5
$\begingroup$

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.

$\endgroup$
3
  • 6
    $\begingroup$ 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. $\endgroup$ Sep 18, 2012 at 18:36
  • $\begingroup$ @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. $\endgroup$ Sep 18, 2012 at 18:56
  • $\begingroup$ @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! $\endgroup$
    – Alekk
    Sep 18, 2012 at 19:51

3 Answers 3

5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $a,b \ge 0$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.