Expectation of X^2 using complement of CDF

Question

For a nonnegative random variable $$E[X]=\int_0^\infty P(X >x)dx$$

This video on a question from IIT JAM 2023 extends it to $$E[X^2]=\int_0^\infty 2x P(X >x)dx$$

How was the result was arrived at? Where can I get references for such results? Books or structured references (rather than single links preferred).

$P(X^{2}>x)=P(X>\sqrt x)$ if $X$ is non-negative. Apply your first formula and make the change of variable $y=\sqrt x$ — geetha290krm, Mar 18 '24 at 08:49

score 1 · Answer 1 · edited Mar 18 '24 at 11:40

1

$$E[X^2]=\int_0^\infty P(X^2 >x)dx=\int_0^\infty P(X>\sqrt{x})dx$$

Substitute $t=\sqrt{x}\implies t^2 =x\implies 2t\text{ } dt=dx$ $$=\int_0^\infty P(X>t) \text{ } 2t \text{ }dt$$

And since the variable of integration is a dummy variable, change back to $x$: $$=\int_0^\infty 2x P(X >x)dx$$

edited Mar 18 '24 at 11:40

S.L.

answered Mar 18 '24 at 11:14

Starlight

1 Answers1