📂Lemmas

Proof of the Expected Form of Jensen's Inequality

Theorem ¹

In the open interval $I$, if the function $\phi$ is convex and twice differentiable, the expected value $X$ of the random variable exists, and $X \subset I $ then $$ \phi [ E(X) ] \le E [ \phi (X)] $$

Other Forms

• Jensen’s Inequality in Finite Form
• Jensen’s Inequality in Integral Form
• Conditional Jensen’s Inequality

It has a form quite similar to the integral form. Upon closer inspection, the finite form can also be considered an expected value in the sense that it is an inequality of the weighted average, even though its terms are not infinite.

Proof

Strategy: In the general proof, $\phi$ does not need to have a second derivative and merely being convex is sufficient. For convenience, we will assume it has a second derivative.

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By Taylor’s theorem, $$ \phi (x) = \phi (\mu) + \phi ’ (\mu) (x - \mu) + \phi ’’ (\xi) {{(x - \mu)^2} \over {2}} $$ there exists $\xi$ between $x$ and $\mu$. Since $\phi$ is convex, $\phi ’’ (\xi) > 0$, $$ \phi ’’ (\xi) {{(x - \mu)^2} \over {2}} > 0 $$ Summarizing, $$ \phi (x) \ge \phi (\mu) + \phi ’ (\mu) (x - \mu) $$ Taking the expected value of both sides, $E$ leads to $E(X-\mu) = 0$, thus $$ E( \phi ( X ) ) \ge \phi ( E (X) ) $$

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