Integral inequality

  • #1
Gaussian97
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Homework Statement:
Prove $$k^2\int_k^\infty g(x) dx\leq\frac{4}{9}\int_0^{\infty}x^2g(x) dx$$ for a non-increasing function ##g(x)##.
Relevant Equations:
The inequality is very easy to prove if ##g(x)## is constant for ##0<x<k+a## and zero otherwise. So I will use it.
I have to prove that, for a non-increasing function ##g(x)## the following inequality is true:
$$k^2\int_k^\infty g(x) dx\leq\frac{4}{9}\int_0^{\infty}x^2g(x) dx$$
This exercise is from the book Mathematical methods of statistics by Harald Cramer, ex. 4 pg 256
Following the instructions of the book I find the following:

For an arbitrary non-increasing function ##g(x)##, I define $$a:=\frac{1}{g(k)}\int_{k}^{\infty}g(x)dx$$ and also I define the function ##h(x)=g(k)## for ##0<x<k+a## (and 0 otherwise). So now, ##h(x)## is a constant non-decreasing function and the inequality $$k^2\int_k^\infty h(x) dx\leq\frac{4}{9}\int_0^{\infty}x^2h(x) dx$$ is very easy to prove. So with that
$$k^2\int_k^\infty g(x)dx\equiv k^2ag(k)=k^2\int_k^{k+a}g(k) dx=k^2\int_k^{\infty}h(x) dx\leq \frac{4}{9}\int_0^{\infty}x^2h(x) dx$$

But now I'm supposed to prove that
$$\int_0^{\infty}x^2h(x) dx\leq \int_0^{\infty}x^2g(x) dx$$
I have tried to reorganize it and I have write
$$\int_0^{\infty}x^2\left(g(x)-h(x)\right) dx\geq 0 \Longrightarrow$$ $$\int_0^{k}x^2\left(g(x)-g(k)\right) dx+\int_{k+a}^{\infty}x^2g(x) dx\geq \int_k^{k+a}x^2\left(g(k)-g(x)\right) dx$$
Where now all the integrals are positive, but I don't know how to continue.

Thank you.
 

Answers and Replies

  • #2
Delta2
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Is it given that ##g(x)## is also non negative(##g(x)\geq 0##)?
 
  • #3
Gaussian97
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Well it's not implicitly said, but if the integral ##\int^{\infty}g(x)dx## converges this implies (I think) that $$\lim_{x\to \infty}g(x)=0$$. So, since ##g(x)## it's non-icreasing this implies ##g(x)\geq0##.
BTW I forgot to say that ##k>0##.
 
  • #4
member 587159
but if the integral ##\int^{\infty}g(x)dx## converges this implies (I think) that $$\lim_{x\to \infty}g(x)=0$$.

This is false. Consider ##f(x)= \sin(x^2)##. Maybe under the additional hypothesis that ##g## is non-increasing it can be correct, but you should justify this step.
 
  • #5
Gaussian97
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This is false. Consider ##f(x)= \sin(x^2)##. Maybe under the additional hypothesis that ##g## is non-increasing, it can be correct, but you should justify this step.
Ok, thank you! I wasn't very sure of that property... In any case, since ##g(x)## is non-increasing, if there exist a value ##x_0## for which ##g(x_0)<0##, then $$g(x)\leq g(x_0), \qquad \forall x>x_0$$ so then,
$$\int_{x_0}^{\infty}g(x)dx\leq \int_{x_0}^{\infty}g(x_0)dx=g(x_0)\int_{x_0}^{\infty}dx\rightarrow -\infty$$
So the integral diverges and therefore, ##g(x)## must be also non-negative.

I think now it's correct, but even if it's not, the main purpose to this is to use it then with probability density functions, so let's assume non-negativity as an assumption if you want :D
 
  • #6
StoneTemplePython
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Homework Statement: Prove $$k^2\int_k^\infty g(x) dx\leq\frac{4}{9}\int_0^{\infty}x^2g(x) dx$$ for a non-increasing function ##g(x)##.


I think now it's correct, but even if it's not, the main purpose to this is to use it then with probability density functions, so let's assume non-negativity as an assumption if you want :D

The thread seems to be bouncing around, and from my vantage, at least, un-motivated. It simply isn't true that PDFs are real non-negative and non-increasing in general. So there seems to be some kind of special structure in mind that is absent from the thread.

Alternatively, I'll point out that being real non-negative and non-increasing is true about all complementary CDFs, and with some care, we can integrate over these to recover moments.

(edited to insert word complementary)
 
Last edited:
  • #7
Gaussian97
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Well... The thread is to prove the inequality... It seems to be bouncing around because we are focusing more on my assumption that ##g(x)## is non-negative than in the thread itself 😅 .

Forget what I've said about PDF, I'm not a mathematician and maybe I haven't been told the most general definition, but for my point of view, a PDF must be real and non-negative. But has nothing to do with the thread.
 
  • #8
StoneTemplePython
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Well... The thread is to prove the inequality... It seems to be bouncing around because we are focusing more on my assumption that ##g(x)## is non-negative than in the thread itself 😅 .

Forget what I've said about PDF, I'm not a mathematician and maybe I haven't been told the most general definition, but for my point of view, a PDF must be real and non-negative. But has nothing to do with the thread.
there are a typically multiple ways to attack these things. The fact that ##g(x)## is supposed to be a density is very relevant and not something you mentioned in the original post. While densities are non-negative, they aren't monotone in general . The "and" in my prior post meant the intersection of the properties was not true in general. Standard methods for bounding the tail of a distribution make use of real-nonegativity, and they get your result without the ##\frac{4}{9}## coefficient. The monotone behavior, which is atypical, allows a slightly sharper estimate at the cost of a lot more symbol manipulation it seems.
 

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