This one is really hard, any help??
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Does that X really belong in that integrand? the idea is that g(X)/g(e)>1 so it should be a 1 in the integrand.
Ever heard of Chernoff's inequality?
What I was saying......... $\displaystyle E(g(X))= \int_0^{\infty}g(x)dP=\int_0^{a}g(x)dP+\int_a^{\in fty}g(x)dP$ $\displaystyle \ge \int_a^{\infty}g(x)dP =g(a)\int_a^{\infty}{g(x)\over g(a)}dP$ $\displaystyle \ge g(a)\int_a^{\infty}dP$
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