Return to Answer

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

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

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

A sufficient condition would be to that show $G(x) = \mathcal{O}(||x||^d)$. This is satisfied if $\pi$ is spherical, representing the decay of the tail. If the distribution is not radial, it could have a long, flat, thin ridge that extends to infinity, upon which $G$ remains constant. It's clearly not a necessary condition.

edit: seems there's already an elegant answer

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

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

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

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

A sufficient condition would be to that show $G(x) = \mathcal{O}(||x||^d)$. This is satisfied if $\pi$ is spherical, representing the decay of the tail. If the distribution is not radial, it could have a long, flat, thin ridge that extends to infinity, upon which $G$ remains constant. It's clearly not a necessary condition.

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

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

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

A sufficient condition would be to that show $G(x) = \mathcal{O}(||x||^d)$. This is satisfied if $\pi$ is spherical, representing the decay of the tail. If the distribution is not radial, it could have a long, flat, thin ridge that extends to infinity, upon which $G$ remains constant. It's clearly not a necessary condition.

edit: seems there's already an elegant answer

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

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

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

If we can show $G(x) = \mathcal{O}(||x||^d)$, then it will follow that $I < \infty$

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

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

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

A sufficient condition would be to that show $G(x) = \mathcal{O}(||x||^d)$. This is satisfied if $\pi$ is spherical, representing the decay of the tail. If the distribution is not radial, it could have a long, flat, thin ridge that extends to infinity, upon which $G$ remains constant. It's clearly not a necessary condition.