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Exponential family has a very good property that could be used to conclude if a statistic is complete:

$X_1,X_2,\ldots,X_n$ are from exponential family which has the form as: $$f(x\mid \theta )=h(x)c(\theta )e^{\sum_{i=1}^k w_i(\theta )t_i(x)},\text{ where }\theta =(\theta_1,\ldots,\theta _d ), d\leqslant k$$

Then $T(X)=\left(\sum_{j=1}^k t_1(X_j), \ldots , \sum_{j=1}^k t_k(X_j)\right)$ is complete if $\{w_1(\theta ), \ldots , w_k(\theta )\}$ contains an open set in $\mathbb{R}^k$

With this property, I'm solving a question as following:

$X_1,X_2,\ldots,X_n$ are iid from $\mathrm{Geo}(\theta ), 0<\theta <1$

$$f(x\mid \theta )=(1-\theta)^{x-1}\theta=\frac{\theta }{1-\theta }e^{x\cdot \ln\frac{\theta}{1-\theta}}$$

My question is:

how to explain $\left\{\ln\frac{\theta}{1-\theta}\right\}$ contains an open set in $\mathbb{R}^1$, isn't $\left\{\ln\frac{\theta}{1-\theta}\right\}$ itself a point in $\mathbb{R}^1$?

Thanks!!

Michael Hardy
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Liz Sugar
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1 Answers1

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Probably what was meant was that $\{(w_1(\theta),\ldots,w_n(\theta) : \theta \in\Theta \}$ contains an open set in $\mathbb R^k$, where $\Theta$ is the set of values of $\theta$ for which the integral of the function $x\mapsto h(x) e^{\sum_{i=1}^k w_i(\theta )t_i(x)}$ is finite. If the integral is finite, then there is a normalizing constant $c(\theta)$ by which the function can be multiplied to get a probability density function.

In the case of the geometric distribution, the "integral" is the sum $$ \sum_{x=1}^\infty e^{x\cdot \ln(\theta/(1-\theta))}. $$ The sum is finite if $\ln(\theta/(1-\theta))<0$, and that happens precisely if $0<\theta<1$. (If $\theta\le0$ or $\theta\ge1$ then the expression is undefined.)

So the set you're looking for is $\left\{ \ln\dfrac \theta{1-\theta} : 0<\theta<1 \right\}$. And that set is actually the whole of $\mathbb R$. And that certainly contains an open subset of $\mathbb R^1$. (I suspect it should have said a non-empty open subset.)

Michael Hardy
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