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

[cutesteph]

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 .

 

[chiro]

Hey cutesteph.

Does Chebychev's inequality work here?

 

[cutesteph]

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)?