why does exact binomial confidence intervals have wider than nominal coverage?

Question

The exact $(1-\alpha)$ level confidence interval lower limit is given by $$ \sum_{k=y}^{n} {n \choose k} {p_L}^k(1-p_L)^{n-k}=\alpha/2 $$ and the upper limit analogously. Why does the resulting C.I. has greater coverage than the nominal $(1-\alpha)$? I thought that the above equation has a valid solution for $p_L$ at any $\alpha$ and so should provide the correct coverage.

Answer 1

Because the distribution is discrete. One might get $$ \sum_{k=y}^n \cdots \ge \frac\alpha2 > \sum_{k=y+1}^n \cdots $$ but there's nothing between $y$ and $y+1$ that you can put there to make it exactly $\alpha/2$, so you err on the side of caution and use the one that makes it more than $\alpha/2$.