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
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
$\displaystyle G_x(n) = k \sum_{x=0}^{\infty} n^x \left( \frac{3}{8} \right)^x = k \sum_{x=0}^{\infty} \left( \frac{3n}{8} \right)^x = \frac{k}{1 - \frac{3n}{8}} = \frac{8k}{8 - 3n}$
since the sum is that for an infinite geometric series (and is not finite unless $\displaystyle 0 < \frac{3n}{8} < 1$).
To get the value of k, you use the infinite geometric series again:
$\displaystyle k \sum_{x=0}^{\infty}\left( \frac{3}{8} \right)^x = 1 \Rightarrow \frac{k}{1 - \frac{3}{8}} = 1 \Rightarrow k = \frac{5}{8}$.
Hello,
If f is the generating function of a random variable X, then :
$\displaystyle f'(1)=\mathbb{E}(X)$ (like you said)
and $\displaystyle f''(1)=\mathbb{E}(X(X-1))=\mathbb{E}(X^2)-\mathbb{E}(X)$, by linearity of the expectation.
This implies : $\displaystyle \mathbb{E}(X^2)=f''(1)+\mathbb{E}(X)=f''(1)+f'(1)$
Knowing that $\displaystyle \text{var}(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2$, you can say :
$\displaystyle \text{var}(X)=f''(1)+f'(1)-\left(f'(1)\right)^2$