I was wondering if the following inequality implies the other.
I have \begin{equation} \int_0^1\int_0^{v_0}\mathbf{1}_{\{v\geq p\}}dvdw>\int_0^1\int_0^{v_0}\mathbf{1}_{\{v+x_w\geq p'\}}\,dv\,dw, \end{equation} where $p$ and $p'$ are two different positive constants and $x_w$ is a non-negative function of $w$.
So basically, I am computing the probability (or just counting) of two events and the above inequality implies that the left is greater than the right.
Now I am wondering if the above inequality implies the following one \begin{equation} \int_0^1\int_0^{v_0}\mathbf{1}_{\{v\geq p\}}(v-p)\,dv\,dw>\int_0^1\int_0^{v_0}\mathbf{1}_{\{v+x_w\geq p'\}}(v+x_w-p')\,dv\,dw, \end{equation} I think intuitively it should be correct, but I am not so sure. Can someone prove or disprove it.
Unless I've miscalculated, the first inequality doesn't imply the second.
Let's be very concrete. Take $v_0=1$, $p=\frac12$, $p'=\frac23$, and $x_w=\frac w3$. Then the integrals in the first inequality are equal, so a tiny perturbation will make the first inequality hold.
For the LHS of the second inequality we can ignore the integral over $w$ for the obvious reason; then we have $\int_p^{v_0}(v-p)\,dv=\frac12(v_0-p)^2=\frac18$. For the RHS we have $\int_0^1 dw\int_{p'-x_w}^{v_0}(v+x_w-p')\,dv=\int_0^1dw\frac12(v_0+x_w-p')^2\,dv=\int_0^1dw\frac12(\frac{1+w}3)^2$ which equals $\frac16\cdot3\cdot[(\frac23)^3-(\frac13)^3]=\frac7{54}$ and this is bigger than $\frac7{56}=\frac18$. So the second inequality fails when the first is right on the edge, and a tiny perturbation will not change that.
(This is really about the convexity of the function mapping $x$ to $x^2$. The values at $w$ and $1-w$ average to more than the value at $w=\frac12$.)