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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!!

Remarks

  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.

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1 Answer 1

up vote 1 down vote accepted

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 $(16+4)(9+1)=200$

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@Anthony: Thanks for the answer. It looks good! I'll check it tomorrow. Thanks, Santiago. –  Santiago Jan 24 '12 at 5:27
    
@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. –  Santiago Jan 24 '12 at 15:14
    
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... –  Anthony Quas Jan 24 '12 at 17:33
    
@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! –  Santiago Jan 24 '12 at 20:34

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