# What could it mean to “Rotate” a distribution? [duplicate]

This question got me thinking about the meaning of variance: Intuition behind standard deviation.

Variance of a set of data is calculated the same way that the moment of inertia is calculated for a physical body. The moment of inertia is related to the energy required to rotate the body at a given speed. A figure skater will rotate faster with arms pulled in than stretched out. So what would be the analogous result of reducing variance, if any. Perhaps the analogy simply breaks down. Are there any publications that have investigated this analogy?

• @whuber I agree that the questions are related, but this one is different in that I am asking explicitly what the probability distribution equivalent of rotation would be. Essentially I would like to discover how far this analogy between probability distributions and distributions of mass can be taken, and what use can be made of this. – Livid Feb 4 '14 at 22:35
• I see that as identical to the several questions that have appeared asking about physical analogs of distributional moments. Check them out with a search. The first (most "relevant") hit has an answer referring to rotation, for instance. – whuber Feb 4 '14 at 22:44
• @whuber I am reading further, but I think that asking about "physical analogs of distributional moments" (other questions), is subtly different from "distributional moment analogs of physical concepts" (this question). Glen_b's answer, for example, appears to claim that the analogy can only be taken so far. I am unconvinced there is not some deeper connection. – Livid Feb 4 '14 at 23:09

The direct analogy is pretty clear:

To make it simple we'll assume it's for a continuous random variable on $(a,b)$. Without loss of generality, let $c=b-a$ and consider the corresponding variable on $(0,c)$; call that random variable $X$.

Now imagine a very thin rod of length $c$, whose density (mass per element of length) is variable in the x-direction (along its length) and consider that the rod happens to have the same material-density as a function of $x$ as the random variable has probability density as a function of $x$.

Then then second moment of inertia of the rod is the variance of $X$.

And hence what it 'means' to rotate a distribution is clear enough - it's quite literally rotating the 'rod' whose density represents probability-density. Variance is how 'hard' it would be to rotate the rod (low variance means 'easy to spin', high variance means it takes more push to spin it ... and stop it, if you spin it).

Think about what inertia (how hard it is to spin) reflects here, which is simply how close the mass is to the mean. The closer the mass is to the mean the easier it is to spin. If you made a physical object whose physical density represents the probability density and the random variable had low variance, the corresponding object would be easy to spin, because most of the mass would be close to the mean - both inertia and variance are how close the mass is to the mean, in a particular (and directly analogous) sense.

You don't actually 'spin' a probability density and imagine that to be physically difficult, any more than electricity is wet because of the water analogy. To expect that level of correspondence is to miss the point of such analogies (the aspects that correspond, correspond, but not every consequence of the correspondence in one realm carries over with it).

The point of saying the 'rod is hard to spin' is to give a pretty direct sense of what high variance is telling you about density. But to insist that the probability density itself spin is to miss the point.

• For clarity, by "continuous random variable on $(a,b)$", do you mean uniform? Otherwise, what are $a$ & $b$? – gung - Reinstate Monica Feb 4 '14 at 21:56
• @gung No, I don't mean uniform, because I explicitly said the density varied (well it could even be uniform, but the discussion is much more general). I mean that the variable is limited to values between $a$ and $b$, and the phrasing I used there is quite common, even standard. The reason is that people find it hard to imagine rotating infinitely long rods and I didn't want any additional strain in doing the imagining - hence the finite length rod, and so a bounded random variable. (Imagine, if you like, dealing with a particular mixture of scaled beta distributions, as an example) – Glen_b -Reinstate Monica Feb 4 '14 at 22:05
• @Glen_b But what does "easy to spin mean" for a probability distribution. – Livid Feb 4 '14 at 22:20
• Please see my edits – Glen_b -Reinstate Monica Feb 4 '14 at 22:39
• @Glen_b I understand your point, do you have any speculation on what "spinning a probability density" could mean? Or you believe that is just taking the analogy too far? – Livid Feb 4 '14 at 23:38

If you have greater variance, then you need to expend a greater effort (meaning, spend more money collecting the data) to obtain a given level of precision. The precision for $\bar X$ is of course is the inverse standard deviation, $\sqrt{n}/\sigma$.

• So to take the analogy further, we would say that estimating the center of mass is more difficult for a larger body. This makes some sense. – Livid Feb 4 '14 at 22:21

In signals, variance is essentially a measure of energy. Of course in this case they are related directly, not inversely.

• Can you expand on this? Increased variance would require greater energy to "change", from that perspective the relationship is not inverse. – Livid Feb 4 '14 at 22:19