Second Order Stochastic Dominance and $\Pr[X \leq Y]\leq1/2$ Let $X$ be a random variables with continuou cdf $F$ in the support $[a,b]$.  Let $Y$ be another r.v. with cdf $G$, which is continuous in $[a,b)$ and has an atom at $b$.  $X$ and $Y$ are independent. We also have $E[X]=E[Y]$ and second order stochastic dominance
$$\int_a^k F(t)\,dt \leq \int_a^k G(t)\,dt, \  \ \  a \leq k \leq b.$$
Can we say the following:
$$\Pr[X \leq Y] \leq \frac 1 2 \text{?}$$
My attempt:  I have tried numerical analysis with different parameter values and the above statement is always true. However, it gets difficult to prove it analytically.
 A: It's not true.
Intuitively, we could concentrate $X$ near the common expectation and spread $Y$ out to the two endpoints.  Provided most of the probability of $Y$ is near the smaller endpoint, $X$ will tend to exceed $Y$, but $Y$ will dominate $X$ (stochastically in second order) because the integral of $G$ rises rapidly at small values while the integral of $F$ does not start to rise above zero until later.

For an explicit counterexample, take $a=0$, $b=1$, and let $0 \lt p \lt 1/2$ be fixed.  Select $0 \lt \epsilon \lt p/2$, then choose $0 \lt \delta \lt p-(1+p)\epsilon$.  Let $X$ have a uniform distribution on $(p + (1-p)\epsilon-\delta, p + (1-p)\epsilon+\delta)$ and let $Y$ be a mixture of a uniform distribution on $(0,2\epsilon)$ (with weight $1-p)$ and an atom at $1$ with weight $p$.

The left hand graphic plots $F$ and $G$.  The right hand plots their integrals.  Graphs associated with $F$ are shown as a blue dashed line while graphs associated with $G$ are shown as a solid red line.
Since the expectation of any uniform distribution is its midrange, compute
$$\mathbb{E}(X) = p + (1-p)\epsilon$$
and since the expectation of a mixture is the weighted combination of the expectations of its components,
$$\mathbb{E}(Y) = (1-p)(\epsilon) + p(1) = p + (1-p)\epsilon.$$
Thus the expectations are equal, as required.
The second-order stochastic dominance is clear, because the $G$ integral is positive for small values of its upper limit while the $F$ integral is zero, then eventually both rise linearly to equal $1-E[X]=1-E[y]$ at $1$.  This is the point at which the graphs meet in the upper right of the right hand graphic.
Finally, since $\Pr(Y=0) = 1-p \gt 1/2$ and $\Pr(X \gt 0) = 1$, $\Pr(X \ge Y) \gt 1/2$.
