Is this a member of an exponential family?

Question

I strongly believe that this distribution does not belong to the exponential family:

$f(x;\theta) = \frac{\theta}{2}^{|x|}(1 - \theta)^{1-|x|}I_{\{-1, 0, 1\}}(x)$.

I have to write $f(x;\theta)$ as $a(\theta)b(x)\exp(c(\theta)d(x))$, I don't think this can be done. All I can do is write:

$f(x;\theta) = \exp(|x|\log(\theta) + (1-|x|)\log(1-\theta))I_{\{-1, 0, 1\}}(x)$.

You can try to factorise the coefficients of $|x|$ in the exponential. — Xi'an, Oct 29 '18 at 17:56
Think in this way: $f(x;\theta) = 1-\theta$ for x = 0, $=\frac \theta 2$ for $x$ = -1 or +1. It is kind of multinomial distribution. — user158565, Oct 29 '18 at 17:56
Or just look at the factorisation $|x|(\log(\theta)-\log(1-\theta))$. — Xi'an, Oct 29 '18 at 18:06

score 1 · Accepted Answer · answered Nov 04 '18 at 19:02

Since $$f(x;\theta) = \frac{\theta}{2}^{|x|}(1 - \theta)^{1-|x|}\mathbb{I}_{\{-1, 0, 1\}}(x)=\frac{1}{2}\left\{\frac{\theta}{(1-\theta)}\right\}^{|x|} \mathbb{I}_{\{-1, 0, 1\}}(x)$$the likelihood associated with a sample $(x_1,\ldots,x_n)$ is $$L(\theta|x_1,\ldots,x_n)=\prod_{i=1}^n f(x_i;\theta)= \frac{1}{2^n}\left\{\frac{\theta}{(1-\theta)}\right\}^{\sum_{i=1}^n|x_i|} $$ which factorises through the unidimensional $$\sum_{i=1}^n|x_i|$$ for all $n$'s. Thus an exponential family.

Is this a member of an exponential family?

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