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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.

$$k^2\int_k^\infty g(x) dx\leq\frac{4}{9}\int_0^{\infty}x^2g(x) dx$$

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.