Variance of a probability distribution with a random (!) permutation

Instead of my general formulation below, I will give an example:

Inside a 2x3x4 inch match box is a female insect F and a male insect M. F is at a corner of the matchbox and M is at a random point. The matchbox has random, stationary "stinky pockets" floating around occupying 1/3 of the total match box. The male is 4 times more likely to be floating around in a "stinky pocket" than in the remaining room of the matchbox, and we don't know how the stinky pockets are assigned in the space within the box.

Now consider a variation of where there are no stinky pockets in the match box (M is equally likely to be anywhere).

Are the variances of the rectilinear distance that F has to cover to reach M the same in the "stinky patches" and in the normal situation?

My intuition tells me yes, because the probability of finding M at each point remains the same.

General formulation

Consider any probability distribution f(x). The support/domain-assignment remains the same, but the y-values/probabilities (weight) are permuted in a random fashion. We do not know the new permutation. Is the variance the same as in the original distribution?

In other words, if we consider all possible permutations (give each one equal weight), on average is the variance the same as in the original distribution?

• The probability distribution of the male's position is the same (uniform) regardless of whether it's determined by "hidden variables" (stinky patches), as long as the stinky patches themselves are uniformly distributed. – mjqxxxx Oct 30 '12 at 5:44
• It seems that you want your example to be an example of the general formulation, but the way they're currently formulated it isn't. In the example, what the general formulation calls the "original distribution" never occurs. In the general formulation, the "original distribution" could be non-uniform (at least that's not excluded), whereas, as @mjqxxxx pointed out, the two distributions in the example are both uniform and thus the same, regardless of the uniformly distributed "hidden variables". So which part of the question describes correctly what you're interested in? – joriki Oct 30 '12 at 7:01
• Wuschel: You are writing your posts MUCH TOO FAST. As somebody said, The problem needs to be formulated precisely. I suggest you ask a question, carefully worded. – Did Oct 30 '12 at 7:27
• I realized that there was a mismatch after I posted it. The "hidden variables" problem is really what interested me. I don't know how to formulate its generalization properly. – Wuschelbeutel Kartoffelhuhn Oct 30 '12 at 8:11

The following might be (or not) a simple case of the situation the OP has in mind. Consider a random variable $X$ with discrete distribution $\mathbb P(X=x_i)=p_i$ for $1\leqslant i\leqslant n$, with $\sum\limits_{i=1}^np_i=1$, for some distinct values $x_i$. For every permutation $s$ of $\{1,2,\ldots,n\}$, consider a random variable $X_s$ with distribution $\mathbb P(X_s=x_{s(i)})=p_i$ for $1\leqslant i\leqslant n$.
Can one compare the characteristics of $X$, for example its mean and variance, to those of $X_\sigma$, where $\sigma$ is uniformly distributed in the symmetric group $\mathfrak S_n$ and independent of $(X_s)_s$?
For every test function $u$, $$\mathbb E(u(X_\sigma)\mid\sigma)=\sum\limits_{i=1}^nu(x_{\sigma(i)})p_i,$$ hence $$\mathbb E(u(X_\sigma))=\frac1{n!}\sum\limits_{s\in\mathfrak S_n}\sum\limits_{i=1}^nu(x_{s(i)})p_i=\frac1n\sum\limits_{i=1}^nu(x_i).$$ Thus, for every initial distribution $(p_i)_{1\leqslant i\leqslant n}$, $X_\sigma$ is uniformly distributed on $(x_i)_{1\leqslant i\leqslant n}$. In particular, $X$ and $X_\sigma$ are identically distributed if and only if $X$ is uniformly distributed.