X is a non-negative continuous random variable with pdf f(x). G(t)=∫∞tf(x)dx. Show that E[X2]=2∫∞0tG(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[X2]<∞. Integration by parts gives
2∫b0tG(t)dt=b2G(b)−∫b0t2G′(t)dt=b2G(b)+∫b0t2f(t)dt.
Taking the limit as b→∞ results in
2∫∞0tG(t)dt=∫∞0t2f(t)dt=E[X2].
To see that limb→∞b2G(b)=0, note that for b>0,
E[X2]=∫b0x2f(x)dx+∫∞bx2f(x)dx≥∫b0x2f(x)dx+b2G(b).
So
0≤b2G(b)≤E[X2]−∫b0x2f(x)dx.
The right-hand side tends to 0 as b→∞, so limb→∞b2G(b)=0.
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