How do we prove $\mathbf{E} \left(\frac{1}{X} \right) \geq \frac{1}{\mathbf{E}(X)}$ for random variable $X$?
Can we use Jensen's inequality ?
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Is $X>0$ almost surely? – Augustin Feb 13 '16 at 09:50
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X is nonnegative – Milan Amrut Joshi Feb 13 '16 at 09:51
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Then yes, use Jensen's inequality, as $x\mapsto\frac{1}{x}$ is convex over $(0,+\infty)$. – Augustin Feb 13 '16 at 09:55
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Please sketch the proof if possible – Milan Amrut Joshi Feb 13 '16 at 09:56
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3Jensen's inequality states that if $\phi$ is a convex function, then $\phi(E(X))\leq E(\phi(X))$. Just take $\phi(x)=\frac{1}{x}$. – Augustin Feb 13 '16 at 10:00
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Jensen's Inequality works since $f(x)=\frac1x$ is convex for $x\gt0$. That is, Jensen says $$ E(f(X))\ge f(E(X)) $$ One can also use Cauchy-Schwarz because $$ E\left(\frac1X\right)E(X)\ge E(1)^2 $$
robjohn
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