Determine the Geometric generating function

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The discussion focuses on determining the geometric generating function for a random variable Y with a uniform distribution on (0, 1). The geometric generating function is expressed as E[(1-tY)^-1], leading to the integral formulation involving the uniform probability density function. The user initially computes the integral and arrives at -ln(|1-t|)/t as the generating function for Y. They seek clarification on whether taking the Taylor series of this result will yield the moments of Y, specifically the third moment E[X^3]. The conversation highlights the need for assistance in confirming these calculations and understanding the process.
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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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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