Inequality of MGF: Show p(tX >s^2 +logM(t)) < e^-s^2

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SUMMARY

The discussion centers on the moment generating function (MGF) of a random variable X, specifically addressing the inequality p(tX > s^2 + log M(t)) < e^-s^2. Participants explore the application of Chebyshev's inequality in this context, where M(t) represents the expected value E[exp(tx)] for t > 0. The conversation also raises questions about the variance being exp(s) and its implications for the inequality.

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  • Understanding of moment generating functions (MGFs)
  • Familiarity with Chebyshev's inequality
  • Knowledge of probability theory and random variables
  • Basic concepts of variance and expectation
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  • Study the properties of moment generating functions (MGFs) in detail
  • Explore advanced applications of Chebyshev's inequality in probability
  • Learn about the derivation and implications of variance in exponential distributions
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Mathematicians, statisticians, and students of probability theory seeking to deepen their understanding of moment generating functions and inequalities in probability.

cutesteph
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MGF of X denote M(x)=E[exp(tx)] exists for every t>0 . For t>0 Show p(tX >s^2 +logM(t)) < e^-s^2 .
 
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Hey cutesteph.

Does Chebychev's inequality work here?
 
Using chebychev's inequality P( | x-u | >= K(sigma) )=< 1/k^2

x=exp(tx) u= M(t) k=exp(s) sigma=exp(s) Is this correct? Why is the variance exp(s)?
 

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