Is a distribution part of the exponential family if the support depends on the parameter? (namely, $f_\theta (x)=e^{-(x-3\theta)}, 3\theta<x<\infty$)

Question

I have trouble verifying is the distribution

$$f_\theta (x)=e^{-(x-3\theta)}\qquad\qquad for \ \ 3\theta<x<\infty$$

Is part of the exponential family. In particular, one can rewrite the above as:

$$f_\theta (x)=e^{-(x-3\theta)}I_{(3\theta,\infty)}(x)$$

If one would like to write $f_\theta(x)$ as

$$f_\theta(x)=c(\theta)h(x)e^{\sum_{j=1}^{p}q_j(\theta)T_j(x)}$$

One should manipulate $I_{(3\theta,\infty)}(x)$ as follows

$$I_{(3\theta,\infty)}(x)=e^{log(I_{(3\theta,\infty)}(x))}$$

Now sustituting

$$f_\theta(x)=e^{-(x-3\theta)}e^{log(I_{(\theta,\infty)}(x))}=e^{-x+3\theta+log(I_{(3\theta,\infty)}(x))}$$

Now, I find it quite dificult to factorize the function $log(I_{(3\theta,\infty)}(x))$ as a product of two functions that depend only on $x$ and $\theta$ respectively. That may indicate that its not an exponential-family-type distribution; however, I don't know how to prove that I am not smart enough to see it.

Thanks in advance for your collaboration.

Answer 1

Suppose it is possible to write $$f_\theta (x)=e^{-(x-3\theta)}\qquad\qquad \text{for} \ \ 3\theta<x<\infty\tag1 $$ in the form $$ f_\theta(x)=c(\theta)h(x)e^{\sum_{j=1}^{p}q_j(\theta)T_j(x)}.\tag2 $$ Then $c(\theta)>0$ for every $\theta$, since every $f_\theta$ is a density. Pick an arbitrary $x\in{\mathbb R}$. Pick $\theta>x/3$. By (1) we have $f_\theta(x)=0$. Comparing with (2) we conclude $h(x)=0$, since all other factors are positive. Since $x$ was arbitrary, we've shown $h(x)=0$ for all $x$ which means $f_\theta(x)=0$ for all $x$ -- a contradiction.