# probability inequality invovling function

• Jan 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
$\displaystyle p(X > \lambda) = p(f(X)>f( \lambda))$, and even somewhere it is written
$\displaystyle p(X > \lambda) \le p(f(X)>f( \lambda))$.

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

Can some explain in detail.

Regards.
• Jan 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
$\displaystyle p(X > \lambda) = p(f(X)>f( \lambda))$, and even somewhere it is written
$\displaystyle p(X > \lambda) \le p(f(X)>f( \lambda))$.

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
• Jan 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... $\displaystyle \{x: x>a\} =\{x: f(x)>f(a)\}$
• Jan 9th 2011, 09:56 PM
amb03

Regards
• Jan 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... $\displaystyle \{x: x>a\} =\{x: f(x)>f(a)\}$

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