Probability theoretic inequality

[winterfors]

Can anyone help me prove under what conditions on the distance function $$d(x_1,x_2)$$ the following inequality holds for any two probability distributions (represented by probability densities) $$p(x)$$ and $$q(x)$$ :

$$2\int{\int{d^2(x_1,x_2)p(x_1)q(x_2)dx_1dx_2} \geq \int{\int{d^2(x_1,x_2)p(x_1)p(x_2)dx_1dx_2} + \int{\int{d^2(x_1,x_2)q(x_1)q(x_2)dx_1dx_2}$$

where $$d^2(x_1,x_2)$$ is the squared distance between $$x_1$$ and $$x_2$$ in some metric space $$\Theta$$. All integrals are over $$\Theta$$.

One can easily verify by insertion that the inequality holds for a Euclidian metric where $$d^2(x_1,x_2)=(x_1-x_2)^2$$, with equality if and only if the expectation of $$p(x)$$ and $$q(x)$$ are the same.

It must surely hold for some more general class of metrics (described by $$d^2(x_1,x_2)$$) - possibly all metrics - but I've so far failed to demonstrate it. Does anyone have an idea of how to prove it in some more general case?

 

[Valkarie]

The inequality is known as the Cauchy-Schwarz-Bunyakovsky inequality. It holds for any metric space and any two probability distributions (represented by probability densities) p(x) and q(x). The proof of this inequality relies on the fact that the integral of d^2(x_1,x_2)p(x_1)q(x_2) over all x_1 and x_2 is non-negative. This is true because d^2(x_1,x_2) is positive by definition, and p(x_1) and q(x_2) are both non-negative. Thus, one can apply the Cauchy-Schwarz inequality to the integrand to obtain the desired result:2\int{\int{d^2(x_1,x_2)p(x_1)q(x_2)dx_1dx_2} \geq \left(\int{\int{d(x_1,x_2)p(x_1)p(x_2)dx_1dx_2}\right)^2 + \left(\int{\int{d(x_1,x_2)q(x_1)q(x_2)dx_1dx_2}\right)^2