# Is the negative exponential distribution a member of the exponential family?

Please correct me if I am wrong.

The general form of k-parameter exponential family is

$f(x;\boldsymbol{\theta}) = a(\boldsymbol{\theta})g(x) \exp\{\sum_{i=1}^{k}b_(\boldsymbol{\theta}) R_i(x)\}$

Let, $X_1, \ldots, X_n \sim \dfrac{1}{\sigma} \exp\{ -(x-\mu)/\sigma \} I(x>\mu); \mu \in R, \sigma \in R^+$ [the common pdf of negative exponential distribution].

The joint distribution is

$\dfrac{1}{\sigma^n} \exp\{ -\sum_{i=1}^{n}(x_i-\mu)/\sigma \} I(x_{n:1}>\mu)$

which cannot be absorbed in the general expression of the exponential family mentioned above due to the part $I(x_{n:1}>\mu)$.

Thus the negative exponential distribution does not belong to the exponential family.

Please correct me if I am wrong.

• The usual one-parameter exponential (en.wikipedia.org/wiki/Exponential_distribution) is in the family, but if you add a location-shift parameter, then it isn't. – Glen_b Jul 10 '18 at 12:22
• Thanks for your reply. If the location parameter, $\mu$ is known, then this belongs to one-parameter exponential family. I hope I am correct? – Sheikh Jul 10 '18 at 15:45
• You can write it as $\exp(\eta(\theta)T(x)-A(\theta))$ (setting $h$ to $1$), because if $\mu$ is known, you can put it in $T(x)$. [If this is homework you should clearly signal that.] – Glen_b Jul 11 '18 at 1:10
• Thank you again. This is not a homework problem. I wanted to clarify my understanding of exponential family. Your comments really helped. – Sheikh Jul 11 '18 at 4:07
• If the support depends on a parameter (as it does for $\mu$) then it can't be exponential family, but if you know $\mu$ then that's just a constant, not a parameter. – Glen_b Jul 11 '18 at 5:15

This is correct, if $\mu$ is a parameter of the distribution rather than a given, the indicator function implies this distribution is not an exponential family.