# Exponential family definition appears vacuous

I am going through Michael Jordan's notes on exponential families and an exponential family is defined w.r.t. functions $$h(\cdot), T(\cdot)$$, and parameter $$\eta$$ such that $$p(x | \eta) = h(x) \exp\bigl\{\eta^\top T(x) - A(\eta)\bigr\}$$ with $$A(\eta) = \log\int h(x) \exp\left(\eta^\top T(x) \right) \, dx.$$ The thing that strikes me as odd about this definition is that it appears too flexible due to the choice of "$$h$$". For instance, say we define our family in terms of $$h_1, \eta_1$$, and $$T_1$$. We could then reparameterize our family in terms of $$h_2, \eta_2$$, and $$T_2$$ such that

1. $$\eta_2 = 0$$
2. or alternatively, $$T_2(x) = 0$$
3. $$h_2(x) = h_1(x) \exp\left(\eta_1^\top T_1(x)\right)$$

so the choices of $$\eta$$ and $$T(\cdot)$$ are not important (at least in terms of defining the density).

Moreover, it seems we might use this trick to mimic any density (not necessarily ones commonly thought of as "exponential families"), simply by letting $$\eta = 0$$ and making a "sneaky" choice for $$h(\cdot)$$.

What am I missing here?

• Suppose I want to consider the family of distributions which are uniform on $[0,\eta]$. So $p(x\mid \eta) = \frac1\eta \mathbb I_{0 \le x \le \eta}$. Can you find suitable $h(x),T(x),A(\eta)$ to show this is an exponential family? Oct 22 '19 at 23:10
• @Henry - is the domain in your example the interval $[0,\eta]$ or $[0,+\infty)$?
– ted
Oct 23 '19 at 0:43
• @Henry - Assuming the domain is $\{ x : x \ge 0\}$, then for $x \in [0,\eta]$ we would let $h_2(x) = \frac{1}{\eta}$ and for $x > \eta$ we would let $h_2(x) = 0$. This should work since the denominator is $\int_0^{\infty} \frac{dx}{\eta} = \int_0^{\eta} \frac{dx}{\eta} = 1$ and the numerator is $h_2(x) = \frac{1}{\eta}$ whenever $x \in [0,\eta]$ (and is zero otherwise).
– ted
Oct 23 '19 at 0:54
• Your definition of $h_2(x)$ seems to depend on the value of $\eta$, and that is precisely what is not allowed for exponential families. Any dependence on the parameter $\eta$ should be in the $\exp(\cdots)$ term. Since that is always positive but here you need it to sometimes to be $0$, you can conclude that this family of uniform distributions is not an exponential family. Oct 23 '19 at 7:52
• @Henry - thanks for helping me see the light! I didn't realize that $h(\cdot)$ couldn't implicitly depend on $\eta$.
– ted
Oct 24 '19 at 4:20

No, the definition is not vacuous. One consequence of the definition is that the support of all the distributions in the family is the same, which excludes the uniform distribution family $$\mathcal{Unif}(0,\theta)$$. This is easy to see, the exponential factor is never zero, and the $$h$$ function is the same for all the family members (your $$h_2$$ example in the post breaks this assumption.)

We can use exponential tilting to generate an exponential family from a random variable (if that random variable admits a moment generating function (mgf), but if we start with a standard uniform random variable, we do not get a family of uniform distributions. What we get is the family $$f(x\,; \theta)\, = \exp(\theta x -k(\theta))\cdot \mathbb{I}(x\in (0,1)\,)$$ where $$k(\theta)=\log( (e^\theta -1)/\theta )$$ (and zero if $$\theta=0$$) is the cgf (cumulant generating function) of the standard uniform. The cgf is the logarithm of the mgf. A few members of this family is shown below:

The code for the plot is below:

k <- function(theta) ifelse(abs(theta)<0.000001, 0.0,
log( (exp(theta)-1) / theta ) )

f_tilted <- function(x, theta) ifelse( (x<1)&(x>0),
exp(theta*x-k(theta)), 0)

library(RColorBrewer)

palette <- brewer.pal(9, "Greens")
theta <- seq(-2.0, 2, length.out=9)
plot(x=c(0, 1),  y=c(0, 2.5), xlab="x", ylab="tilted density", type="n")
for (ind  in 1:9 )  {
plot( function(x)f_tilted(x, theta[ind]),