# Show that an event is improbable for exponential families iff it's improbable for all absolutely continuous distributions

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.

• One way is to show that every a.c. distribution is a member of at least one exponential family.
– whuber
Feb 10 at 14:48
• 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/… Feb 10 at 15:06
• @whuber, do you want to say that the set of all absolutely continuous distributions is same as the set of all exponential families?!! Feb 11 at 8:47
• @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? Feb 11 at 8:48
• 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).
– whuber
Feb 11 at 13:53