- #1
Portella
- 7
- 0
I'm trying to deal with the supremum concept in a specific situation, but I think I'm getting the concept wrong.
A step of a proof I'm going through states:
[itex]
P\ [\sup\limits_{x}\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z]\ \ \leq\ \ \sum_{i=1}^M\ P\ [\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z]\ \ \leq\ \ M\times\ \sup\limits_{x}\ P\ [\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z]
[/itex]
But wouldn't [itex] P\ [\sup\limits_{x}\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z] [/itex] and [itex] \sup\limits_{x}\ P\ [\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z] [/itex] be the same. After all, wouldn't the same x that bears the greatest absolute value also bear the greatest probability in the second case? I just want to understand the need for the step described above and the consequent multiplication by M.
A step of a proof I'm going through states:
[itex]
P\ [\sup\limits_{x}\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z]\ \ \leq\ \ \sum_{i=1}^M\ P\ [\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z]\ \ \leq\ \ M\times\ \sup\limits_{x}\ P\ [\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z]
[/itex]
But wouldn't [itex] P\ [\sup\limits_{x}\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z] [/itex] and [itex] \sup\limits_{x}\ P\ [\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z] [/itex] be the same. After all, wouldn't the same x that bears the greatest absolute value also bear the greatest probability in the second case? I just want to understand the need for the step described above and the consequent multiplication by M.