# Halving a discrete random variable?

Let $X$ be a discrete random variable taking its values in $\mathbb{N}$. I would like to halve this variable, that is, to find a random variable $Y$ such as:

$$X = Y + Y^*$$

where $Y^*$ is an independent copy of $Y$.

• I am referring to this process as halving; this is a made-up terminology. Is there a proper term found in the literature for this operation?
• It looks to me that such $Y$ always exists only if we accept negative probabilities. Am I correct in my observation?
• Is there a notion of best positive fit for $Y$? Aka the random variable that would be the "closest" to solve the equation above.

Thanks!

• In the cases where you can't "halve" exactly, there are multiple possible definitions of "closest"; it depends on what you want to optimize. – Glen_b -Reinstate Monica Oct 19 '15 at 0:11

A notion strongly related to this property (if weaker) is decomposability. A decomposable law is a probability distribution that can be represented as the distribution of a sum of two (or more) non-trivial independent random variables. (And an indecomposable law cannot be written that way. The "or more" is definitely irrelevant.) A necessary and sufficient condition for decomposability is that the characteristic function $$\psi(t)=\mathbb{E}[\exp\{itX\}]$$ is the product of two (or more) characteristic functions.

I do not know whether or not the property you consider already has a name in probability theory, maybe linked with infinite divisibility. Which is a much stronger property of $X$, but which includes this property: all infinitely divisible rv's do satisfy this decomposition.

A necessary and sufficient condition for this "primary divisibility" is that the root of the characteristic function $$\psi(t)=\mathbb{E}[\exp\{itX\}]$$ is again a characteristic function.

In the case of distributions with integer support, this is rarely the case since the characteristic function is a polynomial in $\exp\{it\}$. For instance, a Bernoulli random variable is not decomposable.

As pointed out in the Wikipedia page on decomposability, there also exist absolutely continuous distributions that are non-decomposable, like the one with density$$f(x)=\frac{x^2}{\sqrt{2\pi}}\exp\{-x^2/2\}$$

In the event the characteristic function of $X$ is real-valued, Polya's theorem can be used:

Pólya’s theorem. If φ is a real-valued, even, continuous function which satisfies the conditions

φ(0) = 1,
φ is convex on (0,∞),
φ(∞) = 0,


then φ is the characteristic function of an absolutely continuous symmetric distribution.

Indeed, in this case, $\varphi^{1/2}$ is again real-valued. Therefore, a sufficient condition for $X$ to be primary divisible is that φ is root-convex. But it only applies to symmetric distributions so is of much more limited use than Böchner's theorem for instance.

There are some special cases where this holds true, but for an arbitrary discrete random variable, your "halving" is not possible.

• The sum of two independent Binomial$(n,p)$ random variables is a a Binomial$(2n,p)$ random variable, and so a Binomial$(2n,p)$ can be "halved".
Exercise: figure out whether a Binomial$(2n+1,p)$ random variable can be "halved".

• Similarly, a Negative Binomial$(2n,p)$ random variable can be "halved".

• The sum of two independent Poisson$(\lambda)$ random variables is a Poisson$(2\lambda)$; conversely, a Poisson$(\lambda)$ random variable is the sum of two independent Poisson$(\frac{\lambda}{2})$ random variables. Indeed, as @Xi'an points out in a comment, a Poisson$(\lambda)$ random variable can be "halved" as many times as we like: for each positive integer $n$, it is the sum of $2^n$ independent Poisson$\left(\frac{\lambda}{2^n}\right)$ random variables.

• +1 My recollection is that the discrete uniform is a particular case where it's not possible (I believe there are numerous others, but it's one I have looked at). – Glen_b -Reinstate Monica Oct 18 '15 at 15:33
• Indeed, a uniform distribution is decomposable but not divisible in the above sense. – Xi'an Oct 18 '15 at 15:36
• The Poisson distribution is one example of an infinitely divisible distribution, so can be divided in a sum of an arbitrary number of iid variates. – Xi'an Oct 18 '15 at 15:38

The problem seems to me that you ask for an "independent copy", otherwise you could just multiply with $\frac{1}{2}$? Instead of writing copy (a copy is always dependent), you should maybe write "two independent, but identically distributed random variables".

• what comes closest is maybe the term convolution. For given $X$, you are looking for two iid RV with convolution $X$.
• if you accept negative probabilities, these are no longer random variables, since there is no probability space anymore. There are cases where you can find such $Y,Y^*$ ($X$ $\lambda$-Poisson-distributed, $Y$,$Y^*$ $\frac{\lambda}{2}$-Poisson-distributed), and cases where it is not possible ($X$ Bernoulli, as example).