# probability inequality

#### Zanetti

##### New Member
Let X be a random variable such that |X|<=C (bounded). Let f be an even, non decreasing function over the positive values of x. Prove that:

[E(f(X))-f(a)]/f(C) <= P(|X-E(X)|>=a) <= [E(f(X-E(X))]/f(a).

The 2nd inequality is really easy to prove and it follows from chebyshev inequality...I am having hard time solving the first 1, any help is strongly appreciated. Thanks!

#### fed1

##### TS Contributor
Let a = 0.
Let f(s) = s^2.
Let c = 10.
let Px(x) = Pr[X = x] = u for X >0, l = 1 - u;

Pr[ |X - u| >= a ] = 1.

E[ x^2/100 ] = integral( x^2 * l) from -10 to 0 + integral( x^2 * u) from 0 to 10.

= (1/3)(1/100)(10^3)[ u - l ];

Looks problematic?

#### Zanetti

##### New Member
Thank you for you reply, but you have to prove it for any function f, not for a special case.

#### fed1

##### TS Contributor
I show that it is not true in general.

#### Zanetti

##### New Member
Well obviously it's not gonna work when f(a) = 0 since it appears in the denominator of the Right inequality...