# Thread: Series Expansion of Uniform MGF

1. ## Series Expansion of Uniform MGF

X has uniform dist (0,1).

I found the MGF to be:

$\displaystyle \frac{e^s - 1}{s}$

Now I need to expand the MGF in powers of s up to $\displaystyle s^2$ and use this to find the mean and variance of X. I know the series expansion for $\displaystyle e^s = \Sigma \frac{s^r}{r!}$, but I'm not sure how to apply this in this situation.

2. Hello,

$\displaystyle e^s=\sum_{n=0}^\infty \frac{s^n}{n!}=1+\sum_{n=1}^\infty \frac{s^n}{n!}$

So $\displaystyle \frac{e^s-1}{s}=\sum_{n=1}^\infty \frac{s^{n-1}}{n!}$

by changing the indice, this is $\displaystyle \sum_{n=0}^\infty \frac{s^n}{(n+1)!}$

3. EDIT: I think I understand what you did there.

But, I'm still stuck on another part of this question:

In adding n real numbers, each is rounded to the nearest integer. Assume that the round-off errors, $\displaystyle X_{i}$, i = 1,...,n, are independently distributed as U(-0.5,+0.5). What is the approximate distribution of the total error, $\displaystyle \Sigma X_{i}$, in the sum of the n numbers?

How would I approach this question?