X is a non-negative continuous random variable with pdf f(x). G(t) = \int_{t}^\infty f(x) dx. Show that E[X^2] = 2 \int_{0}^\infty tG(t) dt.
I tried to write out E[X^2] (second moment of X) and see if that corresponds to the G(t) equation. I wasn't able to do that well because I wasn't sure of how the change of variables works.
Answer
I'm assuming E[X^2] < \infty. Integration by parts gives
2\int_0^b t\, G(t)\, dt = b^2 G(b) - \int_0^b t^2 G'(t)\, dt = b^2 G(b) + \int_0^b t^2 f(t)\, dt.
Taking the limit as b \to \infty results in
2\int_0^\infty t G(t)\, dt = \int_0^\infty t^2 f(t)\, dt = E[X^2].
To see that \lim\limits_{b\to \infty} b^2 G(b) = 0, note that for b > 0,
E[X^2] = \int_0^b x^2 f(x)\, dx + \int_b^\infty x^2 f(x)\, dx \ge \int_0^b x^2 f(x)\, dx + b^2 G(b).
So
0 \le b^2 G(b) \le E[X^2] - \int_0^b x^2 f(x)\, dx.
The right-hand side tends to 0 as b\to \infty, so \lim\limits_{b\to \infty} b^2 G(b) = 0.
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