I have stumbled upon a rather intriguing inequality involving the product of the scaled distribution and the scaled density of a random variable. The inequality has a very attractive form, and it resembles, somehow, Cauchy–Schwarz inequality. The problem goes as follows.

The Inequality

Let $X$ be a continuous and non-negative random variable. Denote by $\bar F(x)= \mathbb P \{X>x\}$ the complementary cumulative distribution function (ccdf), and $f(x)$ its probability density function (pdf). Given two scaling factors $a_1,a_2\ge1$; consider the scaled version of the density function $f_i(\cdot) = f(a_i \cdot)$, and the complementary distribution $\bar F_i(\cdot) = \bar F(a_i \cdot)$.

We need to show that

$$ \langle f_1, \bar F_1 \rangle \langle f_2, \bar F_2 \rangle \ge \langle f_1, \bar F_2 \rangle \langle f_2, \bar F_1 \rangle, $$

where $\langle u, v \rangle = \int_0^\infty u(x)v(x)w(x)\,dx$ is the inner product with weight $w(\cdot)\ge0$. We can assume that all functions are square integrable w.r.t. to the weight.

Has anybody problem seen such an inequality? It has quite an appealing form, but I could neither prove it nor find a counterexample. Any reference or ideas would be appreciated!!


  1. The latter holds when the hazard rate of the random variable $X$ is homogeneous, that is, the hazard rate $h(x) = f(x) / \bar F(x)$ satisfies $h(a x) = a^n h(x)$ for some $n$. In this case, it suffices to write $f_i(\cdot) = a_i^n h(\cdot) \bar F_i(\cdot)$ and use Cauchy–Schwarz. The exponential, weibull, and pareto distributions have homogeneous hazard rates.

  2. It tried numerically with other distributions and it seems to hold.

Edit: Counter-example

Anthony Quas has provided an excellent counter-example (see his answer below) for general weights. Actually, I was a looking for a particular weight function, which is given by $$w(x) = x \exp(-c_1 \bar F_1(x) - c_2 \bar F_2(x)).$$ Since, the inequality holds for the case of homogeneous hazard rates, I was hoping that it will hold in full generality. Shame on me! Anyway, hope that somebody has some thoughts on this. It would be much appreciated.


What you're asking translates to the following:

Is it true that for all bounded differentiable decreasing functions $F(x)$ with $F(\infty)=0$, all positive weight functions $w(x)$ and all positive $a$ and $b$, that $$ \int_{-\infty}^\infty F(ax)|F'(ax)|w(x)\,dx\int_{-\infty}^\infty F(bx)|F'(bx)|w(x)\,dx $$

$$ \ge\int_{-\infty}^\infty F(ax)|F'(bx)|w(x)\,dx\int_{-\infty}^\infty F(bx)|F'(ax)|w(x)\,dx $$

For that to be true for all $w(x)$, it would have to be true for all positive measures. We'll get a counterexample taking $w(x)dx=\delta_1+\delta_{10}$. Let's take $a=1$ and $b=2$.

Then the left side is $[F(1)|F'(1)|+F(10)|F'(10)|][F(2)|F'(2)|+F(20)|F'(20)|]$ while the right side is $[F(1)|F'(2)|+F(10)|F'(20)|][F(2)|F'(1)|+F(20)|F'(10)|]$.

Let $F(1)=4$, $F(2)=3$, $F(10)=2$ and $F(20)=1$ (you can scale it later if you want to make it be a reverse cdf. Now choose the density so that $f(1)=3$, $f(2)=4$, $f(10)=1$ and $f(20)=2$.

The left side is then $(12+2)*(12+2)=196$ while the right side is $(16+4)(9+1)=200$

  • $\begingroup$ @Anthony: Thanks for the answer. It looks good! I'll check it tomorrow. Thanks, Santiago. $\endgroup$ – Santiago Jan 24 '12 at 5:27
  • $\begingroup$ @Anthony: Great counter example! I am a little bit disappointed with myself; I was expecting it to be true. Actually, I was looking for a particular weight function, which is given by $w(x) = x \exp(-c_1 \bar F_1(x) - c_2 \bar F_2(x))$. It was wrong of me to assume that the inequality will hold in full generality. Do you have any thoughts for that such a function? Thanks again, Santiago. $\endgroup$ – Santiago Jan 24 '12 at 15:14
  • $\begingroup$ Hi @Santiago, I'm skeptical about the inequality even in that case - I think you still have way too much freedom in the choice of $F$ for the inequality to have a chance of being true in general. I'm not sure I have the stomach to go looking for an example though... $\endgroup$ – Anthony Quas Jan 24 '12 at 17:33
  • $\begingroup$ @Anthony: Haha! Thanks again for the remarks. What threw my off is that I tried -numerically- with a couple of distributions and it always seemed to work out fine. Also, the fact that it works like a charm for the homogeneous hazard rate case was kind of misleading. In addition it holds for $X=\text{Uniform}[0,M]$. Do you have any intuition for what kind of distributions it may work? Thanks! $\endgroup$ – Santiago Jan 24 '12 at 20:34

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