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.


1 Answer 1


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 $\Pr(X \gt 0) = 1$, $\Pr(X \ge Y) \gt 1/2$.


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