Does $E[X] \geq E[Y]$ imply $E[\max(a, X)] \geq E[\max(a, Y)]$ for nonnegative variables? Does $E[X] \geq E[Y]$ imply $E[\max(a, X)] \geq E[\max(a, Y)]$ for nonnegative random variables $X$ and $Y$ and a constant $a >0$?
Here is what I have tried: We know $E[\max(a, X)] \geq E[X]$ and similar for $Y$. This says that $E[X]$ and $E[Y]$ are lower bounds for the respective $\max()$ functions. Therefore $\cdots$.
(This approach can be used to show that $E[X] \geq E[Y]$ implies $E[\max(a, X)] \geq E[\min(a, Y)]$ (note "min"), for example.)
If the statement is true, I think we can use the convexity of the $\max()$ function here, or maybe Markov's inequality.
 A: Try for example

*

*$a=3$,

*$X=2$ with probability $1$,

*$Y=1$ with probability $0.9$ and $Y=4$ with probability $0.1$
Then

*

*$E[X]=2>1.3=E[Y]$,

*$E[\max(a,X)]=3 < 3.1= E[\max(a,Y)] $
A: The condition $\mathbb E[\max(a,X)] \geq \mathbb E[\max(a,Y)]$ for all $a \geq 0 $ and non-negative random variables $X,Y$ is much stronger than the condition $\mathbb E[X] \geq \mathbb E[Y]$ and has deeper meaning. Indeed, it is the root of a stochastic order : a way of comparing random variables , often used in industrial and estimation purposes (modelling appliance lifetimes, risk occurrences etc.).
Indeed, the inequality $\mathbb E[\max(a,X)] \geq \mathbb E[\max(a,Y)]$ is , upon subtraction of $a$ from each side, equivalent to the statement that $\mathbb E[(X-a)_{+}] \geq \mathbb E[(Y-a)_{+}]$ for all $a \geq 0$, where $b_+ = \max(b,0)$. Note that if $a$ is negative then the inequality is true because $\mathbb E[X] \geq \mathbb E[Y]$ is true from setting $a=0$. Therefore, for all $a \in \mathbb R$, it is true that $$\mathbb E[(X-a)_{+}] \geq \mathbb E[(Y-a)_{+}]$$
In the theory of comparison of random variables, this particular inequality can be extended, by density, to the following :$$
\mathbb E[\phi(X)] \geq \mathbb E[\phi(Y)]
$$
for all increasing convex functions $\phi$ (and not just $\phi(x) = x$, which is what would yield $\mathbb E[X] \geq \mathbb E[Y]$).
If this occurs, we say that $X$ dominates $Y$ in the increasing convex order, and write it as $X \geq_{icx} Y$. It's clear that this is far stronger than the mere condition $\mathbb E[X] \geq \mathbb E[Y]$ , when I discuss the equivalent condition :
$X \geq_{icx} Y$ if and only if , for all $x \in \mathbb R$, $$
\int_x^\infty \mathbb P[X > t] dt \geq \int_x^\infty \mathbb P[Y>t] dt \tag{continuous}
$$
$$
\sum_{t \geq x} \mathbb P(X > t) \geq \sum_{t \geq x} \mathbb P(Y > t) \tag{discrete}
$$
Therefore, domination in the increasing convex order is equivalent to an inequality of the distributions functions/pmfs of the relevant random variables : something that is clearly far, far stronger than just $\mathbb E[X] \geq \mathbb E[Y]$.
One can easily construct a counterexample now using discrete random variables. Let's take $X$ constant , say $X=1$ with probability $1$ : this is a simple case, given that the statements have been seen to be very far apart so we expect an elementary counterexample.
Allow $Y$ to take a value larger than $1$, like say $Y = 2$ with probability $0.1$. This ensures that $\mathbb P(Y>1.5) > 0 = P(X>1)$, for example, so $a=1.5$ works and ensures that $X \not \geq_{icx} Y$ with $x=1.5$ in the equivalent condition above. Then one can finish by allowing $Y$ to take a small enough value so that $\mathbb E[Y] \leq \mathbb E[X]$, say $Y = 0.1$ with probability $0.9$. Thus $\mathbb E[X] \geq \mathbb E[Y]$ but $\mathbb E[\max(X,1.5)] \not \geq \mathbb E[\max(Y,1.5)]$.
Nevertheless, while the answer may be easy, it is nice to see the connection with a topic that is not often discussed at a first level in probability : stochastic orders. More on this can be found in books such as "Stochastic Orders" by Shaked and Shanthikumar.

