Upper and lower bounds of the early exercise boundary for American option

In the article about Exercise boundaries of American options by F.AitSahlia and T.L.Lai the closed-form formulas for lower and upper bounds of the exercise boundaries are given as follows: It appears that lower bound is always above the upper bound, which seems to be wrong. Let's assume parameters: rho = 0.5, alpha= 0.1 and s = -2.5. Hence, theta = -0.95. It leads to ln[theta/(theta-1)] entering the second formula as a positive term. Therefore the lower bound is above the upper bound. The question is whether the reasoning above has a flaw?

"Exercise boundaries and efficient approximations to American option prices and hedge parameters" by F.AitSahlia and T.L.Lai

• I don't know if it is relevant, but in the other paper on early exercise by the authors they state the lower boundary as $ln[\theta/(\theta-1)] - [\rho(1-\alpha) - \frac{1}{2}]s$ instead of the above. However, I do think that the case there is for $T = \infty$, but if that implies the order is reversed I don't know. – Forgottenscience Feb 18 '17 at 21:55
• your note seems to solve the problem with the bounds – Олег Бойко Feb 20 '17 at 12:08