# Expectation over a max operation

Let $$X \in \mathbb{R}_{\geq 0}$$ be a "non-negative" random variable and $$c$$ is a "given" strictly positive number. I wonder if the following inequality holds: $$E[\max\{X,c\}] \leq \max\{E[X],c\},$$ where $$E[\cdot]$$ is the expectation.

I suspect from Jensen's inequality the other way around should be true; but since the $$c$$ above is a certain constant, I'm still (naively) hopeful that it could be true.

• $\newcommand{\E}{\Bbb{E}}$For any constant $c$, $x\mapsto \max(x,c)$ is a convex function, so as you noted, Jensen's inequality gives us $\E[\max(X,c)]\ge \max\left(\E[X],c\right)$. Commented Jun 1, 2019 at 12:46

If $$\text{max}(\mathbb{E}[X], c) = c$$, as $$\text{max}(X,c) \geq c$$, we have

\begin{align*} \mathbb{E}[\text{max}(X,c)] &\geq c \\ &\geq \text{max}(\mathbb{E}[X],c) \end{align*}

When $$\text{max}(\mathbb{E}[X],c) = \mathbb{E}[X]$$ then again as $$\text{max}(X,c) \geq X$$ we have

\begin{align*} \mathbb{E}[\text{max}(X,c)] &\geq \mathbb{E}[X] \\ &\geq \text{max}(\mathbb{E}[X],c) \end{align*}

So that the inequality is actually the other way

$$\mathbb{E}[\text{max}(X,c)] \geq \text{max}(\mathbb{E}[X], c)$$

Similar to winperikle's answer, just tightening the arguments a bit: $$\max\{X, c\} \geq X$$ and $$\max\{X, c\} \geq c$$. So, by taking expectation, $$\text{E}\left(\max\{X, c\}\right) \geq \text{E} X$$ and $$\text{E}\left(\max\{X, c\}\right) \geq c$$. Combining, we get $$\text{E}\left(\max\{X, c\}\right) \geq \max \{\text{E} X, c\}$$.

These arguments can be generalized to show that for a sequence of $$\mathcal{L}_1$$ random variables $$(X_n)_{n\geq 1}$$, $$\text{E} \left(\sup_{n \geq 1} |X_n| \right) \geq \sup_{n \geq 1} \text{E}|X_n|$$.

The inequality you have asserted is false: A simple counter-example is $$X \sim \text{Bin}(2,\tfrac{1}{2})$$ and $$c=1$$, which gives you the expectation:

$$\mathbb{E}(\max(X,c)) = \frac{3}{4} \cdot 1 + \frac{1}{4} \cdot 2 = \frac{5}{4}.$$

For this counter-example we have:

$$\frac{5}{4} = \mathbb{E}(\max(X,c)) > \max(\mathbb{E}(X),c) = 1.$$

There is a related inequality that is true: Although the inequality you have asserted is false (or at least, not generally true), the following alternative inequality is true:

$$\mathbb{E}(\max(X,c)) \geqslant \max(\mathbb{E}(X), c).$$

This inequality can easily be proven either for the discrete or continuous (or mixed) case. For a discrete random variable you have:

\begin{aligned} \mathbb{E}(\max(X,c)) &= \sum_{x \in \mathscr{X}} \max(x,c) \cdot p_X(x) \\[8pt] &\geqslant \sum_{x \in \mathscr{X}} x \cdot p_X(x) = \mathbb{E}(X). \\[8pt] \end{aligned}

You also have:

\begin{aligned} \mathbb{E}(\max(X,c)) &= \sum_{x \in \mathscr{X}} \max(x,c) \cdot p_X(x) \\[8pt] &\geqslant \sum_{x \in \mathscr{X}} c \cdot p_X(x) = c. \\[8pt] \end{aligned}

Putting these together gives the inequality.

Let X be uniform in (0, 5) and c=2. Here you have a counterexample with each side of the inequality being 3.5 and 2.5

• Thanks for your reply. But as $X \in \mathbb{R}_{\geq 0}$ your counterexample could not be applied. Commented May 27, 2019 at 13:05
• Oops! My bad! But you can take X uniform in (0,5) and c=2, leading to the same thing. Let's edit the answer Commented May 27, 2019 at 13:21