# PGF of Discrete Random Variable

• Mar 12th 2010, 08:32 AM
sirellwood
PGF of Discrete Random Variable
Hi Everyone,

I am trying to show that the discrete random variable....

P(R=j) = $(j+1)$ $p^j$ ${(1-p)}^2$, j = 0,1,2,.... 0<p<1

Has a P.G.F.....

GR(s) = ( $\frac{1-p}{1-ps}$)^2

Above is meant to be the whole fraction to the power 2, but i was having some Latex Syntax problems showing that (Thinking)

I have tried changing $s^j$ and $p^j$ to $ps^j$and taking it outside the summation when beginning to try and find the PGF but i didnt get anywhere, help needed please! (Nod)

Also maybe $\Sigma$ $(j+1)$ $a^j$ ${(1-a)}^2$ = 1 for any a where $\mid$a $\mid$ < 1 can be used? i cant see how though!
• Mar 12th 2010, 02:17 PM
matheagle
you placed the math OUTSIDE the [tex] tex [ /math]

Quote:

Originally Posted by sirellwood
Hi Everyone,

I am trying to show that the discrete random variable....

P(R=j) = $(j+1)$ $p^j(1-p)^2$, j = 0,1,2,.... 0<p<1

Has a P.G.F.....

GR(s) = $\left(\frac{1-p}{1-ps}\right)^2$

Above is meant to be the whole fraction to the power 2, but i was having some Latex Syntax problems showing that (Thinking)

I have tried changing $s^j$ and $p^j$ to $ps^j$and taking it outside the summation when beginning to try and find the PGF but i didnt get anywhere, help needed please! (Nod)

Also maybe $\Sigma$ $(j+1)$ $a^j$ ${(1-a)}^2$ = 1 for any a where $\mid$a $\mid$ < 1 can be used? i cant see how though!

The trick in these is to differentiate the geometric, that brings downs that power of j.

Inside the interval of convergence you can..........

${d\over dx} \sum_{n=0}^{\infty}x^n=\sum_{n=1}^{\infty}nx^{n-1}$
• Mar 12th 2010, 02:22 PM
sirellwood
Thanks matheagle! any ideas on how to solve the question? :-)
• Mar 12th 2010, 04:21 PM
sirellwood