 at 18:32
  • I don't understand your conterexample, in this case $\mathbb{E}\left[\max_{\substack{0\leqslant n\leqslant N\ 0\leqslant k\leqslant K}} M_{k,n} \right] = \mathbb{E}\left[\max_{\substack{0\leqslant n\leqslant N}} M_{0,n} \right] $ by taking $C=\frac{N}{N+1}$, my clam still holds – 20Xblog8x12 Apr 27 '24 at 19:49
  • Could you check the updated example, in which both row and column expectations are less than $\frac{N}{N+1}$, but the matrix expectation is $\frac{2N}{2N+1}$, larger than $\frac{N}{N+1}$? – Amir Apr 27 '24 at 20:20
  • Yes i see, thank you very much – 20Xblog8x12 Apr 27 '24 at 21:09