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Math Help - Probability/Expectation Bound

  1. #1
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    Probability/Expectation Bound

    Let h: R --> [0, \alpha] be a nonnegative bounded function. Show that for 0\leq a<\alpha,
    P(h(X)\geq a)\geq \frac{E(h(X))-a}{\alpha - a}.

    The only idea I have is that we want a\leq h(X)\leq \alpha. So, P(h(X)\geq a) becomes P(a\leq h(X)\leq \alpha).

    I know that P(h(X) \geq a)\leq \frac{E(h(X))}{a}.

    Then, shouldn't it be that P(a\leq h(X))\leq \frac{E(h(X))}{a} and P(h(X)\leq \alpha)\geq 1 - \frac{E(h(X))}{\alpha}

    However, I'm not sure how to use this all together or if I'm even on the right track.
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  2. #2
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    Quote Originally Posted by azdang View Post
    Let h: R --> [0, \alpha] be a nonnegative bounded function. Show that for 0\leq a<\alpha,
    P(h(X)\geq a)\geq \frac{E(h(X))-a}{\alpha - a}.

    The only idea I have is that we want a\leq h(X)\leq \alpha. So, P(h(X)\geq a) becomes P(a\leq h(X)\leq \alpha).

    I know that P(h(X) \geq a)\leq \frac{E(h(X))}{a}.

    Then, shouldn't it be that P(a\leq h(X))\leq \frac{E(h(X))}{a} and P(h(X)\leq \alpha)\geq 1 - \frac{E(h(X))}{\alpha} In fact, you have \color{red} P(h(X)\leq \alpha)=1 because of the assumption.

    However, I'm not sure how to use this all together or if I'm even on the right track.
    Let Y=h(X). Then Y is a r.v. such that 0\leq Y\leq \alpha. The idea is to procede like for the proof of Markov inequality P(Y>a)\leq \frac{E[Y]}{a}, including the fact that Y\leq\alpha.

    You have E[Y]=E[Y 1_{(Y>a)}]+E[Y 1_{(Y\leq a)}]. Now find an upper bound for the first term using Y\leq \alpha, and an upper bound for the second term (similar to Markov). Then you're almost there... I let you give it a try.
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  3. #3
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    Here's what I tried to do so far:

    <br />
E[Y]=E[Y 1_{(Y>a)}]+E[Y 1_{(Y\leq a)}] (Those are supposed to be a's not alpha's, correct?)


    So, EY\geq E(Y1_{Y>a}) \geq E(a1_{Y>a}) = aE1_{Y>a} = aP(Y>a)

    I'm not sure if this helps or is what you were talking about.

    I tried to find an upper bound on the second term. This is what I did:

    E(Y1_{Y\leq a})\leq E(a1_{Y\leq a})=aE1_{Y\leq a} = aP(Y\leq a) and wouldn't P(Y\leq a) be 0? So the second term is bounded above by 0?
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  4. #4
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    Quote Originally Posted by azdang View Post
    Here's what I tried to do so far:

    <br />
E[Y]=E[Y 1_{(Y>a)}]+E[Y 1_{(Y\leq a)}] (Those are supposed to be a's not alpha's, correct?)


    So, EY\geq E(Y1_{Y>a}) \geq E(a1_{Y>a}) = aE1_{Y>a} = aP(Y>a)

    I'm not sure if this helps or is what you were talking about.
    You gave the upper bound EY\geq E(Y1_{Y>a}) . That's correct, but one can do better: E(Y 1_{Y>a})\leq E(\alpha  1_{Y>a}) because Y\leq \alpha.



    I tried to find an upper bound on the second term. This is what I did:

    E(Y1_{Y\leq a})\leq E(a1_{Y\leq a})=aE1_{Y\leq a} = aP(Y\leq a) and wouldn't P(Y\leq a) be 0? So the second term is bounded above by 0?
    I don't know what this would be zero..... And since the second term is nonnegative, this would mean that it equals zero ; there is no reason for that.

    So, we have EY \leq \alpha P(Y>a) + a P(Y\leq a). Cool. Now, you can express the right-hand side in terms of P(Y>a) only, and conclude.
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  5. #5
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    Quote Originally Posted by Laurent View Post
    You gave the upper bound EY\geq E(Y1_{Y>a}) . That's correct, but one can do better: E(Y 1_{Y>a})\leq E(\alpha  1_{Y>a}) because Y\leq \alpha.
    I am a bit confused about this part.

    If EY\geq E(Y1_{Y>a})\leq E(\alpha)1_{Y>a} how can we definitively say that EY\leq E(\alpha)1_{Y>a}
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  6. #6
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    Nevermind. I completely understand! Thank you SO much!
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