# Why is the continuous uniform distribution not an exponential family?

In class I've been given that a family of distributions $\{P_{\theta} : \theta \in \Theta\}, \Theta \in \mathbb{R}^{k}$ is an exponential family if there exits real-valued functions $\eta_{1}, ... , \eta_{k}, B$ of $\theta$ and $T_{1}, ... , T_{2}, h$ of $x$, such that the pmf or pdf of $P_{\theta}$ can be written as:

$p_{\theta}(x) = exp \left[\sum_{1}^{k}\eta_{i}(\theta)T_{i}(x) \space -B(\theta) \right]h(x)$

I'm asked to determine whether $\mathcal{U}([0,\vartheta]), \space \vartheta > 0$ is an exponential family. I've read online that the uniform distributions are not exponential families, however based on the definition I've been given it seems fairly easy to show that it is.

Let $\eta_{1}(\vartheta) = 0, \space T_{1}(x) = 0, \space B(\vartheta) = ln(\vartheta), \space h(x) = 1$ and we have that:

$exp \left[\sum_{1}^{k}\eta_{i}(\theta)T_{i}(x) \space -B(\theta) \right]h(x) = exp[-\ln(\vartheta)] = \frac{1}{\vartheta}$ gives us our desired pdf.

So what is the problem here?

Is it that we have to restrict $x \in [0,\vartheta]$ for it to truly be our desired pdf? Because we have to restrict $x$ for the Bernoulli distribution as well and that is still an exponential family.

Has it just been decided that the uniform distributions aren't exponential families by convention since they don't vary anyways so it's not worth thinking of them as exponentials?

If someone could explain this I'd be very grateful.

• Simply because the support depends on $\theta$ – Xi'an Dec 9 '16 at 10:12
• To put a little differently, the density is not $1/\theta$, but $1/\theta\cdot I_{x\in[0,\theta]}$. – Christoph Hanck Dec 9 '16 at 10:13
• To clarify: at the start you say $B$ is a function of $\theta$, but then later in your definition of $p_\theta(x)$ you have $B$ as a function of $x$, and then again later it changes back to $\theta$, and then again back to $x$. Which is right? – Ruben van Bergen Dec 9 '16 at 10:14
• @RubenvanBergen Apologies, those were just typos. I've edited the post to fix them. – bones_mccoy Dec 9 '16 at 10:20
• @ChristophHanck Of course! I hadn't thought of how restricting the support is the same as including the indicator function as a factor. This makes things much more clear and also makes the restriction of the support in the Bernoulli example seem much less arbitrary. – bones_mccoy Dec 9 '16 at 10:29

Since the indicator function $\mathbb{I}_{(0,\theta)}(x)$ is part of the definition of the Uniform density, one cannot enter it inside the exponential product part $\exp\{\eta_1(\theta)T_1(x)+\ldots\}$.