Say I have a random variable $X$ and I am interested in finding a good bound on $\mathbb{P}( X \geq a)$ for $a \in \mathbb{R}$. I could do this through the the one sided Chebyshev inequality or I could rely on the moment generating function and use the Chernoff bounds. Assuming you have the $s$ that minimizes the MGF, my question is, which would you expect to be tighter to the actual probability $\mathbb{P}( X \geq a)$ for a given $a$? It seems like it would be Chernoff since the MGF contains information about all the moments while the Chebyshev is just using the first two.
Chernoff bound vs One-Sided Chebyshev
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probability
statistics
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0Chebyshev's inequality is more general; consequently, weaker. Chebyshev's inequality for the normal distribution is very poor, for instance. So definitely Chernoff bounds if you need to be accurate. – 2017-02-02
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0... or an explicit computation if that is viable, of course. – 2017-02-02
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0Cheers. Quick question, is there a way to recover the CDF (and hence the probability) directly from the MGF? – 2017-02-02
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0That is an instance of the moment problem (have a look at Wikipedia) that can be solved through the (inverse) Laplace and Fourier transforms. – 2017-02-02