proof or disproof of inequality for probability distributions

Question

Let there exist 3 valid probability distributions: $p_1(x), p_2(x), p_3(x)$

does the following inequality hold: $\int_{-\infty}^{\infty}{\sqrt{p_1p_2}+\sqrt{p_2p_3}} \ \ dx \leq 1 + \int_{-\infty}^{\infty}{\sqrt{p_1p_3}} \ \ dx$

Any help with this would be greatly appreciated.

Answer 1

No, it does not generally hold.

Let $p_1(x) = \delta(0) $, $p_2(x) = 0.5 \delta(0) + 0.5 \delta(1) $, $p_3(x) = \delta(1) $, then you should have

$$ \int_{-\infty}^{\infty}{\sqrt{p_1p_2}+\sqrt{p_2p_3}} \ \ dx = 2 \sqrt{0.5} \leq 1 + \int_{-\infty}^{\infty}{\sqrt{p_1p_3}} \ \ dx = 1 $$ which is wrong.

Answer 2

Ok, here you go: take $p_1 = U(0,0.5)$, $p_2 = U(0,1)$ and $p_3 = U(0.5,1)$. Then $$\int_{-\infty}^{\infty}{\sqrt{p_1p_2}}\ dx = \sqrt{2} / 2$$ $$\int_{-\infty}^{\infty}{\sqrt{p_2p_3}}\ dx = \sqrt{2} / 2$$ $$\int_{-\infty}^{\infty}{\sqrt{p_1p_3}}\ dx = 0$$ so the inequality does not hold.

Now, a bit on how to come to it: take $x = \sqrt{p_1}$, $y = \sqrt{p_1}$, $z = \sqrt{p_1}$, and let $<\!\cdot\,,\cdot\!>$ be a standard scalar product on $L^2$. Your problem is rephrased the following way:

given $x,y,z$ are unit vectors in $L^2$, prove(?) following: $$<\!x\,,y\!>+<\!y\,,z\!> \ \le\ 1\, + <\!x\,,z\!>.$$ Now, if you write $1$ as $<\!x\,,x\!>$ and apply scalar product rules you get: $$<\!y\,,x+z\!> \ \le\ <\!x\,,x+z\!>.$$ One can easily find lots of examples where this equality doesn't hold.