# Thread: PGF of Discrete Random Variable

1. ## PGF of Discrete Random Variable

Hi Everyone,

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

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

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

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

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

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

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

2. you placed the math OUTSIDE the [tex] tex [ /math]

Originally Posted by sirellwood
Hi Everyone,

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

P(R=j) = $\displaystyle (j+1)$$\displaystyle p^j(1-p)^2, j = 0,1,2,.... 0<p<1 Has a P.G.F..... GR(s) = \displaystyle \left(\frac{1-p}{1-ps}\right)^2 Above is meant to be the whole fraction to the power 2, but i was having some Latex Syntax problems showing that I have tried changing \displaystyle s^j and \displaystyle p^j to \displaystyle ps^jand taking it outside the summation when beginning to try and find the PGF but i didnt get anywhere, help needed please! Also maybe \displaystyle \Sigma \displaystyle (j+1)$$\displaystyle a^j$$\displaystyle {(1-a)}^2$ = 1 for any a where $\displaystyle \mid$a$\displaystyle \mid$ < 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..........

$\displaystyle {d\over dx} \sum_{n=0}^{\infty}x^n=\sum_{n=1}^{\infty}nx^{n-1}$

3. Thanks matheagle! any ideas on how to solve the question? :-)