What Is the PDF Given This DF?

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The discussion revolves around determining the probability density function (pdf) from a given distribution function (DF). The DF is defined piecewise, with specific values for different ranges of x, and there is a correction suggested for continuity at x=1. It is clarified that the pdf is indeed the derivative of the DF, leading to the correct pdf expressions for each range of x. The probability of x being between 0 and 1 can be directly obtained from the DF without needing the pdf, as F(1) equals 1/2. The conversation concludes with a confirmation of the correct value of F(1/2).
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I have been asked to determine the pdf given a DF.

F(x) = 0 for x<0
F(x) = x^2/x for x 0<= x < 1
F(X) = x/2 for 1<=x<2
F(x) = 1 for x>= 2

Is the pdf the derivative of the df

So if you wanted the probability of x between 0 and 1
The pdf would just be x ?


regards
Brendan
 
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brendan said:
I have been asked to determine the pdf given a DF.

F(x) = 0 for x<0
F(x) = x^2/x for x 0<= x < 1
Surely this isn't right? did you mean x^2/2, so that F is continuous at x=1?

F(X) = x/2 for 1<=x<2
F(x) = 1 for x>= 2

Is the pdf the derivative of the df

So if you wanted the probability of x between 0 and 1
The pdf would just be x ?


regards
Brendan
You don't need the pdf at all to answer that question. F(x) is the probability that the random variable is between 0 and x. The probablity that x is between 0 and 1 is just F(1)= 1/2.

The pdf is the function
f(x)= 0 for x< 0
f(x)= x for 0\le x&lt; 1
f(x)= 1/2 for 1\le x&lt; 2
f(x)= 0 for 2\le x
 
Thanks alot. You are right it is x^2/2 my mistake.
So does that mean if x = 1/2 , F(1/2) = 1/2 ?
 
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