# Name/significance of integral of the square of a probability density function

### Background/Motivation

Given a probability density function $f(x)$, the mean of the corresponding random variable is the $x$-coordinate of the centroid of the region under the graph of $f$. I wondered: "does the $y$-coordinate have any significance in probability?". Of course, the $y$-coordinate is given by $\frac12\int_{-\infty}^\infty f^2$, so without worrying about the $1/2$, I wondered if the integral of $f^2$ itself has any significance. An internet search brought up only this physics forum thread and this yahoo answer, neither of which seemed to have any useful information.

### My thoughts

One thing I thought about was the discrete case. Then $\Sigma_n p(n)^2$ is a probability, analogous to the probability of rolling doubles. Unfortunately, $\int f^2$ can easily be more than 1, so it does not have such a nice interpretation.

I also considered uniform distributions. A uniform distribution of probability density equal to $p$ has square-integral $(1/p)*p^2=p$, so that maybe this square integral is something like "the density of the uniform distribution $f$ is most like". Flipping this idea on its head, a uniform distribution over an interval of length $1/p$ has square-integral $p$, so that the reciprocal of the square-integral is like the length of the uniform distribution with the same$\ldots$"clumpiness"? (It reminded me of curvature, being the reciprocal of the radius of the circle with the same curviness.) But I don't know if these interpretations are useful in any way.

### The question

Does the integral of the square of a pdf (or half of it) have a name (aside from "the square of the $L^2$ norm")? Is it used for anything? Is there a better angle from which to think about it?

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Nice question. As you say, this is the $L^2$ norm of $f$ in the space $L^2(\lambda)$ where $\lambda$ is the Lebesgue measure. This is also $E(f(X))$ where the random variable $X$ has density $f$ and, as such, this is the $L^1$ norm of $f$ in the space $L^1(\mu)$ where $\mu$ is the measure with density $f$. "Is it used for anything?" Not that I would be aware of. – Did Feb 16 '14 at 8:36
@Did Thank you for your comment. $E(f(X))$ is somewhat satisfying as it is an interpretation that is easy to explain/sounds meaningful in the discrete case, but still applies in the continuous case. Do you want to turn your comment into an answer? – Mark S. Feb 22 '14 at 22:57

One thing I thought about was the discrete case.

Actually that works here too. One just has to be careful of dimensions.

Notice that

$$I = \int_{\mathbb R} (f(x))^2 \, d x$$

has dimensions of $[x]^{-1}$. Therefore, in order to interpret this as a (dimensionless) probability, it's necessary to multiply it by something else with dimensions of $[x]$. Call this $\delta x$, and assume it's small. Then $I \delta x$ can be interpreted as the (approximate) probability of "rolling doubles" but with a tolerance of $\delta x$.

Why is this so? As it turns out, $I \delta x$ is just the limiting case of this integral:

$$\iint_{\mathbb R^2} f(x) f(x') \Delta(x - x') \, d x' \, d x$$

where $\Delta$ is some sort of distribution function sharply peaked around zero (e.g. a very narrow Gaussian). If the width of $\Delta$ is narrow enough and $f$ is sufficiently smooth, then one may simply approximate the inner integral as a product of the width and the height.

Of course, even $I \delta x$ can exceed one – that's simply because the approximation has broken down by that point.

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I wondered if the integral of $f^2$ itself has any significance.

As you say, this is the $L^2$ norm of $f$ in the space $L^2(λ)$, where $λ$ is the Lebesgue measure. This is also $E(f(X))$, where the random variable $X$ has density $f$. As such, this is the $L^1$ norm of $f$ in the space $L^1(μ)$ where $μ$ is the measure with density $f$.

Is it used for anything?

Not that I would be aware of.

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