Let X be a non-negative continuous random variable with probability density function f(x). Let $$G(t) = \int_{t}^{\infty} f(x)dx$$ Show that$$E(X^{2}) = 2\int_{0}^{\infty} tG(t)dt$$
My thoughts: I know that $$E(X^{2}) = \int_{-\infty}^{\infty} x^{2}f(x)dx$$, so here I need to prove $$x^{2}f(x)dx = tG(t)dt$$ I'm stuck in this step. Can someone explain how to prove this problem? Any clue would be appreciated. Many thanks.
It's a bit more complicated than that. Change the order of integration: $$\begin{align*} E(X^2) &= \int_0^{\infty} x^2 f(x) \, dx \\ &= \int_0^{\infty} \int_0^x 2t f(x) \, dt \, dx \\ &= \int_0^{\infty} \int_t^{\infty} 2t f(x) \, dx \, dt \\ &= \int_0^{\infty} 2t G(t) \, dt, \end{align*}$$ as required.
(It may be more obvious if you read the string of equalities starting from the bottom.)