probability inequality invovling function

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January 9th 2011, 09:36 PM
amb03

probability inequality invovling function

Dear All.

I have read in somewhere that if f is non decreasing function then
$p(X > \lambda) = p(f(X)>f( \lambda))$ , and even somewhere it is written
$p(X > \lambda) \le p(f(X)>f( \lambda))$ .

My question is this how this inequality is true, and which inequality is correct?

Can some explain in detail.

Regards.
January 9th 2011, 09:44 PM
snowtea

Quote:

Originally Posted by amb03

Dear All.

I have read in somewhere that if f is non decreasing function then
$p(X > \lambda) = p(f(X)>f( \lambda))$ , and even somewhere it is written
$p(X > \lambda) \le p(f(X)>f( \lambda))$ .

My question is this how this inequality is true, and which inequality is correct?

Can some explain in detail.

Regards.

Neither is true. Consider f(x) = 0 for all values of x.

Do you mean f is strictly increasing?
In which case both are true since f(x) > f(y) iff x > y
January 9th 2011, 09:45 PM
matheagle

It has more to do with set theory

If a<b and f is strictly increasing then f(a)<f(b)

If f is strictly increasing then the two sets are equal... $\{x: x>a\} =\{x: f(x)>f(a)\}$
January 9th 2011, 09:56 PM
amb03

Thank for reply, but may you please check the following link, this inequality is stated here

http://www.cs.cmu.edu/afs/cs/academi...ribes/lec9.pdf

Regards
January 27th 2011, 01:46 AM
amb03

Quote:

Originally Posted by matheagle

It has more to do with set theory

If a<b and f is strictly increasing then f(a)<f(b)

If f is strictly increasing then the two sets are equal... $\{x: x>a\} =\{x: f(x)>f(a)\}$

Would you please explain me how is it possible?. I could not understand