Stochastic dominance relationship between two symmetric random variables Suppose $X$ is distributed symmetrically on the support $[a,b]$ 
with the cdf $F_X$ and $Y$ is distributed symmetrically on the support $[c,d]$ with the cdf $F_Y$. Suppose it is known that $F_X$ and $F_Y$ are distinct and $X$ first-order stochastically dominates $Y$. 
I think that it is clear that we should have $a \geq c$. Am I right to conclude though the following: that necessarily either 1) $a > c$ or 2) $a=c$ and $b>d$? 
Or can $X$ and $Y$ have the same support? 
 A: For simplicity assume $X$ and $Y$ have symmetric PDFs over $[0,1]$ with no impulses, and different CDF functions. The symmetry implies one cannot be stochastically greater than or equal to the other. (A similar argument can be done for an interval $[a,b]$ by shifting and scaling). 
Suppose $X$ is stochastically greater than or equal to $Y$ (we reach a contradiction).  Then $F_X(t) \leq F_Y(t)$ for all $t \in [0,1]$. By symmetry we have $F_X(1/2) = F_Y(1/2)=1/2$. 
Case 1:  Suppose there is a $t \in (0,1/2)$ such that $F_X(t) < F_Y(t)$.  Then: 
$$F_X(1/2) - F_X(t) = \int_t^{1/2} f_X(x) dx \underbrace{=}_{symmetry} \int_{1/2}^{1-t} f_X(x)dx = F_X(1-t) - F_X(1/2)  $$
Likewise: 
$$F_Y(1/2) - F_Y(t) = \int_t^{1/2} f_Y(y) dy \underbrace{=}_{symmetry} \int_{1/2}^{1-t} f_Y(y)dy = F_Y(1-t) - F_Y(1/2)  $$
But $F_X(t)  < F_Y(t)$ and $F_X(1/2)=F_Y(1/2)$ implies: 
$$ \underbrace{F_X(1/2) - F_X(t)}_{F_X(1-t)-F_X(1/2)} = F_Y(1/2) - F_X(t) > \underbrace{F_Y(1/2)-F_Y(t)}_{F_Y(1-t)-F_Y(1/2)} $$
And so: 
$$ F_X(1-t) - \underbrace{F_X(1/2)}_{1/2} > F_Y(1-t) - \underbrace{F_Y(1/2)}_{1/2} $$
and so $F_X(1-t) > F_Y(1-t)$, a contradiction. 
Case 2:  $F_X(t) = F_Y(t)$ for all $t \in [0,1/2]$.  By symmetry it holds that $F_X(t) = F_Y(t)$ for all $t \in [0,1]$, contradicting the fact that $X$ and $Y$ have different CDFS.
