Determine the Geometric generating function

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SUMMARY

The geometric generating function RX(t) = E[(1 − tX)−1] is established for the random variable Y, which follows a uniform distribution on (0, 1). The correct formulation for the geometric generating function of Y is derived as -ln(|1-t|) / t. To compute the third moment E[X^3], one must take the Taylor series expansion of -ln(|1-t|) / t and extract the coefficient corresponding to the third term.

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Homework Statement



Suppose RX(t) = E[(1 − tX)−1] is called the geometric generating function
of X. Suppose the random variable Y has a uniform distribution on (0, 1); ie
fY (y) = 1 for 0 < y < 1. Determine the geometric generating function of Y .

Homework Equations





The Attempt at a Solution



E[(1-tY)^-1] = \int (1-ty)^-1 f_Y(y) dy

-ln(|yt-1|) / t

Do I than take the Taylor series of the result to give the geometric generating function for Y?
 
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I have been checking the integral should have been:
-ln(|1-t|) / t

and I believe that this is the generation function of Y.

To find the value E[X^3] of -ln(|1-t|) / t
apparently I have to take the taylor series of -ln(|1-t|) / t
and read off the 3rd moment.

I'm a bit lost Any Help greatly appreciated

regards
Brendan
 

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