Setting u n+1 = 1, from Figure 2, we can find the market coverage of θ j , j = 1, 2, …, n, as x j = u j+1 – u j , which can be summed over j to obtain u i = 1 – G n j =i x j. Substituting this into (2) for i = 1 and noting that p 0 = θ 0 = 0, we find (A1) p G u G x j j n 1 11 1 1 1 = = −       =  θ θ We will now prove the lemma by induction. It is clear that (3) reduces to (A1) for i = 1...