f(alpha) exercise 3 solution

For r = 0.5, p1 = 0.05, p2 = 0.2, p3 = 0.3, and p4 = 0.05, the formula

alpha = (log(p1)*p1q + log(p2)*p2q + log(p3)p3q + log(p4)*p4q)/ (log(r)*(p1q + p2q + p3q + p4q))

and the formula

tau(q) = -Log(p1q + p2q + p3q + p4q)/Log(r)

yield

alpha(-1) = (Log(.05)*(.05-1) + Log(.2)*(.2-1) + Log(.3)*(.3-1) + Log(.45)*(.45-1))/(Log(.5)*(.05-1 + .2-1 + .3-1 + .45-1)) = 3.48

tau(-1) = -Log(.05-1 + .2-1 + .3-1 .45-1)/Log(.5) = 4.93

f(alpha) = alpha*q + tau(q), so f(3.48) = 3.48*(-1) + 4.93 = 1.45

alpha(1) = -(Log(.05)*(.051) + Log(.2)*(.21) + Log(.3)*(.31) + Log(.45)*(.451))/(Log(.5)*(.051 + .21 + .31 + .451)) = 1.72

tau(1) = -Log(.051 + .21 + .31 + .451)/Log(.5) = 0

f(1) = 1.72*1 + 0 = 1.72

alpha(3) = -(Log(.05)*(.053) + Log(.2)*(.23) + Log(.3)*(.33) + Log(.45)*(.453))/(Log(.5)*(.053 + .23 + .33 + .453)) = 1.35

tau(3) = -Log(.053 + .23 + .33 + .453)/Log(.5) = -2.99

f(3) = 1.35*3 - 2.99 = 1.06

Return to f(alpha) exercise 3.