Hcfyk

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.

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 $(X_n)_{ngeq 1}$, $text{E} left(sup_{n geq 1} |X_n| right) geq sup_{n geq 1} text{E}|X_n|$.

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

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.