Weighted Right Convex Function (WRCF) Theorem

Theorem. Suppose that $f : I \to \mathbb{R}$ be a function on an interval $I$ and that there exists $s \in I$ such that $f$ is convex on $I \cap [s,\infty)$. Then for any $p_1,\cdots,p_n \in (0, 1]$ satisfying $\sum_{i=1}^{n} p_i = 1$ and $p = \min\{p_1,\cdots,p_n\}$, the followings are equivalent:

1. For any $x_1, \cdots, x_n \in I$ such that $\sum_{i=1}^{n} x_i p_i \geq s$, we have $$f\left(\sum_{i=1}^{n} x_i p_i \right) \leq \sum_{i=1}^{n} f(x_i) p_i.$$
2. For any $x, y \in I$ satisfying $x \leq s \leq y$ and $p x + (1-p)y = s$, we have $$p f(x) + (1-p)f(y) \geq f(s)$$

The following proof is an adaptation of the argument of [CVB11] in probabilistic language.

Proof. The direction $1 \Rightarrow 2$ is straightforward. So we prove the converse. We need the following easy lemma:

Lemma. Let $f : [a, b] \to \mathbb{R}$ be convex. Suppose that $\mu$ and $\nu$ are finite measures on $[a, b]$ such that $\mu([a, b]) = \nu([a, b])$, $\int_{[a,b]} x \, \mu(dx) = \int_{[a,b]} x \, \nu(dx)$ and $\operatorname{supp}(\nu)\subseteq \{a, b\}$. Then $$\int_{[a,b]} f(x) \, \mu(dx) \leq \int_{[a,b]} f(x) \, \nu(dx).$$

Indeed, the convexity of $f$ tells that the line $\ell$ joining $(a, f(a))$ and $(b, f(b))$ satisfies $f(x) \leq \ell(x)$ for all $x \in [a, b]$. Then $$\int_{[a,b]} f(x) \, \mu(dx) \leq \int_{[a,b]} \ell(x) \, \mu(dx) = \int_{[a,b]} \ell(x) \, \nu(dx) = \int_{[a,b]} f(x) \, \nu(dx)$$ The second equality follows from the first two assumptions and the last equality follows since $\nu$ is supported on $\{a, b\}$ and $\ell \equiv f$ on $\{a,b\}$. This completes the proof of Lemma.

Let $\phi$ and $\tilde{f}$ be functions on $I$ defined by $$\phi(x) = \frac{s-px}{1-p}, \qquad \tilde{f}(x) = \begin{cases} f(x), & x \geq s \\ \frac{f(s) - (1-p)f(\phi(x))}{p}, & x < s \end{cases}$$ Notice that item 2 of the theorem is equivalent to the inequality $f(x) \geq \tilde{f}(x)$ for all $x \in I$. Now consider a random variable $X$ taking values in $\{x_1,\cdots,x_n\} \subseteq I$ with $P[X = x_i] = p_i$ such that $E[X] \geq s$. We want to prove that $E[f(X)] \geq f(EX)$. Write $$E[f(X)] \geq E[\tilde{f}(X)] = E\left[ \frac{f(s) - (1-p)f(\phi(X))}{p} ; X < s\right] + E[f(X); X \geq s]$$ and notice that $E[f(X) \mid X \geq s] \geq f(E[X \mid X \geq s])$ by the Jensen's inequality. Combining altogether, if we let $m_+ = E[X \mid X \geq s]$, then it suffices to prove that $$\frac{P[X < s]}{p}f(s) + P[X \geq s] f(m_+) \geq \frac{1-p}{p} E[ f(\phi(X)) ; X < s] + f(E X) \tag{*}$$ The following claim is useful for our purpose:

Claim. If $j$ is any index such that $x_j < s$, then $\phi(x_j) \in (s, m_+]$.

Indeed $\phi(x_j) > s$ is easy to check. To prove the upper bound, let $\tilde{X}$ be a random variable such that $$P[\tilde{X} = x_i] = \begin{cases} p_i/(1-p), & i \neq j \\ (p_i - p)/(1-p), & i = j \end{cases}$$ Then we easily check that $E[\tilde{X} \mid \tilde{X} \geq s] = m_+$ and hence $$\phi(x_j) \leq \frac{(EX) - x_j p}{1-p} = E[\tilde{X}] \leq E[\tilde{X} \mid \tilde{X} \geq s] = m_+.$$ This completes the proof of Claim. Therefore, if we write $$\mu = \frac{1-p}{p} P[\phi(X) \in \cdot \, ; X < s] + \delta_{E X}, \qquad \nu = \frac{P[X < s]}{p} \delta_s + P[X \geq s]\delta_{m_+}$$ then by the claim above both $\mu$ and $\nu$ are finite measures on $[s, m_+]$ and $\text{(*)}$ is equivalent to the inequality $$\int_{[s,m_+]} f(x) \, \mu(dx) \leq \int_{[s,m_+]} f(x) \, \nu(dx).$$ Finally, this inequality follows from the lemma above since $$\mu([s,m_+]) = \frac{p[X < s]}{p} + P[X \geq s] = \nu([s,m_+])$$ and $$\int_{[s,m_+]} x \, \mu(dx) = \frac{P[X < s]}{p}s + E[X ; X \geq s] = \int_{[s,m_+]} x \, \nu(dx)$$ and $\nu$ is supported on $\{s, m_+\}$. Therefore the desired claim $E [f(X)] \geq f(EX)$ follows. ////

References.

• [CVB11] Cirtoaje, Vasile, and Alina Baiesu. "An extension of Jensen's discrete inequality to half convex functions." Journal of Inequalities and Applications 2011, no. 1 (2011): 101.