3
$\begingroup$

Are there notable cases where analytic continuations of cumulative distribution functions to complex arguments have a meaningful interpretation or are otherwise useful?

If a one dimensional CDF is analytic, then it can be continued to some complex arguments. The resulting values would not be directly meaningful as probabilities, but perhaps for some natural families of probability distributions, they have some other useful meaning. The meaning would presumably be specific to a distribution family (as opposed to being some general theory of complex probability) because an $ε$-change to CDF (or PDF) can have an unbounded effect on analytic continuation of CDF. (Plus, the continuation might not be one-to-one, and it may depend on the branch.)

This question was inspired by a question Complex (i.e., Imaginary) Probability. Let me know if it is too elementary and should be migrated to Math Stack Exchange.

Improve this question
$\endgroup$
3
  • $\begingroup$ (I do not think the question is too elementary, but I was not sure.) $\endgroup$
    – Dmytro Taranovsky
    Commented Sep 27 at 1:14
  • $\begingroup$ The phrase "cumulative density functions" is an oxymoron, since "cumulative" contradicts "density." There are cumulative distribution functions "c.d.f."s and there are probability density functions "p.d.f."s $\endgroup$
    – Michael Hardy
    Commented Sep 27 at 19:01
  • 1
    $\begingroup$ @MichaelHardy Good point. I fixed the post. $\endgroup$
    – Dmytro Taranovsky
    Commented Sep 27 at 19:12

1 Answer 1

Reset to default
7
$\begingroup$

Analytic continuation of probability densities and PDF to the complex plane has no direct probabilistic meaning but this is an important tool in proving many results which do have probabilistic meaning. An example is Marcinkiewicz theorem:

If the sum of two independent random variables is normally distributed, then each summand is normally distributed.

To prove this one notices that the characteristic function of the sum is entire, of order $2$ and without zeros in the complex plane, and then a simple argument shows that the summands must have the same property.

Similar results hold for Poisson distributions, and for sums of Gaussian and Poisson distributions.

Ref. Linnik, Ostrovskii, Decomposition of random variables and vectors, AMS, 1977.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you. I was hoping for applications of analytic continuations of PDFs or CDFs, but your answer is also helpful. Note that the theorem does not assume that the probability distribution is analytic since we are using characteristic functions, which are entire for distributions with $e^{-ω(n)}$ tails. $\endgroup$
    – Dmytro Taranovsky
    Commented Sep 27 at 18:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .