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We know that Chebyshev's inequality is:

$\Pr[|X-\mathbb E(X)| \ge t] \le \frac{\mathrm Var (X)}{t^2}$

And for all $r>0$ and $t>0$,

$\Pr[|X-\mathbb E(X)| \ge t]= \Pr[|X-\mathbb E(X)|^r \ge t^r]$

So by Markov's inequality,

$\Pr[|X-\mathbb E(X)| \ge t] \le \frac{\mathbb E[|X-\mathbb E(X)|^r] }{t^r}$

My first question is: in what situation, we can get a better bound that the Chebyshev's bound using the above bound? The second question is: how large should we I choose $r$ to get a best bound?

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    **Related**: T. K. Phillips and R. Nelson (1995), [The moment bound is tighter than Chernoff's bound for positi$v$e tail probabilities](http://www.jstor.org/pss/2684633), *The American Statistician*, vol 42, no. 2., 175-178.2011-09-21

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