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I would like to demonstrate that when Y>X, this inequality is true.

$E_{y}+\left(\int_{a}^{+\infty}\left(y-E_{y}\right)^{m}f_y(y)dy\right)^{\frac{1}{m}}\geq E_{x}+\left(\int_{b}^{+\infty}\left(x-E_{x}\right)^{m}f_x(x)dx\right)^{\frac{1}{m}}$

where a > b and $E_{y}>E_{x}$. Where $E_{y}$ and $E_{x}$ represented the expected utility and $f_x$ and $f_y$ the pdf of X and Y.

Do you have any idea where I can start? It seems quite difficult.

Otherwise, if that is not possible, do you know a way to show a counterexample?

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  • $\begingroup$ What are $E_x, E_y$ and $f$ ? $\endgroup$
    – Sassatelli Giulio
    Feb 3 at 10:00
  • $\begingroup$ E_x and E_y are the non-conditional average of X and Y. f is the density function of x and y $\endgroup$
    – Silvia F.
    Feb 3 at 10:03
  • $\begingroup$ So your $X$ and $Y$ are two identically distributed (but manifestly non-independent) random variables with distribution $f$ ? $\endgroup$
    – Sassatelli Giulio
    Feb 3 at 10:04
  • $\begingroup$ Yes, X and Y are two i.i.d. disitribution whose density function is f(x) for X and f(y) for Y and we just know that X>Y in each scenarios (that is why E[y] > E[x]) $\endgroup$
    – Silvia F.
    Feb 3 at 10:07
  • $\begingroup$ You are right. I am used to write i.i.d. and I did not think it thorugh. I will change the notation for the two pdfs. Do you have an hint on how can i prove it? $\endgroup$
    – Silvia F.
    Feb 3 at 10:12

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