What Is the PDF Given This DF?

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The discussion centers on determining the probability density function (pdf) from a given distribution function (DF). The DF is defined as follows: F(x) = 0 for x < 0, F(x) = x^2/2 for 0 ≤ x < 1, F(x) = x/2 for 1 ≤ x < 2, and F(x) = 1 for x ≥ 2. The pdf is confirmed to be the derivative of the DF, resulting in f(x) = 0 for x < 0, f(x) = x for 0 ≤ x < 1, f(x) = 1/2 for 1 ≤ x < 2, and f(x) = 0 for x ≥ 2. The probability of x being between 0 and 1 is directly derived from F(1) = 1/2.

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brendan
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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 probability 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 [itex]0\le x< 1[/itex]
f(x)= 1/2 for [itex]1\le x< 2[/itex]
f(x)= 0 for [itex]2\le x[/itex]
 
Thanks a lot. 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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