1. Probability Generating Functions

Can I please have some help on this subject?

I have a continous random variable given by

f(x)=k(3/8)^x

and have been asked to find the probabilty generating function Gx(N) and so far have:

Gx(N)=SUM(0 -> infinity) k(n^x)(3/8)^x

Can I please have some help on this subject?

I have a continous random variable given by Mr F says: Surely you mean discrete random variable ......

f(x)=k(3/8)^x

and have been asked to find the probabilty generating function Gx(N) and so far have:

Gx(N)=SUM(0 -> infinity) k(n^x)(3/8)^x
$G_x(n) = k \sum_{x=0}^{\infty} n^x \left( \frac{3}{8} \right)^x = k \sum_{x=0}^{\infty} \left( \frac{3n}{8} \right)^x = \frac{k}{1 - \frac{3n}{8}} = \frac{8k}{8 - 3n}$

since the sum is that for an infinite geometric series (and is not finite unless $0 < \frac{3n}{8} < 1$).

To get the value of k, you use the infinite geometric series again:

$k \sum_{x=0}^{\infty}\left( \frac{3}{8} \right)^x = 1 \Rightarrow \frac{k}{1 - \frac{3}{8}} = 1 \Rightarrow k = \frac{5}{8}$.

3. Thanks, with a follow up question;
The wiki page on this is confusing me, to find the expectation;
do I differentiate and put n=1?
and if so is it the same approach for variance?

4. Hello,

Thanks, with a follow up question;
The wiki page on this is confusing me, to find the expectation, do I differenttiate and put n=1? and if so is it the same approach for Variance?
If f is the generating function of a random variable X, then :

$f'(1)=\mathbb{E}(X)$ (like you said)

and $f''(1)=\mathbb{E}(X(X-1))=\mathbb{E}(X^2)-\mathbb{E}(X)$, by linearity of the expectation.
This implies : $\mathbb{E}(X^2)=f''(1)+\mathbb{E}(X)=f''(1)+f'(1)$

Knowing that $\text{var}(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2$, you can say :
$\text{var}(X)=f''(1)+f'(1)-\left(f'(1)\right)^2$

5. One further thing, its just a clarification on notation
my question literally says:

f(x)= xexp(x)

does that mean f(x) = x*2.718^x ? as in the expotential?