# Integral inequality

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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

Delta2
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Is it given that ##g(x)## is also non negative(##g(x)\geq 0##)?

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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##.

• Delta2
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.

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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

• Delta2
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:
Homework Helper
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.

StoneTemplePython
Science Advisor
Gold Member
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.

• Delta2