Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 36 / 2 / d / Solution

Write $H_k={\sum }_{j=1}^{k}1/j$, with $H_0=0$. The truncated uniform prior gives the posterior distributionIts posterior mean isEach term in the numerator of the expectation is $N/N=1$, which explains the exact count $M-y+1$.