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This should just be a quick clarification, but I couldn't find a clear answer on the MSE or elsewhere online.

It is clear that if $X\sim Beta(\alpha,\beta)$, then $X$ we can write the pdf of $X|\theta$ as

$$f(x|\theta) = g(\theta)h(x)exp[\eta(\theta)\cdot t(x)]$$ with $g(\theta) = \frac{1}{B(\alpha,\beta)}, h(x)=1, \eta(\theta)=(\alpha-1,\beta-1)^T$, and $t(x) = (ln(x), ln(1-x))^T$. We also know that the support of $X$, $(0,1)$, does not depend on the vector of parameters. Thus, $X$ belongs to the exponential family.

It is also clear that in general if $Y \sim U(0,\theta)$, then the support of $Y$ depends on the parameter $\theta$, and thus $Y$ cannot belong to the exponential family.

But in the case that $Y \sim U(0,1)$, then this is just a special case of the beta distribution with $\alpha = \beta = 1$. This seems to indicate to me that the standard uniform should be contained in the exponential family, despite the fact that the general uniform is not.

Is my intuition correct here? Should we instead say that $Y \sim U(0,\theta)$ does not belong to the exponential family for $\theta \neq 1$?

theauthor
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  • $Y \sim U(0,\theta)$ does not belong to the exponential family of beta distributions when $\theta \neq 1$, rather obviously. But for given $\theta>0$, there will be another exponential family of distributions that it does belong to. – Henry Jan 20 '25 at 14:56

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The language is loose, and this can cause confusion.

An exponential family of distributions is a set of parameterised distributions with the properties you describe; the functions $g,h,\eta,t$ may be different in different exponential families. One such is the set of beta distributions with the parameter $\theta=(\alpha,\beta)$. Another is the set of normal distributions. There is also the set of exponential distributions, as well as (with a wider parameter space) the set of gamma distributions. Yet another is the set of binomial distributions for a given fixed sample size $n$ with the parameter $\theta=p$. But the set of binomial distributions for a given fixed $p$ and parameter $\theta=n$ is not an exponential family, and in particular has a support which varies with $n$.

The exponential family of distributions is typically used to mean the set of exponential families of distributions, i.e. a set of sets. So you cannot just say $Bin(5,\frac13)$ is a member of the exponential family; instead you can say $Bin(5,\frac13)$ is a member of the $Bin(5,p)$ exponential family of distributions, but it is also a member of the $Bin(n,\frac13)$ non-exponential family of distributions and the wider $Bin(n,p)$ non-exponential family of distributions.

Similarly you can say $U[0,1]$ is is a member of the $Beta(\alpha, \beta)$ exponential family of distributions, but it is also a member of the $U[0,\theta]$ non-exponential family of distributions and the wider $U[\theta_1,\theta_2]$ non-exponential family of distributions.

The same is true of any other uniform distribution. For example $U[0,7]$, is a member of the exponential family of distributions for random variables which are seven times a beta-distributed random variable, but it is not a member of the same exponential family as $U[0,1]$ or indeed the same exponential family as $\mathcal N(0,1)$ or as $Bin(5,\frac13)$.

Henry
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