Think about probability distributions supported on the positive integers. Not all of them have a "square root" -- that is, given a random variable

*X* supported on the positive integer, there do not exist independent, identically distributed variables

*Y*_{1},

*Y*_{2} such that X has the same distribution as Y

_{1} + Y

_{2}.

You might be wondering why it's natural to refer to this as a "square root". Well, let

*p*_{k} be the probability of the event

*X* =

*k*. Then the probability generating function for X is

.

Similarly, let

*q*_{j} be the probability of the event

*Y* =

*j*, and define the probability generating function

.

Let

*Y*,

*Y*_{1},

*Y*_{2} be equidistributed. Then

*X* is equidistributed with

*Y*_{1} + Y

_{2} if and only if f(z) = g(z)

^{2}, since addition of independent random variables corresponds to

convolution of distributions and multiplication of their generating functions.

Conversely, the random variable

*X* has a square root in this sense if and only if its generating function

*f*(

*z*) has a square root which has a Taylor series at

*z* = 0 with all coefficients positive.

The simplest case is when

*X* has finite support. Then

*X* has a square root if and only if its generating function

*f*(

*z*) is the square of a polynomial. For example, the random variable which takes values 0, 1, 2 with probabilities 1/4, 1/2, 1/4 has a square root; its generating function is

.

But the random variable taking values 0, 1, 2 with probability 1/3 each is not, since 1/3 + z/3 + z

^{2}/3 is not a square.

But what about distributions with infinite support? Some of these are easy -- the square root of the

Poisson distribution with mean λ is Poisson with mean λ/2. This can easily be seen since the probability generating function of a Poisson(λ) random variable is

.

(In fact, the Poisson distribution is

infinitely divisible; in the nomenclature used in this post one might say it has roots of all orders.)

Now consider a random variable

*X*, with

geometric distribution with parameter 1/2 supported on {0, 1, 2, 3, ...}; this has P(

*X* =

*k*) = 2

^{-(k+1)}. This is a special case of the

negative binomial distribution, which is infinitely divisible. We have f(z) = 1/2 + z/2

^{2} + z

^{2}/2

^{3} + ... = 1/(2-z). So the square root distribution has generating function

and in general the coefficient of z

^{n} is, by the binomial theorem,

That binomial coefficient is ${1 \over 2} {2n \choose n}$; we have ${2n \choose n} \sim 4^n/\sqrt{\pi n}$, so the coefficient of z

^{n} in our probability generating function, which we'll call q

_{k}, is asymptotic to

In particular, the "square root" distribution decays just a bit faster than the distribution that it's the square root of, the main difference being the additional factor of n

^{-1/2}. This is reasonable, if you think about the process of convolution. We have

p

_{n} = q

_{0} q

_{n} + q

_{1} q

_{n-1} + q

_{2} q

_{n-2} + ... + q

_{n-1} q

_{1} + q

_{n} q

_{0}and each of the

*n* terms is roughly 1/(n2

^{n}). This is just another negative binomial distribution. (I actually didn't realize that until I started writing this post; the post was originally titled "what's the name of this distribution?" and then I did some research.)