# Finding the pmf from the distribution function

• Apr 22nd 2008, 09:53 PM
cirrus74
Finding the pmf from the distribution function
http://i235.photobucket.com/albums/e...4/untitled.jpg

• Apr 23rd 2008, 04:42 AM
CaptainBlack
Quote:

Originally Posted by cirrus74
http://i235.photobucket.com/albums/e...4/untitled.jpg

I think we will have to assume that the possible values of $x$ are integers, then:

(a) well obviously we have:

$p_X(x)=0,\ x<-1$

and

$p_X(x)=0,\ x \ge 1$,

so that leaves only $p_X(-1)$ and $p_X(0)$ to be determined, can you do that?

RonL
• Apr 23rd 2008, 09:25 PM
cirrus74
Well actually, the answer that was given to us was:

http://i235.photobucket.com/albums/e...untitled-1.jpg

But I don't know how they got it?
• Apr 23rd 2008, 09:36 PM
CaptainBlack
Quote:

Originally Posted by cirrus74
Well actually, the answer that was given to us was:

http://i235.photobucket.com/albums/e...untitled-1.jpg

But I don't know how they got it?

When $F$ jumps by $d_i$ at $x_i$, then $p(x_i)=d_i$

RonL