What's the null hypothesis in a one-sided Kolmogorov-Smirnov test? Suppose I use K-S to figure out if the CDF of $X$ is greater than the CDF of $Y$. I get the statistic $D^+ = \max_u\{C_x(u) - C_y(u)\}$ where $C_x$ is the ECDF for $X$ and similarly for $C_y$.
There are various tables which translate this into a p-value. Intuitively, if I'm testing if $C_x < C_y$ it seems like the null hypothesis should be $C_x \geq C_y$. (If it were $C_x = C_y$ then I don't understand how it would differ from a two-sided test.)
Yet for my data set I find low p-values for both the proposition that $C_x < C_y$ and $C_x > C_y$, leading me to believe that the null hypothesis is not the complement of the alternative.
Am I completely misunderstanding something?
 A: I think most of the tables providing p-values for the K-S statistic are based on a two-sided test. The null hypothesis assumed by the values in the table is that the two samples are drawn from the same distribution (ie, that $C_x=C_y$). So really the table is only concerned with the absolute value of the difference between $C_x$ and $C_y$ and not the sign. That's why it does not matter if your result shows $C_x<<C_y$ or $C_x>>C_y$. Both are considered strong evidence against the null hypothesis, with a small p-value. 
Let's say your null hypothesis is $C_x \leq C_y$ and your desired criticality level is $\alpha$. You could adapt the values in the table by finding the critical value of $D+$ corresponding to $2\alpha$ and using that instead. This works because the table is splitting up the probability densities into the two tails, so by doubling the specified total tail density, you are "tricking" it into allocating $\alpha$ into the upper tail, which is what you want in the one-sided test.
