*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*be the probability of the event

_{k}*X*=

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

.

Similarly, let *q*be the probability of the event

_{j}*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.)

## 4 comments:

Your comment about roots of all orders got me thinking--is there much use to the notion of the log of a distribution?

Cooper,

I can't think of one off the top of my head, but I'm not trying that hard. One might start by thinking of whether the convolution

exponentialof a distribution (which could be defined as exp(X) = 1 + X + X*X/2 + X*X*X/6 + ...) is useful; the logarithm would then be its inverse.Post a Comment