The binomial distribution is written as

$$p(r|n,\theta )=\binom{n}{r}\theta ^r(1-\theta )^{n-r}$$

where $n$ is a positive integer, $0\leq\theta\leq1$, and $r$ is an integer taking values from $0$ to $n$.

I'm trying to find changes of variable that leave this distribution invariant. To illustrate, a general change of variables from $n,r,\theta$ to $m,s,\psi$ would have the form

$$\begin{align*}m&=f(n,r,\theta)\\ s&=g(n,r,\theta)\\ \psi&=h(n,r,\theta)\end{align*}$$

for some functions $f,g,h$. Here we require that $m,s$ be non-negative integers with $s\leq m$ and that $\psi$ be a real number in the interval $0\leq\psi\leq1$. Moreover the transformation should be a bijection and $h$ should be continuous. Then I say that the binomial distribution is invariant with this change of variables if


An example would be the following change of variables

$$\begin{align*}m&=n\\ s&=n-r\\ \psi&=1-\theta\end{align*}$$

Substituting it is easy to verify that the invariance condition holds.

Are there any more such transformations that leave invariant the binomial distribution? Thanks.

  • $\begingroup$ Thanks for editing the tags; this is a very interesting question. I'd be very interested in a real answer that specifies the group (my transformation is not part of a group; it is just a bizarre non-invertible projection). $\endgroup$ – Jack Schmidt Aug 19 '11 at 4:43
  • $\begingroup$ Note that preimages of $p(\cdot|\cdot,\cdot):X\to[0,1]$ will be invariant under the desired transformations $(n,r,\theta)\to(m,s,\psi)$. Thus if you can reparametrize $X$ as $x(\alpha,\beta,\gamma)$, with $\{x\}$ always determining a level surface for $\gamma$ held fixed, then all desired transformations can be decomposed into maps $(\alpha,\beta)\to(\alpha',\beta')$ satisfying appropriate continuity restrictions. $\endgroup$ – anon Aug 19 '11 at 8:36
  • $\begingroup$ @Jack Schmidt: After giving it more thought I decided to put the other conditions as well. I'm editing the question again. If then there turn out to be no more transformations other than the trivial one I give as an example ($m=n$, $s=n-r$, $\psi=1-\theta$), then the conditions could be relaxed. $\endgroup$ – becko Aug 20 '11 at 4:21
  • $\begingroup$ @anon: as explained in the answer below, for continuous invertible transformations the problem is equivalent to the number-theoretic question of solving $p(n,r,r/n) = p(m,s,s/m)$. For discontinuous invertible transformations one can take any fiber of $p$ and permute it. $\endgroup$ – zyx Aug 20 '11 at 5:15

There will be many such functions if there is no constraint on the nature of the transformation. The function $f(p,q,\theta) = \binom{p+q}{q} \theta^p (1-\theta)^q : \Bbb{N} \times \Bbb{N} \times [0,1] \to \Bbb{R}$ is not one-to-one, and any "transformation" that permutes each fiber (or maps each fiber into itself, as in Jack Schmidt's answer), is a solution of the problem.

Continuous invertible transformations are equivalent to non-negative integer solutions of $M(p,q) = M(r,s)$ where $M(p,q)=f(p,q,p/(p+q))$ is the maximum value of $f$ with respect to $\theta$. The trivial solutions are the identity transformation $(r,s)=(p,q)$ and the involution that exchanges $\theta$ and $1 - \theta$ from $(r,s)=(q,p)$. Finding or precluding non-trivial solutions is an interesting problem in number theory.

That's for the case of non-negative integer parameters in the binomial coefficients, and a continuous solution in that case would provide a solution for more general parameters (extending it by the identity transformation).

[update: in the case where $n$ and $r$ (and thus $p,q,r,s$ in the notation of the previous paragraph) are allowed to be real, there are continuous solutions by identifying the graphs of $A(\theta) = f(p,q,\theta)$ and $B(\psi)=f(r,s,\psi)$ in the case where the maxima of these unimodal functions are equal, $M(p,q)=M(r,s)$. There will in fact be infinite-dimensional families of continuous invertible solutions of this type due to existence of arcs in $(p,q)$ parameter space on which $M(p,q)$ is constant. For each arc one has the freedom to choose a reparametrization (a homeomorphism) and a partition of the arc into pieces on which to use the $\theta$ or the $(1-\theta)$ variant, so the dimensionality of the space of "good" transformations is infinite, and possibly uncountable if continuity is required but not smoothness.]


Here is a bizarre transformation to show one probably wants some extra requirement. Set:

  • $m = f(n,r,\theta) = 1$
  • $s = g(n,r,\theta) = 1$
  • $\psi = h(n,r,\theta) = p(r|n,\theta)$

Then $p(s|m,\psi) = \binom{1}{1} \psi^1 (1-\psi)^{1-1} = \psi = p(r|n,\theta)$.

I think one should require the change of variable to be invertible, and that h should be continuous; implicitly we require m, s to be non-negative integers, presumably sm, and ψ to be a real number, 0 ≤ ψ ≤ 1.

I'm not sure if there should be more conditions, and I'm not sure if there are any more "reasonable" changes of variable.

  • $\begingroup$ I agree that the transformation should be invertible and that $h$ should be continuous. Also, $s,m,\psi$ should all be real (no complex numbers here). As to the other conditions I am not so sure. All that is required is that $\left(\begin{array}{c} m \\ s\end{array}\right)\psi ^s(1-\psi )^{m-s}$ be well defined (using Gamma functions $m,s$ are not integers). I'm editing the question to add the conditions on which we agree. $\endgroup$ – becko Aug 19 '11 at 5:36
  • 2
    $\begingroup$ I don't see this example as bizarre at all. For $\phi:V\to W$ surjective, the algebraic structure of all maps $\tau:V\to V$ for which $\phi$ is $\tau$-invariant will generally be a monoid, not necessarily a group. $\endgroup$ – anon Aug 19 '11 at 7:30

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