Exercise 1.40 (Complex Jensen inequality)

Question

Let $f: {\bf C} \rightarrow {\bf R}$ be a convex function, and let $X$ be a complex random variable with $X$ and $f(X)$ both absolutely integrable. Show that

$$\displaystyle f( {\bf E} X ) \leq {\bf E} f(X).$$

Elu Thingol · Accepted Answer

We make use of the supporting hyperplane theorem:
\begin{theorem} (Supporting hyperplane, complex version) 
Let $f:\mathbb{C}	o\mathbb{R}$ be a convex function. Then, for each $x_0 \in \mathbb{C}$ there exists an affine function $l(x)= ax+b$ such that
$$f(x) \geq l(x)$$
holds for all $x \in \mathbb{R}^n$ and $l$ supports $f$ at the point $x_0$:
$$f(x_0) = l(x_0).$$
\end{theorem}

We apply the theorem to our assertion by setting $x_0 = {\bf E} X$; i.e, we find an affine $l_{{\bf E}X}(x) = ax + b$ such that $l_{{\bf E}X}({\bf E}X) = f({\bf E}X)$. In particular,
$$\displaystyle f(X) \geq aX + b.$$
Taking expectations and using linearity of expectation, we conclude
$$\displaystyle {\bf E} f(X) \geq {\bf E}(aX+b) = a{\bf E}X + b = l_{{\bf E}X}({\bf E}X) = f({\bf E}X).$$

We now provide the proof for the supporting hyperplane theorem that we have used in this theorem.
\begin{proof} (Supporting hyperplane)
Let $\Delta > 0$ be arbitrary. Notice that the point $f(x_0)$ is in the middle of $f(x_0+\Delta)$ and $f(x_0 - \Delta)$; thus,
$$f(x_0) = f\left(\frac{1}{2}(x_0 + \Delta) + \frac{1}{2}(x_0-\Delta)ight) \leq \frac{1}{2}f(x_0+\Delta) + \frac{1}{2}f(x_0-\Delta).$$
From this we immediate
$$\frac{1}{2}f(x_0) - \frac{1}{2}f(x_0-\Delta) \leq \frac{1}{2}f(x_0 + \Delta) - \frac{1}{2}(x_0)$$
and thus,
$$\frac{f(x_0)-f(x_0-\Delta)}{\Delta} \leq \frac{f(x_0+\Delta)-f(x_0)}{\Delta}.$$
We can take limits while preserving inequality
$$\sup_{\Delta>0}\frac{f(x_0)-f(x_0-\Delta)}{\Delta} \leq \inf_{\Delta>0}\frac{f(x_0+\Delta)-f(x_0)}{\Delta}.$$
Let $a$ be an arbitrary number between the two limits. Define
$$l:\mathbb{C}	o\mathbb{R}, \quad x\mapsto a(x-x_0) + f(x_0).$$
We then have
$$l(x_0) = f(x_0)$$
and
\begin{itemize}
\item If $x \geq x_0$ then with $\Delta:=x-x_0$ we have $f(x) - f(x_0) = f(x+\Delta)-f(x_0) \geq a\Delta + f(x_0)$ by construction of $a$. Therefore, $f(x)\geq l(x)$.
\item If $x \leq x_0$ we similarly obtain $f(x) \geq l(x)$.
\end{itemize}
Thus, $l$ satisfies all of the desired properties.
\end{proof}

Exercise 1.40 (Complex Jensen inequality)

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Exercise 1.40 (Complex Jensen inequality)

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