geometric generating function

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 .


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

= -ln(|yt-1|) / t

Not sure where to go from here to find the generating function of Y.
Any help would be greatly appreciated.

I've never heard of the geometric generating function of an RV but it would seem like you've basically got it. Using that definition, RY(t) = E[(1-tY)^-1] = what you have, which is "called the geometric generating function of" Y. One thing you've forgotten to do is to take the definite integral over your domain. Aside from that, though....

I have been checking the integral should have been:
-ln(|1-t|) / t (No y in it )

and I beleive 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.
But I'm not sure what the value of t is supposed to be ?

I've taken the taylor series on my calculator

and get :

1 + t + t/2 + t^2/3 + t^3/4

regard
Brendan

Yeah, my point was that you had forgotten to get rid of the y in your answer by using a definite instead of indefinite integral. The y drops out of you take the integral with respect to y from 0 to 1.

I'm sorry, I don't understand the business about the 3rd moment and Taylor series. Asking your teacher/prof is probably the best plan. I'm interested in the connection (but don't wanna try deriving it, sorry).

Actually, I do understand it. Generating functions in general yield a polynomial with coefficients of interest. The one you've gotten is a moment-generating function for X (which I'm assuming is uniformly distributed on 0 to 1, as Y is above). You can use it to easily find the moments of X about 0. For example, the first moment of X about 0 is

Integral of (x-0)^1*1 for x between 0 and 1,

which is 1/2, which is the coefficient on the t^1 term of your generating function expansion. The second moment of X about 0, equivalently E[X^2], is

Integral of (x-0)^2*1 = x^2 for x between 0 and 1,

which happens to be 1/3, which is the coefficient on the t^2 term of your generating function expansion. In general, the nth moment of X about zero, equivalently E[X^n], is

Integral of (x-0)^n*1 = x^n for x between 0 and 1,

which happens to be 1/(1+n). If you look at your Taylor series, looking at the t^n term, your coefficient is 1/(1+n). I won't literally say the answer in this case, since if you followed this it should be blindingly obvious. (If not feel free to ask questions.) The upshot is that t is for all intents and purposes a dummy variable, and we're just interested in the coefficients of the generating function. I've heard generating functions described as a "clothesline" where our coefficients of interest are hung at each point on the clothesline, with the polynomial providing nothing more than a convenient algebraic construct to stick some coefficients on.

Thanks for your help!
The moments are
E[x^0] = 1
E[x^1] = 1/2
E[x^2] = 1/3
E[x^3] = 1/4

I'm going to read up on
"Generating functions in general yield a polynomial with coefficients of interest"
I think that would give me a better idea.

Could you please explain where the "Integral of (x-0)^1*1" came from?
regards

Certainly. The nth moment of any random variable X with probability density function p(x) about a particular point c is given by

\int (x-c)^n*p(x) dx.

That's just the definition, copied off the page I linked above. Here, you're ignoring the words "about a particular point" and just assuming that point is 0. Then the nth moment of any random variable X with pdf p(x) is given by

\int (x-0)^n*p(x) dx = \int x^n*p(x) dx.

Here your random variable is uniformly distributed on 0 to 1, so p(x) is just 1 the entire time. The limits of integration could be thought of in two ways, but I think it's easier just to say they go from 0 to 1 since those are the only values X can take on. This gives

\int x^n*1 dx = \int x^n dx, all with limits of integration 0 to 1. [equation 2]

Subbing in n=1, c=0, p(x)=1 [uniform] to my first equation (which is what we've been doing step-by-step) we get

\int (x-c)^n*p(x) dx = \int (x-0)^1*1 dx = \int x^1 dx = "1st moment of X".

I was wondering, I've just picked a certain PDF normal PDF = (1/sqrt(2(pi)))e-(x2 / 2)
Using your information.
The nth moment of any random variable X with probability density function p(x) about a particular point c is given by

\int (x-c)^n*p(x) dx.

If I want the 4th moment I should do

\int (x-0)^4*(1/sqrt(2(pi)))e-(x^2 / 2)dx

= 0.056
I'm relying on this being uniformly distributed on 0 to 1
Is this right?

Your formula "\int (x-0)^4*(1/sqrt(2(pi)))e-(x^2 / 2)dx" is good, but your limits of integration are off. Imagine the graph of the normal curve you've given. You're integrating from 0 to 1--but wait, that's cutting off a ton of your PDF. You could think of it in a programming-esque fashion like this: to perform the integral above, integrating as you did from 0 to 1, you'd never have to evaluate your PDF at points outside the range 0 to 1. Your integral can't take into account the information at the ends of the PDF since it never sees that info, but this is just wrong--the entire PDF should really matter, intuitively.

So, the resolution to it is this: integrate from -inf to inf. Why? Because the normal PDF you've listed is non-zero on that range and contributes information at each non-zero point. Why did we integrate from 0 to 1 before and -inf to inf now? Because the uniform PDF (uniform on 0 to 1) is by definition zero at any point outside of 0 to 1. Those points contribute no information so we ignore them. Algebraically, you could write the integral in three pieces-- -inf to 0, 0 to 1, and 1 to inf. Since p(x) = 0 on the first and the last range if p is uniform on 0 to 1, those integrals cancel out, and we don't write them.

The upshot is, integrate over the region where your probability density function is non-zero. Then it'll turn out the 4th moment in your example is 3.

Thats makes a whole lot of sense I re-did the integration and saw that the result was 3.
Thanks alot for your explanation,
Brendan