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Suppose $g$ is a function from $\mathbb{R}\to\mathbb{R}$ with the following properties:

1) $g$ is continuous and increasing

2) $g(x)\leq 1$ for all $x>0$

3) $g(0)=0$

Suppose $X$ is a random variable. Show that $P(|x| > b) \geq E[g(x)] - g(b)$ for all $b\geq0$

I've thought about the Chebychev inequality, but that doesn't help since the inequality sign is in the wrong direction over there. I also tried changing the probability term on the left hand side to Expectations of appropriate indicator functions. Doesn't seem to be going anywhere. Any ideas?

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1 Answer

Notice that

$$E[g(X)]=\int_{|X|\leq b}g(X)dP + \int_{|X|>b}g(X)dP\leq g(b)+P(|X|>b)$$

where we used that $g(b)\geq g(y)$ for $y\leq b$, $P(|X|\leq b)\leq 1$ and $g(y)\leq 1$

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IS the second step $\int g(X)dp < \int g(b)dp=g(b) \int dp< g(b)$? –  Alex Oct 20 '12 at 6:13
    
Not quite. The first integral is over $|X|\leq b$, otherwise you can't conclude what you wrote as $X$ could be bigger than $b$ –  Alex R. Oct 20 '12 at 12:26
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