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$a(x),b(x),c(x),$ and $d(x)$ are positive function of $x$. $\frac{a(x)}{b(x)}$ and $\frac{c(x)}{d(x)}$ increases in $x$. Moreover, we have $\frac{a(x)}{b(x)}<\frac{c(x)}{d(x)}$ holds for all x $\in \Omega=[0,\bar{x}]$. Can we show the following inequality?

\begin{equation} \frac{\int_\Omega a(x)\,\mathrm{d}x}{\int_\Omega b(x)\,\mathrm{d}x} <\frac{\int_\Omega c(x)\,\mathrm{d}x}{\int_\Omega d(x)\,\mathrm{d}x} \end{equation}

If not, what is the sufficient condition?

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    $\begingroup$ What is $\overline{x}?$ $\endgroup$ Jul 23 '13 at 19:01
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    $\begingroup$ $\bar{x}$ is the upper bound, basically it says $x$ is finite. $\endgroup$
    – wilson
    Jul 23 '13 at 19:10
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There is no reason whatsoever (other than pure wishful thinking) for the inequality to hold true.

Hint: The discrete case is well known to be false. Look at Simpson's Paradox.

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  • $\begingroup$ Thanks Calvin, the reference is very helpful. $\endgroup$
    – wilson
    Jul 23 '13 at 22:22
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With not much care in picking the "best" counterexample, take, for instance, $a(x)=x(1+x)$, $b(x)=x$, $c(x)=1$ and $d(x)=e^{-x}$. Then $a(x)/b(x)=1+x$ is increasing and $c(x)/d(x)=e^x$ is increasing as well. Additionally, $a/b\leq c/d$ since $1+x\leq e^x$. All your specified conditions are met.

Now compute

$$RHS = \frac{\int_0^a x+x^2 dx}{\int_0^a x dx}=\frac{\frac{1}{2}a^2+\frac{1}{3}a^3}{\frac{1}{2}a^2}=1+\frac{2}{3}a$$ and $$LHS=\frac{\int_0^a 1 dx}{\int_0^a e^{-x}dx}=\frac{a}{1-e^{-a}}.$$

If we pick $a=1$, then numerical computations show that $RHS \approx 1.66$ and $LHS \approx 1.59$, which violates the inequality.

I do wonder if you required $a(x),b(x),c(x)$ and $d(x)$ to be increasing as well, maybe even bounded below, if your inequality will hold. In this case, the trick I used to get a counterexample would not work.

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  • $\begingroup$ thanks. The function of a(x),b(x),c(x) and d(x) themselves are complex, but their ratio is very simple, that's why I focus on the ratios. In fact I know, $\frac{a(x)}{b(x)}=\gamma_1 x$ and $\frac{c(x)}{d(x)}=\gamma_2 x$ and $0<\gamma_1 \le 1< \gamma_2 $. Will the linearity of these ratios be sufficient for the inequality? $\endgroup$
    – wilson
    Jul 24 '13 at 17:02
  • $\begingroup$ No, it is not. Take $\gamma_1 = 1$ and $\gamma_2 = 1.5$. Then let $a(x)=x^{n+1}\gamma_1$, $b(x)=x^n$, $c(x)=\gamma_2 x^{n+1}$ and $d(x)=x^n$, then you'll want $$\frac{n+1}{n+2} \gamma_1 a \leq \gamma_2 \frac{1}{2} a$$ to hold up. But picking $n$ large and $a=1$ will make this not the case. $\endgroup$
    – abnry
    Jul 24 '13 at 17:15
  • $\begingroup$ You mean $n=0$ for $c(x)$ and $d(x)$, right? It seems that restrictions on the ratio won't be sufficient. $\endgroup$
    – wilson
    Jul 24 '13 at 18:03
  • $\begingroup$ Oh, yes, of course. It is true $n=1$ was meant. $\endgroup$
    – abnry
    Jul 24 '13 at 18:57
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I know this is pretty late, but just came across a similar situation in research... Also, this answer will not be rigorous, coming from a physicist... But,


If $a(x), b(x), c(x), d(x)$ are functions over $x\in[l, u]$, satisfying:

  1. $b(x) > 0, d(x) > 0 ~~~\forall x$
  2. $\frac{a}{b} \leq \frac{c}{d} ~~~\forall x$
  3. $\frac{c}{d}$ is non-decreasing in $x$
  4. $\frac{d}{b}$ is non-decreasing in $x$

Then (assuming all four integrals are finite), $$\frac{\int\limits_l^u~dx~a(x)}{\int\limits_l^u~dx~b(x)} \leq \frac{\int\limits_l^u~dx~c(x)}{\int\limits_l^u~dx~d(x)}$$


"Proof": $$LHS = \frac{\int\limits_l^u~dx~b(x)~~a/b}{\int\limits_l^u~dx~b(x)}$$ $$\leq \frac{\int\limits_l^u~dx~b(x)~~c/d}{\int\limits_l^u~dx~b(x)}~~~~(\text{since } a/b \leq c/d)$$ $$\leq \frac{\int\limits_l^u~dx~b(x)~~d/b~~c/d}{\int\limits_l^u~dx~b(x)~~d/b}~~~~(\text{explained below})$$ $$=\frac{\int\limits_l^u~dx~c(x)}{\int\limits_l^u~dx~d(x)} = RHS$$ To see why the third line works, note that the second line represents the expectation value of $c(x)/d(x)$ over $x$ sampled according to a probability distribution function proportional to $b(x)$. This interpretation is allowed because $b(x) \geq 0$ and has a finite and positive integral.

The third line represents the weighted expectation value of the same quantity $c(x)/d(x)$ over $x$ sampled according to the same probability distribution function. The only difference is that now $c/d$ is weighted by $d/b$ (allowed because $d$ is also non-negative with a finite and positive integral).

Because c/d and d/b are both non-decreasing, this weighted average assigns higher weightage to higher values of c/d (sloppy phrasing). So the weighted average in line 3 is $\geq$ the unweighted average in line 2.


The result will continue to hold if condition (3) is replaced by

3*. $a/b$ is non-decreasing in $x$

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A sufficient condition, which isn't very practical, is to say that ${a(x_1)}/{b(x_2)} < {c(x_3)}/{d(x_4)}$ for any $x_1,x_2,x_3,x_4 \in \Omega$. The result then follows from the first order Talyor expansion of the functions $A(x) = \int_0^x a(t) dt$, etc.

A little silly, I know, but it works!

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  • $\begingroup$ A sufficient condition is if they hold for all $x$ (as opposed to 4 differing values of $x_i$). $\endgroup$
    – Calvin Lin
    Nov 1 at 16:56

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