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);
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.

  • $\begingroup$ i think the 2nd term on the lhs is bigger by itself, unless you are keenly interested in the strictness of the inequality. $\endgroup$ – mike Dec 23 '13 at 21:49
  • $\begingroup$ @mike I just ran the simulation for your comment. The inequality you described fails for roughly $m<0.2$ in the case of above mentioned family of curves. $\endgroup$ – ubaabd Dec 23 '13 at 22:02

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 $$\int_{-\infty}^{\infty}\phi(u-\mu)\Phi(-u-\mu)du = \int_{-\infty}^{\infty}\phi(u-\mu)\Big[1-\Phi(u+\mu)\Big]du$$


$$=1 - \int_{-\infty}^{\infty}\phi(u-\mu)\Phi(u+\mu)du $$

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.

  • $\begingroup$ Though your conclusion is right, your example is not.. $f(u)=1/(b-a)$ but not $1$ when you plug in. $\endgroup$ – MoonKnight Dec 24 '13 at 0:14
  • $\begingroup$ @MoonKnight Thanks for spotting the mistake -the habit of working all the time with $U(0,1)$... Since your answer showed that the inequality does not hold in general, I changed mine to show that it does hold for the normal distribution. $\endgroup$ – Alecos Papadopoulos Dec 24 '13 at 2:29
  • $\begingroup$ Thanks @AlecosPapadopoulos. That really helped. $\endgroup$ – ubaabd Dec 24 '13 at 5:52
  • $\begingroup$ Nice to hear. One can also show that the inequality will hold if the distribution is a uniform $U(a,b)$, $a<0<b,\; b>-a$. $\endgroup$ – Alecos Papadopoulos Dec 24 '13 at 14:42
  • $\begingroup$ @AlecosPapadopoulos Can you please explain how you came up with the second step after "break and rearrange ... " and the second last step with $\Phi(-\sqrt{2})$. Secondly, I think the same proof holds for any symmetric distribution, not necessarily Gaussian. Can you comment on this as well ? It is because, if I am correct, you have used the symmetry property of Gaussian distribution about its mean in the second step ? $\endgroup$ – ubaabd Dec 24 '13 at 19:00

Hmm, I think your statement is not true in general. It might be true for some specific distribution you tried. But the condition you give is not sufficient.

Here is a counter example. $f(x)=0.55\delta(x-2)+0.05\delta(x+1)+0.4\delta(x+3)$

$$ \int_0^\infty \left[du \int_{-u}^0f(x)f(u)dx\right]=0.55*0.05=0.0275 $$

$$ \int_0^\infty \left[du \int_{-\infty}^0f(x)f(u)dx\right]=0.55*(0.05+0.4)=0.2475 $$

$$ \int_{-\infty}^0 \left[du \int_{-\infty}^{-u}f(x)f(u)dx\right]=(0.4+0.05)*(0.4+0.05)+0.4*0.55=0.4225 $$

Your inequality does not hold since $0.0275+0.2475<0.4225$.

More illustration, the first integral takes the space enter image description here

The second integral takes the space enter image description here

The third integral on the RHS takes the space enter image description here

The red points above are the only possible points that my example can allow.

  • $\begingroup$ thanks for your detailed reply. You are right. I think the inequality would be true if we restrict the distributions further to only symmetric distributions, i.e., in above mentioned example, $f(x) = e^{-x} f(-x)$. Could you please comment on this or should I ask a separate question ? This symmetry condition is available in my problem, but I thought I might not need it. $\endgroup$ – ubaabd Dec 23 '13 at 23:25
  • $\begingroup$ @ubaabd, I prefer you start a new question. btw, your symmetry here looks weird because if $f(x)=e^{-x}f(-x)$, then $\forall x>0, f(x)<f(-x)$. $\endgroup$ – MoonKnight Dec 23 '13 at 23:33
  • $\begingroup$ you are right its $f(x) = e^x f(-x)$. Sorry about that. $\endgroup$ – ubaabd Dec 24 '13 at 0:03
  • $\begingroup$ @ubaabd. I'm not sure how to construct a distribution function from this "symmetry", but if you only want to prove for the Gaussian you mentioned above. Here is the proof. $\endgroup$ – MoonKnight Dec 24 '13 at 0:08
  • $\begingroup$ Yes. A proof for the Gaussian would work for me. There are no links under "Here is the proof". Can you please explain what do you mean by "Here is the proof" ? $\endgroup$ – ubaabd Dec 24 '13 at 0:21

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