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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:
<br /> 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]<br />
But wouldn't P\ [\sup\limits_{x}\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z] and \sup\limits_{x}\ P\ [\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z] 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:
<br /> 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]<br />
But wouldn't P\ [\sup\limits_{x}\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z] and \sup\limits_{x}\ P\ [\ |f(x)\ -\ f'(x)|\ >\ y\ |\ z] 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.