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]. Here, $I$ is an indicator function.

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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Jul 10, 2018 at 12:22
  • $\begingroup$ Thanks for your reply. If the location parameter, $\mu$ is known, then this belongs to one-parameter exponential family. I hope I am correct? $\endgroup$
    – Sheikh
    Jul 10, 2018 at 15:45
  • $\begingroup$ 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.] $\endgroup$
    – Glen_b
    Jul 11, 2018 at 1:10
  • $\begingroup$ Thank you again. This is not a homework problem. I wanted to clarify my understanding of exponential family. Your comments really helped. $\endgroup$
    – Sheikh
    Jul 11, 2018 at 4:07
  • 2
    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Jul 11, 2018 at 5:15

1 Answer 1


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.

  • $\begingroup$ Thanks for your reply. Thanks for writing the logic of not belonging to the exponential family. $\endgroup$
    – Sheikh
    Jul 10, 2018 at 15:40
  • $\begingroup$ Hi Xi'an. I'm trying to understand your reasoning here and have posted this stats.stackexchange.com/q/520049/163242 question. $\endgroup$ Apr 16, 2021 at 15:17
  • 2
    $\begingroup$ Any parametric family that enjoys a variable (i.e., parameter dependent) support is not an exponential family, because the indicator function cannot be turned into the exponential of a function of the parameter times a function of the sample. $\endgroup$
    – Xi'an
    Apr 16, 2021 at 16:41

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