Are they using different "1-b" in the trailing/leading P/E formula?

The formula for trailing P/E in Schweser is (1-b)(1+g)/(r-g). The “b” must be different because the leading P/E formula is (1-b)/(r-g). Deriving this formula: P=(D1/(r-g)), so Leading P/E=P/E1=(D1/E1)/(r-g) so D1/E1 must be the “1-b” they are talking about. But when you plug in this “1-b” to trailing P/E formula =P/E0=(D1/E1)(1+g)/(r-g) and this doesn’t equal P/E0! Are they using two different "b"s or two different “1-b”?

you have put a lot of thought into this… I am having a tough time following your derivation, but i am certain that the 1-b are the same thing in both formulas…

rellison what are you talking about? P=D1/(r-g) = E1(1-b)/(r-g) so P/E1 = (1-b)/(r-g) Now another way: E1 = E0 * (1+g) so P/E0 = (1-b)(1+g)/(r-g)

Case1: Trailing P0 = D1/(r-g) P0/E0 = [D1/E0]/(r-g) P0/E0 = [D0/E0]*(1+g)/(r-g) P0/E0 = (1 - b)*(1 + g)/(r-g) 1 - b = D0/E0 Case2: Leading P0 = D1/(r-g) P0/E1 = [D1/E1]/(r-g) P0/E1 = [D1/E1]/(r-g) P0/E1 = (1 - b)*/(r-g) 1- b = D1/E1 Eventually they both are Retention Rates

oh ok they’re just multiplying a 1+g to the denominator of the left side, and multiplying the numerator of the right side to balance it out. Got it.