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Given a physical random variable x, $E(x)$ and $E((x-<x>)^2)$ defines mean and variance. From a statistical point of view variance represents the statistic error (isn't it?). If variance is not finite, can i say that there is no characteristic scale because the error on the mean is not finite?

First Edit: Reading the comments i think that the core of question can be formulated as: which is the rigorous definition of characteristic scale from the statistical point of view?

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No, I would say it is wrong to immediately conclude that there is no scale just because the variance diverges. Only functions of $x$ of the form $x^n$, a power law, have a chance to be considered scale-free; none of these functions may be considered a probability distribution because the integral diverges. Any other function – and therefore any normalizable probability distribution – includes some scale (or many scales if it is a complicated enough function).

For example, take the probability distribution function $$ \rho(x) = \frac{1}{\pi} \cdot \frac{1}{x^2+1} $$ I normalized it for the integral over real $x$ to be equal to one. The mean value is $x=0$ but the variance is computed from the integral $$ \frac{1}{\pi}\cdot \int_{-\infty}^{\infty} dx\frac{x^2}{x^2+1} $$ which clearly diverges. However, the original distribution has a width that is of order one which defines the characteristic length scale. This width may be decoded from many other detailed parameters that play a similar role as the variance but they're not variance.

For example, for the resonance curve above, the natural length scale may be defined as the width $\Delta x$ of the function at one-half of the height. One-half of the height occurs when $x^2+1=2\times (0^2+1) = 2$ which means $x=\pm 1$. That's why the width is $2$ and that defines a typical length scale.

You may also define the characteristic scale from quantities more similar to the variance such as the expectation values of some more complicated functions. For example, you could say – almost universally – that the typical length scale $L$ of a distribution is such that $$\int_{-\infty}^{\infty} dx\,\rho(x)\,\exp((x-x_0)^2/L^2) = \frac{1}{2} $$ Note that the exponential reduces the integral and if the exponential were absent, you would get one. So we want to reduce $1$ to $1/2$. The required $L$ can be neither too long because the reduction would be too modest, nor too short because the reduction would be too huge. This definition yields a unique finite $L$ in all examples I can think of.

Of course, the length scale $L$ you get from this method – or another method – only coincides with the length scale obtained from the variance – or other methods – only up to numerical constants of order one. Only the "order of magnitude" estimate of the length scale has a method-invariant meaning and for extreme/singular enough distributions, the gap may grow large (or very large).

This returns me to the point that functions may depend on several scales. For example, consider the sum of two Gaussians of some width $w$ each and some relative distance between them $\delta x_0$. These two parameters, $w$ and $\Delta x_0$, may be vastly different but they still define some scales needed to define the distribution. Your variance method as well as my "Gaussian reduction to $1/2$ method" pick the $\Delta x_0$ scale but there are other methods, those that depend on high-frequency behavior of the distribution, that would pick $w$ instead. So one must carefully avoid certain sloppy ideas such as the claim that "every distribution only depends on one scale" and "it is encoded to all quantities one may calculate". For sufficiently generic or "natural" functions, this may be right, but it's always easy to find counterexamples.

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The Lorentzian distribution you introduced is indeed perfect to illustrate the question asked. However, I just want to point out that its mean value is also undefined. – Oli Jun 5 '12 at 12:12
@Lubos Moti ...i don't agree. Imagine a random variable y that is power law distributed (You say that there is not a characteristic scale due scaling property). If i take a sum of N random power law distributed variables, the resulting distribution lacks the power law shape. From your point of view this means that the sum of N scale free random variable can generate a non scale free random variable. mmm I don't think so. – Emanuele Luzio Jun 5 '12 at 12:33
Dear @Oli, fine, it depends how cleverly you define it. A straightforward integral has two canceling log divergences. By principal value integrals, symmetric ones, the mean value is zero as the distribution is even. – Luboš Motl Jun 5 '12 at 13:36
Dear Emanuele, it makes no sense, due to the dimensional analysis, to add up variables distributed along different power laws in scale-invariant contexts because they have different units (different powers of a meter, if you wish, with the exponents given by the exponents from the power law). To add them in accordance with the dimensional analysis, you have to put relative coefficients in between the terms you mention. The ratios of these coefficients then define the length scale. – Luboš Motl Jun 5 '12 at 13:38
@LubošMotl: of course it's a matter of definitions, but there is at least one good, pragmatic reason to say that the Cauchy distribution does not have an expectation: an iid sequence of such random variables will not satisfy the law of large numbers (the sample mean is again Cauchy-distributed)... – Yvan Velenik Jun 5 '12 at 14:03

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