$$\bullet \text{ Let} \quad X_1,X_2,…,X_n \text{ iid }X_1\sim \operatorname{Ge}\big( \theta\big)$$
$$\text{Goal: } \mathbf X =(X_1,…,X_n)^T \in \text{ Exponential Family of distributions.}$$
$$\text{ Its enough to show that :}$$
$$\exists \phi \in \Phi \subseteq \mathbb R^k,k\in\mathbb N, \text{parametric space s.t pmf:}$$
$$f(\mathbf X;\phi)=h(x)\cdot \exp{\bigg\{ \big< s(\mathbf X),\phi \big>_{\mathbb R^k} -K(\phi)\bigg\}},\mathbf X\in \mathcal X ,\phi\in\Phi . $$
$$\text{Note : $\big< \cdot \big>$ Inner product on $\mathbb R^k$}.$$
$$\text{pmf of Geometric distr: }\Omega =\mathbb N,\operatorname{P}\big( X=x\big)=f_X(x)=(1-\theta )^{x-1}\theta .$$
$$\text{cdf: }F_X(x)=\operatorname{P}\big( X\le x \big)=1-(1-\theta )^x $$
$\text{So,}$
$$F_{\mathbf X}(\mathbf x)=\operatorname{P}\big(X_1\le x_1,X_2\le x_2,…,X_n\le x_n \big)=$$
$$ \color{black}{\underbrace{ \operatorname{P}\big(X_1\le x_1\big) \operatorname{P}\big(X_2\le x_2\big)\cdots \operatorname{P}\big( X_n\le x_n \big)= }_{\text{independent} }}$$
$$=\color{black}{\underbrace{ \operatorname{P}\bigg(X_1\le x_1 \bigg)^n }_{ \text{identically }}}\implies F_{\mathbf X}'(\mathbf x)=n\operatorname{P}\big( X_1\le x_1\big)^{n-1}\cdot f_{X_1}(x_1) $$
$$=n\big( 1-(1-\theta )^{x_1}\big)^{n-1}\cdot \theta \big( 1-\theta \big)^{x_1-1}=f_{\mathbf X}(\mathbf x)$$
So, how can i continue the proof from this point?
This is as far as i can get…
Any ideas?
Thank you.
Best Answer
let $\theta = p$ and $1-\theta = q$, then $$ P_{\mathrm{X}}(x_1,...,x_n) = \prod_{i=1}^n q^{x_i-1}p=q^{\sum x_i - n}p^n, $$ applying $x = g(x) = \exp\{\ln x\}$ you have $$ P_{\mathrm{X}}(x_1,...,x_n) = \exp\Big\{(\sum x_i - n)\ln q+n\ln p\Big\} $$