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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. This integral appeared while studying a Metropolis-Hastings Markov chain.

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The answer must be no, and in fact you must have the integral bounded by the volume of a ball of radius $\lVert X\lVert_d$. This is because you can swap regions of equal volume in $R^d$ about to move the large probability regions closer to the origin, which doesn't change the integral but can only decrease $\lVert X\lVert_d$. So, it reduces to the radially decreasing density case. –  George Lowther Sep 18 '12 at 18:36
    
@George: very nice! I'm kicking myself for not thinking of it. @Alekk: the buzzwords for what George is describing are "symmetric decreasing rearrangement". See Chapter 3 of Analysis by Lieb and Loss for more information. –  Mark Meckes Sep 18 '12 at 18:56
    
@George: great! I can't believe that I missed that. I did notice that the whole thing was invariant by rearrangement but did not see that $E \|X\|^d$ was decreasing. Many thanks! –  Alekk Sep 18 '12 at 19:51
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3 Answers

up vote 4 down vote accepted

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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One easy sufficient condition (though not necessarily useful or natural, depending on what you know about $\pi$) is $\int \sqrt{\pi(u)} \ du < \infty$, since $\min(a,b) \le \sqrt{ab}$ for $a,b \ge 0$.

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Not an answer yet, but some thoughts that may lead to one...

Let $$G(z) = \int \mathbb{1}_{\pi(y)\geq z} ~dy$$

then $$I = \int \int_{x,y} \min( \pi(x), \pi(y) )~dx~dy = 2 \int \pi(x) G(\pi(x))~dx$$

$$I = 2 \int_{0}^{\pi_{\max}} z G(z) \int \mathbb{1}_{\pi(x)=z} dx dz$$

$$I = 2 \int_{0}^{\pi_{\max}} zG(z)G'(z) dz$$

edit: seems there's already an elegant answer

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