Two probability density functions $f$ and $g$ are known to have distribution functions $F$ and $G$ respectively with $F(y)>G(y)$ for all $y$, say on $\mathbb{R}$.
It is known that if we convolve $f$ with itself and $g$ with itself in the range $[-\infty,\infty]$ then the distribution functions of $f*f$, ($F^*$) and $g*g$, ($G^*$) will again satisfy $F^*(y)>G^*(y)$ for all $y$ (sums of independent random variables preserve stochastic ordering). Let us assume that the expected value calculated from both $f$ and $g$ be negative. Consider the following two convolutions $$h_0(x):=\int_{b}^{a}f(x-t)f(t)\mbox{d}t\quad\quad h_1(x):=\int_{b}^{a}g(x-t)g(t)\mbox{d}t$$ where $b<0$ and $a>0$.
Given
$$H_0(y):=\int_{-\infty}^{y}h_0(x)\mbox{d}x\quad\quad H_1(y):=\int_{-\infty}^{y}h_1(x)\mbox{d}x$$
Is it also true that $H_0(y)>H_1(y)$ for all $y$?
No. For a counterexample, consider $f$ with mean at $u\lt b$ and $g$ with mean at $v$ in $(b,a)$, then $H_0(y)=P[X+X'\leqslant y,b\leqslant X\leqslant a]$ and $H_1(y)=P[Y+Y'\leqslant y,b\leqslant Y\leqslant a]$ where $X$ and $X'$ are i.i.d. with density $f$ and $Y$ and $Y'$ are i.i.d. with density $g$. When the variances of $f$ and $g$ go to zero, $H_0(y)\leqslant P[b\leqslant X\leqslant a]\to0$ and $H_1(y)\sim P[Y+Y'\leqslant y]$, in particular $H_0(y)\lt H_1(y)$ for all small enough variances and every $y\geqslant2v$.