Prove or disprove the following inequality?

Question

Let $Y \geq X \geq 0$ be two random variables with cdf $F_X$ and $F_Y$ such that $var(Y) \leq var(X) < \infty$. Is

$$ \int_{z + \mathbb{E}[X]}^{\infty}\mathbb{E}[(X - t)^+]F_X(t)dt \geq \int_{z + \mathbb{E}[Y]}^{\infty}\mathbb{E}[(Y - t)^+]F_Y(t)dt $$ true for all $z \in \mathbb{R}$?

Edit: Moreover, assume that $esssup(X) = esssup(Y)$

Answer 1

The inequality is false.

Let us first give the expressions in a different form (this part is not essential though): $$ f(X,z):=\int_{z + \mathbb{E}[X]}^{\infty}\mathbb{E}[(X - t)^+]F_X(t)dt = \int_{z + \mathbb{E}[X]}^{\infty}\mathbb{E}[(X - t)^+]\int_0^t F_X(ds)\, dt\\ = \int_0^\infty \mathbb{E}\left[\int_{(z + \mathbb{E}[X])\vee s}^\infty (X-t)^+ dt\right] F_X(ds) = \frac12\int_0^\infty \mathbb{E}\left[\left((X-(z + \mathbb{E}[X])\vee s)^+\right)^2 \right] F_X(ds)\\ =\frac12 \mathbb{E}\left[\left((X-(z + \mathbb{E}[X])\vee X')^+\right)^2 dt\right], $$ where $X'$ is an independent copy of $X$.

In particular, for $z=0$ the expression is invariant under additions of non-random constants. This means that the condition $0\le X\le Y$ is not really important: we can always add large constants to both to make them (almost) positive and (almost) dominated.

Example $X = 0$ or $1$ with probability $\frac12$, $Y=2$ or $3$ with probabilities $\frac23$ and $\frac13$ respectively. Then $0\operatorname{var}(Y)$, but $2f(X,0) = \frac{1}{4}(1-\frac12)^2= \frac{1}{16}$ (the expression under expectation is non-zero only for $X=1,X'=0$), while $2f(Y,0) = \frac{2}{9}(3-\frac{7}{3})^2 = \frac{8}{81}>2f(X,0)$.

This answers the question as originally posted; with the extra $\operatorname{ess\,sup}$ condition take $X = 0$, $1$, or $3$ with probabilities $\frac12$, $\frac12-\varepsilon$, $\varepsilon$ respectively with $\varepsilon = 0.0001$.