5
$\begingroup$

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

$$\binom{n}{r}\theta^r(1-\theta)^{n-r}=\binom{m}{s}\psi^s(1-\psi)^{m-s}$$

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.

  • 0
    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).2011-08-19
  • 0
    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.2011-08-19
  • 0
    @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.2011-08-20
  • 0
    @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.2011-08-20

2 Answers 2