# Is an exponential family closed under convolution?

Let $$y = x_1 + x_2 + x_3$$ where $x_1, x_2, x_3$ are draws of random variables of the same family and different parameters: $p(x_1| \theta_1), p(x_2| \theta_2), p(x_3| \theta_3)$

We know that, for some distributions that are closed under convolution, then the distribution over $y$ belongs to the same distribution.

My question concerns the case where $x_i$ belong to some exponential family of distributions: does the sum of random variables from an exponential family also belong to some exponential family?

• Use the MGF uniqueness theorem
– Sycorax
Oct 6 '17 at 16:05
• @kjetilbhalvorsen this is not a all self-study. I've been searching in texts about the Exponential Family but I haven't seen any mention to it. Oct 6 '17 at 16:08
• @alberto you still aren't grasping. The exponential distribution is a member of the exponential family. As I said, it is an example. Your assertion is wrong. In math, a counter example suffices to disprove a claim, so your "other examples" should have discredit the thought immediately. Oct 6 '17 at 16:27
• Well, the sum of (independent) variables pertaining to some exponential family belongs to some other exponential family! This is proved in (from memory) Bickel&Doksum, for instance Oct 6 '17 at 16:30
• @AdamO You and kjetil and alberto appear to be talking at cross-purposes: the issue concerns whether alberto means the family of Exponential distributions or an exponential family of distributions. Notice the distinctions between "the" and "an," as well as the different capitalization. The meanings are quite different! Alberto, please edit your question to clarify your meaning.
– whuber
Oct 6 '17 at 16:37

The property that you describe refers to stable distributions

You can study these convolutions of pdf's by products of characteristic functions. For instance:

• The Cauchy-Lorentz distribution is a stable distribution $$E(e^{itX}) = e^{it\mu_X+\sigma_X\vert t\vert}$$ and $$E(e^{itX+Y}) = e^{it\mu_X+\sigma_X\vert t\vert} e^{it\mu_X+\sigma_X\vert t\vert} =e^{it(\mu_X+\mu_Y)+(\sigma_X+\sigma_Y)\vert t\vert} = e^{it\mu_Z+\sigma_Z\vert t\vert}$$ with $\mu_Z=\mu_X+\mu_Y$ and $\sigma_Z=\sigma_X+\sigma_y$

• but, for instance, the Gamma distribution is not a stable distribution $$E(e^{itX}) = (1-it\theta_X)^{k_X}$$ and $$E(e^{itX+Y}) = (1-it\theta_X)^{k_X} (1-it\theta_Y)^{k_Y} \neq (1-it\theta_Z)^{k_Z}$$ where we can not make this final step (unless $\theta_X = \theta_Y$)

So for a particular distribution that is in the exponential family it is not (generally) true that is a stable distribution.

Still you may wonder whether the sum of any two distributions in the exponential family could be any other distribution in the exponential family.

• There is no general description of the characteristic function of an exponential family distribution $f(x) = h(x)g(\theta) e^{\eta(\theta) \cdot T(x)}$, and trying to work out the convolution for the general case is messy (and I don't think it works).

Possibly one could work out a particular case to show that a convolution is yielding some distribution that is not a exponential family.

• If the sufficient statistic T(x) is a linear function then at least the cumulant generating function for T might be used to proof some stability property:

$$k(u \vert \eta) = A(\eta + u) + A(\eta)$$

by using the property $k(X+Y) = k(X) + k(y)$. However, I am not sure about the inverse step, whether the cumulant generating function $k(X) + k(y)$ is necessarily referring to a valid distribution from the exponential family. But at least this is an interesting notion.