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


No answer given on [math.stackexchange](https://math.stackexchange.com/questions/198079/minimum-of-two-probability-densities). This integral appeared while studying a Metropolis-Hastings Markov chain.



minimum of two probability densities
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.


