Show f is a probability density function

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Homework Help Overview

The discussion revolves around demonstrating that a given function, f(x) = (1 + ux)/2 for -1 ≤ x ≤ 1 and 0 otherwise, qualifies as a probability density function. The participants are exploring the necessary conditions for f to be a valid density function.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the requirement that f(x) must be non-negative and explore different methods to show this. One participant mentions having already verified that the integral of f equals 1, while another suggests examining the boundaries of the interval to determine the maximum and minimum values of f. There are also attempts to derive inequalities to confirm the non-negativity of f.

Discussion Status

The discussion is active, with participants providing various approaches to confirm that f(x) is non-negative. Some have proposed checking the boundaries of the function, while others have suggested using inequalities to establish the non-negativity condition. There is no explicit consensus yet, but multiple lines of reasoning are being explored.

Contextual Notes

Participants are working within the constraints of the defined interval for x and the parameter u, which is bounded between -1 and 1. The discussion includes verification of conditions necessary for f to be a probability density function.

saizen21
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Homework Statement



Let f(x) = (1 + ux)/2 for -1<= x <= 1 and 0 otherwise . where -1<= u <= 1

a) show f is a density


Homework Equations



TO show
1. f(x) >= 0

2. intergeral f (from -infinity to infinity) = 1

The Attempt at a Solution



I have done 2. and proved that it is 1 by taking the intergeral of f.

However, how do u show f >= 0.

i have found the dervavtive of f to be u / 2 for -1<= x <= 1 and 0 otherwise.

I dun know where to go?
 
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If you've shown f'(x)=u/2 then that means the max and min of f(x) must be at x=(-1) or x=1, the boundaries of your interval. Can you take it from there?
 
is this also possible?

1<=x<=1
-1<=ux<=1
0<=1+ux<=2
0<=(1+ux)/2<=1
therefore its always >= 0
 
saizen21 said:
is this also possible?

1<=x<=1
-1<=ux<=1
0<=1+ux<=2
0<=(1+ux)/2<=1
therefore its always >= 0

That works too.
 

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