# Finding the pmf from the distribution function

Printable View

• Apr 22nd 2008, 08:53 PM
cirrus74
Finding the pmf from the distribution function
http://i235.photobucket.com/albums/e...4/untitled.jpg

No idea how to do part (a). Can someone please help me? Thanks!
• Apr 23rd 2008, 03:42 AM
CaptainBlack
Quote:

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

No idea how to do part (a). Can someone please help me? Thanks!

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

(a) well obviously we have:

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

and

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

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

RonL
• Apr 23rd 2008, 08:25 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, 08:36 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 \$\displaystyle F\$ jumps by \$\displaystyle d_i\$ at \$\displaystyle x_i\$, then \$\displaystyle p(x_i)=d_i\$

RonL