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For a continuous real random variable $X$ with CDF $F_X(x)$ and PDF $f_X(x)$ I want to prove the following $$\int\limits_{-\infty}^{\infty}x(2F_x(x)-1)f_x(x)dx\geq 0$$ I was thinking of integration by parts but $x$ complicates it. Any hints?

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Let $X$ and $Y$ denote i.i.d. random variables whose cdf are given by $F$.

Then $P(\max{\{X,Y\}}\leq x) = P(X\leq x, Y \leq x) = P(X \leq x)P(Y \leq x) = F(x)^2$.

Therefore the cdf of $\max\{X,Y\}$ is given by $\frac{d}{dx}F(x)^2 = 2F(x)f(x)$.

We clearly have that $X \leq \max\{X,Y\}$, so that $E[X] \leq E[\max\{X,Y\}]$ which means that $$\int xf(x)dx \leq \int 2xF(x)f(x)dx$$

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