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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.

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  • $\begingroup$ $\newcommand{\E}{\Bbb{E}}$For any constant $c$, $x\mapsto \max(x,c)$ is a convex function, so as you noted, Jensen's inequality gives us $\E[\max(X,c)]\ge \max\left(\E[X],c\right)$. $\endgroup$ – Minus One-Twelfth Jun 1 '19 at 12:46
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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

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

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    $\begingroup$ How did you obtain "$c \ge \max(\mathbb{E}[X], c)$"? $\endgroup$ – whuber May 27 '19 at 13:44
  • $\begingroup$ If you speak about the third line of my answer, I simply substituted $c$ by $\text{max}(\mathbb{E}[X],c)$ $\endgroup$ – winperikle May 27 '19 at 14:01
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    $\begingroup$ Sorry about that comment--I figured it out immediately after writing it and thought I had deleted it. (+1 already.) $\endgroup$ – whuber May 27 '19 at 14:53
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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 $(X_n)_{n\geq 1}$, $\text{E} \left(\sup_{n \geq 1} |X_n| \right) \geq \sup_{n \geq 1} \text{E}|X_n|$.

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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:

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

You also have:

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

Putting these together gives the inequality.

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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

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  • $\begingroup$ Thanks for your reply. But as $X \in \mathbb{R}_{\geq 0}$ your counterexample could not be applied. $\endgroup$ – Navid Noroozi May 27 '19 at 13:05
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    $\begingroup$ 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 $\endgroup$ – David May 27 '19 at 13:21

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