Approximation of Increasing Convex Functions
I'll present only brief details : I found this material in "Convex functions and their applications" by Nicolaescu and Persson. We'll call a function $f: [a,b] \to \mathbb R$ a piecewise affine function, if there exist $a=x_0 < x_1<\ldots < x_n = b$ such that restricted to each $[x_i,x_{i+1}]$, the graph of $f$ is linear.
Lemma (Adapted from lemma 1.9.1) : Every increasing convex function $f : [a,b] \to \mathbb R$ is the pointwise limit of a sequence of piecewise linear increasing convex functions.
Proof : Consider $f_n(x)=y$ where $(x,y)$ is the unique point lying on the graph formed by joining consecutive points of the form $\left(\frac{(n-k)a+kb}{n}, f\left(\frac{(n-k)a+kb}{n}\right)\right) , 0 \leq k \leq n$ by straight lines. Then, this $f_n$ is convex, increasing and piecewise linear : it will also converge pointwise, as one can clearly see from continuity of $f$.
Lemma(Adapted from Lemma 1.9.2) : Let $f : [a,b] \to \mathbb R$ be a piecewise affine increasing convex function with the piece endpoints described by $a=x_0 <x_1 < \ldots 
< x_n = b$. Then, for suitable $\alpha, c_i \geq 0$ and $\beta$ real, $$
f(x) = \alpha x + \beta + \sum_{i=1}^n c_i (x-x_i)_+
$$
on $[a,b]$.
Proof : Let $\alpha , \beta$ be chosen such that $\alpha x + \beta$ is the affine function corresponding to the piece on $[x_0,x_1]$. Note that $\alpha \geq 0$ as $f(x_1) \geq f(x_0)$ so the line will have a non-negative slope. Then , $f(x) - (\alpha x + \beta)$ is an increasing convex function that vanishes on $[x_0,x_1]$. Now, think about the function $f(x) - (\alpha x + \beta)$ on $[x_1,x_2]$. At $x_1$, it's equal to $0$. Let's say that it's equal to some $D$ at $x_2$. Then the slope of this function from $x_1$ to $x_2$ is constant, and it rises by $D$. Therefore, it follows that the slope is $C_1 = \frac{D}{x_2-x_1}$, and it follows that on $[x_0,x_2]$, $f(x) = (\alpha x + \beta) + c_1 (x-x_1)_+$. By now considering this function on $[x_2,x_3]$ and getting $c_2$, we can get all the $c_i$ : which are non-negative because on each occasion there is a rise by some $D_k \geq 0$ so the ratio representing $c_k$ is positive.
Theorem : Every increasing convex function $g$ is the limit of a sequence of piecewise linear functions , extended suitably by straight lines to be piecewise linear on $\mathbb R$.
Proof : Let $g_n$ be the restriction of $g$ to $[-n,n]$ and let $h_{n^2}$ be the $n^2$th iteration of the piecewise linear approximation considered in the first lemma, extended by straight lines of appropriate slope in order to retain continuity and their increasing nature on $\mathbb R$. One can extend the previous lemma to include $x_{-1} = -\infty, x_{l+1} = +\infty$ in the partition of $h_{n^2}$ and see that $h_n^2$ is still of the form described.

It follows that every increasing convex function is a limit of a non-negative linear combination of functions of the form $(x-a)_+$ and the function $x$ (along with a constant $\beta$, but that can be ignored as we see later).
However, it's quite clear that if $\mathbb E[(X-a)_+] \geq \mathbb E[(Y-a)_+]$ for all $a \in \mathbb R$ and $\mathbb E[X] \geq \mathbb E[Y]$, then $\mathbb E[f(x)] \geq \mathbb E[f(Y)]$ for every piecewise affine function $f$ on $\mathbb R$. Indeed, the $\beta$ cancels out, and the rest of the domination carries over because $\alpha$ and each of the $c_i$ are non-negative.
Now, using an appropriate limit theorem such as the $DCT$, one sees that $\mathbb E[\phi(X)] \geq \mathbb E[\phi(Y)]$ for every $\phi$ increasing convex on $\mathbb R$.
