I know that for nonnegative continuous R.V, $E[X]=\int_0^\infty P(X>t)dt$. Is there a formula for $E[X^2]$ when we only have $P(X>t)$?
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1Could this work $E[X^2] = \int P(X^2 > t)dt = \int P(X > \sqrt{t}) + P(X< -\sqrt{t})dt$ – Student Jan 17 '20 at 15:40
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@nls would $P(X<-\sqrt t)=0$ – johnson Jan 17 '20 at 16:50
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1This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post. It is generalized to any $E[X^p]$ in the answer post there. – Lee David Chung Lin Nov 20 '21 at 18:57
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Using integration by parts one may show: $$EX^2=\int_0^{\infty}2xP(X>x)dx$$
the tails should go to zero fast enough though :-)
Addendum
\begin{align} \int_0^{\infty}2xP(X>x)dx&=\int_0^{\infty}2x\left(1-F(x)\right)dx\\ &=x^2\left(1-F(x)\right)|_0^{\infty}+\int_0^{\infty}x^2f(x)dx \end{align} Note now that we need $\lim_{x\to\infty}x^2(1-F(x))=0$, i.e. the tails should vanish faster than $x^2$.
Math-fun
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I understand now what you mean. Nonetheless I would like you to explain why I get a contradiction when I use $P(X\le x) $as the derivative instead of $P(X>x)$ – johnson Jan 17 '20 at 19:36
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1just added more details. Just let the community know if this is not helping :-) – Math-fun Jan 17 '20 at 20:24
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I understand that derivation as it has been showed to me in an actuarial course. Could you tell me why this does not work $\int_0^\infty x^2f(x)dx=[x^2F(x)]^\infty_0-2\int^\infty_0xF(x)dx$. This is also done using integration by parts but I get inifity - infinity. – johnson Jan 17 '20 at 20:32
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By the way what about nls's suggestion given in the comments above. Thank you very much. – johnson Jan 17 '20 at 20:33
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Would be great to ask all the "new" questions, as a new question :-) it does attract enough attention from the community and it is not easy to answer them in comments cleanly. – Math-fun Jan 17 '20 at 20:45