Since all the exponential families are absolutely continuous, if part is trivial. However, I could not prove the only if part. My idea is to prove by contradiction, i.e. given an event $A$ such that $P(X\in A)=0$ for $X\sim F$, where $F$ is some distribution from the exponential families. We wish to show that $P(Y\in A)=0$ if $Y$ follows any absolutely continuous distribution. My idea is to assume the contrary and then create a random variable $Z$ from exponential family such that $P(Z\in A)=0$. My problem actually is that, I can not find a way to utilize the given information to get a contradiction. Please help.

  • $\begingroup$ One way is to show that every a.c. distribution is a member of at least one exponential family. $\endgroup$
    – whuber
    Feb 10 at 14:48
  • 1
    $\begingroup$ Maybe change the title to say probabiity zero white it now says improbable. As for the hint by @whuber, look into exponential tilting for instance stats.stackexchange.com/questions/432688/… $\endgroup$ Feb 10 at 15:06
  • $\begingroup$ @whuber, do you want to say that the set of all absolutely continuous distributions is same as the set of all exponential families?!! $\endgroup$
    – Martund
    Feb 11 at 8:47
  • $\begingroup$ @kjetilbhalvorsen, Isn't improbable same as probability 0? Also, I couldn't understand exponential tilting from the link you've given. What do we do for uniform distribution? $\endgroup$
    – Martund
    Feb 11 at 8:48
  • $\begingroup$ That would be a confusing thing to maintain, because a family is a set of distributions. What is at issue here mathematically is the question whether a point in a the plane (say) lies on any line. Distributions are points in the space of all distributions and a distribution family is a curve (or higher dimensional manifold) in that space. Exponential families are special kinds of curves (just like lines are special kinds of curves). $\endgroup$
    – whuber
    Feb 11 at 13:53

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