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 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 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 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 at 4:07
  • 1
    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 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.

  • Thanks for your reply. Thanks for writing the logic of not belonging to the exponential family. – Sheikh Jul 10 at 15:40

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.