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Why is the following $$f(x|\theta) = \theta^{-1} \exp(1-(x/\theta)), \ \ 0 < \theta < x <\infty$$ not an exponential family? We know that $$f(x| \theta) = h(x)c(\theta) \exp(w(\theta)t(x))$$ where $h(x) = e, \ c(\theta) = \theta^{-1}, w(\theta) = \theta^{-1}$ and $t(x) = -x$.

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up vote 3 down vote accepted

It depends on your definition of exponential family, but the problem is that the support of the distribution depends on $\theta$. You cannot enforce that $f(x|\theta)=0$ when $x\le \theta$ with functions that depend solely on $x$ or $\theta$ and not both.

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In other words, this $h(x)c(\theta)\exp(w(\theta)t(x))$ is NOT $f(x\mid\theta)$ for every $(x,\theta)$. –  Did Jan 28 '12 at 17:08

A more formal version of deinst's argument: Assume $f(x|\theta)=h(x)c(\theta)\exp(w(\theta) t(x))$. For any fixed $\theta$ and $x<\theta$ we have $$f(x|\theta)=h(x)c(\theta)\exp(w(\theta) t(x))=0,$$ but on the other hand $f(x|\theta)>0$ for all $x>\theta$. This implies $c(\theta)>0$ and $h(x)=0$ for all $x<\theta$. Since $\theta$ was arbitrary, we get $h(x)\equiv 0$, a contradiction.

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