1. ## 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.

2. Originally Posted by azdang
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.

3. Here's what I tried to do so far:

$
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?

4. Originally Posted by azdang
Here's what I tried to do so far:

$
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.

5. Originally Posted by Laurent
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 $Y\leq \alpha$.
If $EY\geq E(Y1_{Y>a})\leq E(\alpha)1_{Y>a}$ how can we definitively say that $EY\leq E(\alpha)1_{Y>a}$