soxumis saxelmwifo universitetis S r o m e b i VII

Therefored lim ln L .d The continuous functiond lndLchanges the sign and thereforedthere exists a point ˆ such that ln L 0 . Let us verify that thedˆsecond derivative at this point is negative.We write the second derivative in the formn 2n 22dln L 2dttss1t sln ts 1t s31t snn 2 2 ts t s t s t s221 t s2ts 1 t sln t s31 t sWe have to show that these four fractions, when summed, have anegative value. The first fraction is negative. For the third fraction2 1 ts1 n t s2 1t s 2we have a negative value since it is obvious that ts 1. Itremains to show that2 01 t sln t s ttss1 t s2ts .Since for 0 x 1we have 1tlnt 0 , we need to prove thatttss1 t s 2ts 2 0 . (7).40