# Thread: probability inequality invovling function

1. ## 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.

2. 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

3. 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... $\displaystyle \{x: x>a\} =\{x: f(x)>f(a)\}$

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

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