Does $E[X] \geq E[Y]$ imply $E[\max(a, X)] \geq E[\max(a, Y)]$ for nonnegative variables

Question

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.

Answer 1

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)] $