Determining the PDF Given a DF

[brendan]

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

 

[HallsofIvy]

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 $0\le x< 1$
f(x)= 1/2 for $1\le x< 2$
f(x)= 0 for $2\le x$

 

[brendan]

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 ?