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If $X$ is a continuous random variable, how to show that $$E[X|X>c] = \int_c^{\infty} (1-F_X(x))dx?$$

I showed

$$EX = E \int_0^ \infty 1_{X>x}dx = \int_0^\infty P(X>x)dx = \int_0^{\infty} (1-F_X(x))dx.$$

If it is given that $X>c$, does that mean that $E(X|X>c) = \int_c^{\infty} (1-F_X(x))dx$? Not sure why.

Vika
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  • The equation is false. RHS is a constant and LHS need not be. In particular the equation fails when $X$ has an exponential distribution. – geetha290krm Oct 04 '22 at 05:12
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    Hint: When $F$ is the unconditional CDF of $X$ then $G(x)=F(c<X\le x)/(1-F(c))$ is the conditional CDF. – Kurt G. Oct 04 '22 at 10:11

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