Bsrgvty

Let $X in mathbbR_geq 0$ be a "non-negative" random variable and $c$ is a "given" strictly positive number. I wonder if the following inequality holds:
$$
E[maxX,c] leq maxE[X],c,
$$
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 $textmax(mathbbE[X], c) = c$, as $textmax(X,c) geq c$, we have

beginalign*
mathbbE[textmax(X,c)] &geq c \
&geq textmax(mathbbE[X],c)
endalign*

When $textmax(mathbbE[X],c) = mathbbE[X]$ then again as $textmax(X,c) geq X$ we have

beginalign*
mathbbE[textmax(X,c)] &geq mathbbE[X] \
&geq textmax(mathbbE[X],c)
endalign*

So that the inequality is actually the other way

$$
mathbbE[textmax(X,c)] geq textmax(mathbbE[X], c)
$$

Similar to winperikle's answer, just tightening the arguments a bit:
$maxX, c geq X$ and $maxX, c geq c$. So, by taking expectation, $textEleft(maxX, cright) geq textE X$ and $textEleft(maxX, cright) geq c$. Combining, we get $textEleft(maxX, cright) geq max textE X, c$.

These arguments can be generalized to show that for a sequence of $mathcalL_1$ random variables $(X_n)_ngeq 1$, $textE left(sup_n geq 1 |X_n| right) geq sup_n geq 1 textE|X_n|$.

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 $X sim textBin(2,tfrac12)$ and $c=1$, which gives you the expectation:

$$mathbbE(max(X,c)) = frac34 cdot 1 + frac14 cdot 2 = frac54.$$

For this counter-example we have:

$$frac54 = mathbbE(max(X,c)) > max(mathbbE(X),c) = 1.$$

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:

$$mathbbE(max(X,c)) geqslant max(mathbbE(X), c).$$

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

$$beginequation beginaligned
mathbbE(max(X,c))
&= sum_x in mathscrX max(x,c) cdot p_X(x) \[8pt]
&geqslant sum_x in mathscrX x cdot p_X(x) = mathbbE(X). \[8pt]
endaligned endequation$$

You also have:

$$beginequation beginaligned
mathbbE(max(X,c))
&= sum_x in mathscrX max(x,c) cdot p_X(x) \[8pt]
&geqslant sum_x in mathscrX c cdot p_X(x) = c. \[8pt]
endaligned endequation$$

Putting these together gives the inequality.