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Thread: Bernstein's inequality (please help)

  1. #1
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    Bernstein's inequality (please help)

    A) Prove Bernstein's inequality: If X is a random variable for which Mx(t) exists for some t>0, the P(X>=x) <= Mx(t)e^(-tx). Assuming X is continuous.

    B) If X~N(0,1), then Mx(t)=e^((t^2)/2) for all t.
    Thus, by A), P(X>=x)<=e^((t^2)/2 - tx).
    For x=2, find the value of t for which this inequality is sharpest.
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  2. #2
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    I'm assuming $\displaystyle M_x(t)$ is the moment generating function. You might want to write out the expression for the moment generating function explicitly in integral form. Similarly, express $\displaystyle P(X\geq x)$ as an integral. Compare the two and play around with it to get the desired inequality.
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  3. #3
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    Quote Originally Posted by ninano1205 View Post
    A) Prove Bernstein's inequality: If X is a random variable for which Mx(t) exists for some t>0, the P(X>=x) <= Mx(t)e^(-tx). Assuming X is continuous.

    B) If X~N(0,1), then Mx(t)=e^((t^2)/2) for all t.
    Thus, by A), P(X>=x)<=e^((t^2)/2 - tx).
    For x=2, find the value of t for which this inequality is sharpest.
    For A), you can write, for $\displaystyle t>0$, $\displaystyle P(X\geq x)=P(e^{tX}\geq e^{tx})$ and apply Markov inequality.
    In B), you are asked for the $\displaystyle t$ for which the given right-hand side is minimum, so you have to study the variations of the function $\displaystyle t\mapsto e^{(t^2)/2 - tx}$ (differentiate, study the sign of the derivative,...) to find which value of $\displaystyle t$ minimizes it.
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