Let X be geometrically distributed with parameter $0<p<1$. Therefore $E[X] = 1/p$. Im trying to solve the fisher information but im stuck on how
$-E(1/p^2 + x(-1)/(1-p)^2)$ which = $1/p^2 + E[X]/(1-p)^2$ simplifies to become $$1/p^2 +1/p(1-p)$$. which equals $$ 1/p(1-p)$$
How does it become $1/(p(1-p))$? And why does the $1/p^2$ disappear?
Would really appreciate your help.
There are two Distributions named Geometric. They appear to be using the "from-zero" supported sequence.
The distribution of failures before the first success in an indefinite sequence of independent Bernoulli trials with identical success rate $p$ has a Geometric Distribution with mean $(1-p)/p$.$$\begin{align}X&\sim\mathcal{Geo}_0(p)\\[1ex] \mathsf P(X{=}k)&=(1-p)^k p~\mathbf 1_{k\in\{0,1,2,\ldots\}}\\[1ex] \mathsf E(X)&=\dfrac{1-p}p\end{align}$$
Then, however, we should have $$\begin{align}\dfrac 1{p^2}+\dfrac 1{p(1-p)} &= \dfrac{(1-p)+p}{p^2(1-p)}\\[1ex]&=\dfrac{1}{p^2(1-p)}\end{align}$$