Consider a smooth probability density $\pi(x)$ on $\mathbb{R}^d$. I am looking for natural conditions that ensure that the integral $\iint_{u,v} \ \min\big(\pi(u), \pi(v) \big) \ du \ dv$ is finite. If $\pi$ is radially decreasing, this is equivalent to the condition $\mathbb{E}\big[ \|X\|^{d} \big] < \infty$. Are there smooth densities verifying this moment condition such that $\iint_{u,v} \min\big(\pi(u), \pi(v) \big) \ du \ dv = \infty$?


1 Answer 1


This question has been asked and answered on MathOverflow. I have replicated the accepted answer by George Lowther below.

If $\mathbb{E}\left[\lVert[ X\rVert^d\right]$ is finite then the integral in the question is necessarily finite. As mentioned, this holds whenever $\pi$ is radially decreasing. However, in the general case, you can swap regions with equal volume in $\mathbb{R}^d$ about in order to move the large probability regions closer to the origin. Doing this has no effect on the integral in question, but can only decrease $\mathbb{E}\left[\lVert[ X\rVert^d\right]$. So, it reduces to the radially decreasing case.

It isn't hard to make this idea more rigorous. If $\pi(x)$ is 'radially decreasing', so that it is a decreasing function of $\lVert x\rVert$, then

$$ \begin{align} \iint \min\left(\pi(x),\pi(y)\right)\,dxdy&=2\iint_{\lVert y\rVert\le\lVert x\rVert}\pi(x)\,dxdy\cr &=2K\iint\lVert x\rVert^d\pi(x)\,dx=2K\mathbb{E}\left[\lVert X\rVert^d\right] \end{align} $$

where $K$ is the volume of the unit ball in $\mathbb{R}^d$.

In the general case, we have

$$ \mathbb{E}\left[\lVert X\rVert^d\right]=\int_0^\infty\int_{\pi(x)\ge p}\lVert x\rVert^d\,dxdp. $$

Letting $S_p=\lbrace x\colon\pi(x)\ge p\rbrace$ then, for a given volume $V_p$ for $S_p$, the integral $\int_{S_p}\lVert x\rVert^ddx$ is minimized when $S_p$ is a ball about the origin. In particular, if we define a radially decreasing function $\tilde\pi\colon\mathbb{R}^d\to\mathbb{R}$ by

$$ \tilde\pi(x)=\sup\lbrace p\in\mathbb{R}\colon K\lVert x\rVert^d\le V_p\rbrace $$

then $\lbrace x\colon \tilde\pi(x)\ge p\rbrace=\lbrace x\colon K\lVert x\rVert^d\le V_p\rbrace$ has volume $V_p$. So, $\mathbb{E}\_{\tilde\pi}\left[\lVert X\rVert^d\right]\le\mathbb{E}\_\pi\left[\lVert X\rVert^d\right].$

Also, the integral $\iint\min(\pi(x),\pi(y))\,dxdy$ is unchanged by passing to $\tilde\pi$. This follows from the fact that $\tilde\pi=\pi\circ f$ (almost everywhere) for some (Lebesgue) measure preserving Borel isomorphism of $\mathbb{R}^d$. Alternatively, the equality can be seen by showing that the integral only depends on $V_p$,

$$ \begin{align} \iint \min\left(\pi(x),\pi(y)\right)\,dxdy &=2\int\pi(x)V_{\pi(x)}\,dx\cr &=-2\int pV_p\,dV_p=\int_0^\infty V_p^2\,dp. \end{align} $$

So, better than just finiteness of the integral, we have the inequality

$$ \begin{align} \iint \min\left(\pi(x),\pi(y)\right)\,dxdy&=\iint \min\left(\tilde\pi(x),\tilde\pi(y)\right)\,dxdy\cr &=2K\mathbb{E}\_{\tilde\pi}\left[\lVert X\rVert^d\right]\cr &\le2K\mathbb{E}\_{\pi}\left[\lVert X\rVert^d\right], \end{align} $$

which is an equality whenever $\pi$ is radially decreasing.


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