A Cauchy–Schwarz Type Inequality Involving Scaled Distributions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T09:19:15Z http://mathoverflow.net/feeds/question/86503 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86503/a-cauchyschwarz-type-inequality-involving-scaled-distributions A Cauchy–Schwarz Type Inequality Involving Scaled Distributions Santiago 2012-01-24T01:38:32Z 2012-01-24T15:23:18Z <p>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.</p> <h3>The Inequality</h3> <p>Let $X$ be a continuous and non-negative random variable. Denote by <code>$\bar F(x)= \mathbb P \{X&gt;x\}$</code> 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 <i>scaled</i> version of the density function $f_i(\cdot) = f(a_i \cdot)$, and the complementary distribution $\bar F_i(\cdot) = \bar F(a_i \cdot)$.</p> <p>We need to show that</p> <blockquote> <p><code>$$\langle f_1, \bar F_1 \rangle \langle f_2, \bar F_2 \rangle \ge \langle f_1, \bar F_2 \rangle \langle f_2, \bar F_1 \rangle,$$</code></p> </blockquote> <p>where <code>$\langle u, v \rangle = \int_0^\infty u(x)v(x)w(x)\,dx$</code> is the inner product with weight $w(\cdot)\ge0$. We can assume that all functions are square integrable w.r.t. to the weight.</p> <p>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!! </p> <h3>Remarks</h3> <ol> <li><p>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.</p></li> <li><p>It tried numerically with other distributions and it seems to hold.</p></li> </ol> <h3>Edit: Counter-example</h3> <p>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 <code>$$w(x) = x \exp(-c_1 \bar F_1(x) - c_2 \bar F_2(x)).$$</code> 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.</p> http://mathoverflow.net/questions/86503/a-cauchyschwarz-type-inequality-involving-scaled-distributions/86508#86508 Answer by Anthony Quas for A Cauchy–Schwarz Type Inequality Involving Scaled Distributions Anthony Quas 2012-01-24T04:31:10Z 2012-01-24T04:31:10Z <p>What you're asking translates to the following:</p> <p>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 $$\int_{-\infty}^\infty F(ax)|F'(ax)|w(x)\,dx\int_{-\infty}^\infty F(bx)|F'(bx)|w(x)\,dx$$</p> <p>$$\ge\int_{-\infty}^\infty F(ax)|F'(bx)|w(x)\,dx\int_{-\infty}^\infty F(bx)|F'(ax)|w(x)\,dx$$</p> <p>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 $b=2$. </p> <p>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 $[F(1)|F'(2)|+F(10)|F'(20)|][F(2)|F'(1)|+F(20)|F'(10)|]$.</p> <p>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 $f(20)=2$. </p> <p>The left side is then $(12+2)*(12+2)=196$ while the right side is $(16+4)(9+1)=200$</p>