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Inequality of mgf

  1. Nov 3, 2012 #1
    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 .
     
  2. jcsd
  3. Nov 3, 2012 #2

    chiro

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    Hey cutesteph.

    Does Chebychev's inequality work here?
     
  4. Nov 4, 2012 #3
    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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