When is $\Pr(U\leq v|V=v)$ not an increasing function of $v$? Suppose $U$ and $V$ are random variables. I'm trying to find an example for which $\Pr(U\leq v|V=v)$ it not an increasing function of $v$.
Current thoughts: when $U$ and $V$ are independent, $\Pr(U\leq v|V=v)=\Pr(U\leq v)$, which is increasing in $v$. So this doesn't work. I have also considered the case in which
$$
\begin{pmatrix}U\\V\end{pmatrix}\sim N\left(\begin{pmatrix}0\\0\end{pmatrix},\begin{pmatrix}1 & \rho \\ \rho & 1\end{pmatrix}\right)
$$
and it turns out that regardless of what $\rho$ is in $[-1,1]$, the conditional probability $\Pr(U\leq v|V=v)$ is increasing in $v$. Thus this doesn't work either.
So next, I assume $U$ and $V$ have joint density $f$ and $V$ has marginal density $g$. Then the Leibniz's Rule gives:
$$
\frac{\partial}{\partial v}\int_{-\infty}^v\frac{f(u,v)}{g(v)}du=\int_{-\infty}^v\frac{\partial}{\partial v}\frac{f(u,v)}{g(v)}du+\frac{f(v,v)}{g(v)}.
$$
The second term above is non negative whereas the first term, I think, can take on either sign. So it seems I should look at cases for which the first term above is negative, but I haven't been able to come up with a concrete example.

Let $\Omega = \{1,2,3,4\}$, with $\mathbb{P}(\{v\} ) = \frac{1}{4}$.
Now define $U (v) = v$ and $V(v) = \left\{ \begin{matrix} 2 & v \le 2 \\ 3 & v \ge 3 \end{matrix} \right.$. Then we have 
\begin{eqnarray}
\mathbb{P}(U \le 2 \mid V =2) &=& \frac{ \mathbb{P}(\{1,2\})}{\mathbb{P}(\{1,2\})} &=& 1 \\
\mathbb{P}(U \le 3 \mid V =3) &=& \frac{ \mathbb{P}(\{3\})}{\mathbb{P}(\{3,4\})} &=&  \frac{1}{2}
\end{eqnarray}
For simplicity, let $\Omega = \{1,2,3\}$ with equal probability.
Now, let $U(v) = v $ and $V(v) = \left\{ \begin{matrix} 1 & v \le 2 \\ 2 &v=3\end{matrix} \right. $, then we have
\begin{eqnarray}
\mathbb{P}(U \le 1 \mid V =1) &=& \frac{ \mathbb{P}(\{1\})}{\mathbb{P}(\{1,2\})} &=& \frac{1}{2} \\
\mathbb{P}(U \le 2 \mid V =2) &=& \frac{ \mathbb{P}(\emptyset)}{\mathbb{P}(\{3\})} &=&  0
\end{eqnarray}
