Consider a probability distribution $f(x)$ with the assumption $\int_0^{\infty}f(x) > 1/2$. I am trying to prove the following inequality:
\begin{align} \left( \int_0^{\infty}\int_{-u}^{0}f(x)f(u)dxdu +\int_0^{\infty}\int_{-\infty}^{0}f(x)f(u)dxdu \right) > & \int_{-\infty}^{0}\int_{-\infty}^{-u}f(x)f(u)dxdu. \end{align}
My attempt is as follows:
\begin{align} LHS &= \left( \int_0^{\infty}\int_{-u}^{0}f(x)f(u)dxdu +\int_0^{\infty}\int_{-\infty}^{0}f(x)f(u)dxdu \right) \\ &= \left( \int_{-\infty}^{0}\int_{-x}^{\infty}f(x)f(u)dudx +\int_{-\infty}^{0}\int_0^{\infty}f(x)f(u)dudx \right) \\ &= \int_{-\infty}^{0} f(x) \left( \int_{-x}^{\infty}f(u)du + \int_{0}^{\infty}f(u)du \right)dx \\ &= \int_{-\infty}^{0} f(u) \left( \int_{-u}^{\infty}f(x)dx + \int_{0}^{\infty}f(x)dx \right)du. \\ \end{align}
I am stuck at this step. I used the following MATLAB simulation to prove the last step using a Gaussian family of distributions
$$ f(x,u) = f(x)f(u) = \frac{1}{2 \pi (2m)} e^{\frac{-(x-m)^2}{(2m)}} e^{\frac{-(u-m)^2}{(2m)}}. $$ The constant $\frac{1}{2 \pi (2m)}$ is ignored in the simulation.
close all; clear all;clc;
mvals = (1e-5:.1:5);
q1 = zeros(1, length(mvals));
q2 = q1;
q3 = q1;
for i=1:length(mvals);
mval = mvals(i);
sigma2 = 2*abs(mval);
% Function
fun1 = @(x,y) exp(-(x-mval).^2./(2*sigma2)) .* exp(-(y-mval).^2./(2*sigma2));
ymax = @(x) -x;
% LHS Integrals
q1(i) = integral2(fun1, -Inf, 0, ymax, Inf);
q2(i) = integral2(fun1, -Inf, 0, 0, Inf);
% RHS Integral
q3(i) = integral2(fun1, -Inf, 0, -Inf, ymax);
end
figure('Color', [1 1 1]);
plot(mvals, q1, 'b'); hold on;
plot(mvals, q2, 'r'); hold on;
plot(mvals, q1+q2, 'g'); hold on;
plot(mvals, q3, 'k');
I would appreciate any help/hints.
As another answer showed, this inequality does not hold in general. But I will show that it holds for the normal distribution. Assume that $f(x)$ is a normal $N(\mu,1)$ density, therefore , $f(x) =\phi(x-\mu) $and denote $F(x) = \Phi(x-\mu)$ the cumulative distribution function. $\phi()$ and $\Phi()$ are the standard normal density and cdf respectively. $\mu>0$ to reflect the assumption that the probability mass is greater in the positive orthant. Then
\begin{align} \left( \int_0^{\infty}\int_{-u}^{0}f(x)f(u)dxdu +\int_0^{\infty}\int_{-\infty}^{0}f(x)f(u)dxdu \right) > & \int_{-\infty}^{0}\int_{-\infty}^{-u}f(x)f(u)dxdu \end{align}
$$\Rightarrow \int_0^{\infty}\phi(u-\mu)\left[\Phi(-\mu)-\Phi(-u-\mu)\right]du +\int_0^{\infty}\Phi(-\mu)\phi(u-\mu)du \;>\; \int_{-\infty}^{0}\Phi(-u-\mu)\phi(u-\mu)du $$
Break and rearrange to obtain $$\Rightarrow 2\Phi(-\mu)\int_0^{\infty}\phi(u-\mu)du \;>\; \int_{-\infty}^{\infty}\phi(u-\mu)\Phi(-u-\mu)du \qquad [1]$$
Now, regarding the LHS, by the properties of integration we can subtract $\mu$ from the limits of integration and add it to the argument. So we get
$$\int_0^{\infty}\phi(u-\mu)du = \int_{-\mu}^{\infty}\phi(u)du = 1-\Phi(-\mu) = \Phi(\mu) \qquad [2]$$ the last equality from the symmetry properties of the normal.
For the RHS we can write $$\int_{-\infty}^{\infty}\phi(u-\mu)\Phi(-u-\mu)du = \int_{-\infty}^{\infty}\phi(u-\mu)\Phi(-(u+\mu))du$$ and using the symmetry properties of the normal we can write further $$\int_{-\infty}^{\infty}\phi(u-\mu)\Phi(-u-\mu)du = \int_{-\infty}^{\infty}\phi(u-\mu)\Big[1-\Phi(u+\mu)\Big]du$$
$$=1\cdot\int_{-\infty}^{\infty}\phi(u-\mu)du-\int_{-\infty}^{\infty}\phi(u-\mu)\Phi(u+\mu)du$$
$$=1 - \int_{-\infty}^{\infty}\phi(u-\mu)\Phi(u+\mu)du $$
since the first integral equals unity. Using again integral properties, we subtract $\mu$ from the limits of integration and add it to the integrating variable to obtain
$$\int_{-\infty}^{\infty}\phi(u-\mu)\Phi(-u-\mu)du = 1 - \int_{-\infty}^{\infty}\phi(u)\Phi(u+2\mu)du \qquad [3] $$
Inserting results $[2]$ and $[3]$ into $[1]$ we want to prove the following inequality:
$$...\Rightarrow 2\Phi(-\mu)\Phi(\mu)\;>\; 1- \int_{-\infty}^{\infty}\phi(u)\Phi(u+2\mu)du\qquad [4]$$
Regarding the remaining integral in $[4]$ the following general result holds (see for example here):
$$\int_{-\infty}^{\infty}\phi(z)\Phi(a+bz)dz = \Phi\left(\frac {a}{\sqrt{1+b^2}}\right)$$ In our case, $a= 2\mu$ and $b=1$ so
$$1-\int_{-\infty}^{\infty}\phi(u)\Phi(u+2\mu)du = 1-\Phi\left(\frac {2\mu}{\sqrt{2}}\right) = 1-\Phi(\sqrt2 \mu) = \Phi(-\sqrt2 \mu) \qquad [5]$$ using again the symmetry property. Inserting $[5]$ into $[4]$ we have
$$...\Rightarrow 2\Phi(-\mu)\Phi(\mu)\;>\; \Phi(-\sqrt2 \mu)$$
$$\Rightarrow 2\Phi(\mu)\;>\; \frac {\Phi(-\sqrt2 \mu)}{\Phi(-\mu)}$$
This inequality holds $\forall \mu>0$ since $\Phi(\mu) >1/2 \Rightarrow 2\Phi(\mu)>1$ and $\Phi(-\sqrt2 \mu) < \Phi(-\mu)$.
It can be reasonably conjectured that the inequality will hold for any distribution with a cdf that satisfies the symmetry property $F(x) = 1-F(-x)$, but this needs a proof on its own.