For the geometric distribution p(y) = y (1 ), y = 0, 1, 2,... , show that the tail method for constructing a confidence

For the geometric distribution p(y) = π^y(1– π), y = 0, 1, 2,... , show that the tail method for constructing a confidence interval [i.e., equating P(Y ≥ y) and P(Y ≤ y) to α/2] yields [(α/2)^1/y, (1 – a/2)^1/(y+1)]. Show that all π between 0 and 1 – a/2 never fall above a confidence interval, and hence the actual coverage probability exceeds 1 – a/2 over this region.