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I'm trying to prove that $e^{-k-1} \leq \frac{1}{k^2}$ for $k > 1$ but I feel I'm missing something (maybe an standard inequality?) Could anyone give me a pointer like "Use such fact"?


For $k > 1$ and the random variable $X \sim \mbox{Exponential}(\beta)$:

$P(|X-\mu_X| \geq k \sigma_X) = e^{-k-1}$

And using Chebyshev's inequality:

$P(|X-\mu_X| \geq k \sigma_X) \leq \frac{1}{k^2}$


(This is my first question in the site, so please let me know if I did something wrong.)

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As a remark, König's theorem in a special case gives a standard inequality $k<2^k$ and so $k^2<e^{k\cdot\log(4)}$ follows. This is slightly weaker than what you want, however. – NikolajK Sep 21 '12 at 16:51
up vote 1 down vote accepted

Put $$ f(k)=k^2 e^{-k-1} $$ and compute $$ f'(k)=-e^{-k-1}k(k-2). $$ You can easily check that $k=2$ is a maximum point at which $f$ is less than $1$.

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Thanks, that worked for me. – Franco Victorio Sep 21 '12 at 14:57

$e^{-k-1} \leq \frac{1}{k^2} \Leftrightarrow \frac{1}{e}k^2 \leq e^k $ and then taylor

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You have to prove that $e^{k+1}\geq k^2$ for $k\geq 1$. Since for $k\geq 3$ we have $\frac{k^3}6\geq\frac{k^2}2$, we have, using the power series defining the exponential, $$e^{k+1}\geq e^k\geq \frac{k^2}2+\frac{k^3}6\geq k^2.$$

Edited: If $1\leq k\leq 2$, $k^2\leq 4\leq e^2\leq e^{k+1}$; and if $2\leq k\leq 3$, $e^{k+1}\geq e^3\geq 9$.

So Chebyshev doesn't give a sharp bound.

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If I'm understanding well, you're assumming $k$ is an integer, but I think it's not the case. (But maybe I'm not understanding your answer.) – Franco Victorio Sep 21 '12 at 15:01
I indeed assumed it was an integer (maybe the notation $k$ for an undiscrete variable is not common, but it's allowed). i've corrected accordingly. – Davide Giraudo Sep 21 '12 at 15:17

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