Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor

    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)?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...