Global Journal of Science Frontier Research, F: Mathematics and Decision Science, Volume 21 Issue 4

⟹ ( 0 ; , . ) ≤ = 0 ⟹ ( 0 ; , . ) = This contradict the fact that ≠ 0 . Thus we must have that there is at least one ∈ + (for such ∈ + , = 0 ) that satisfies the maximization problem. This completes the proof. We now proceed to solve the maximization problem of equation (6.3) and equation (6.4) which is equivalent to the maximization problem of equation (6.1) and equation (6.2). Clearly ( ; ( 0 ), . ( 0 ), 0 ) is differentiable in the given subset of + and by classical optimization theory of calculus, a necessary condition for existence of maximum (extreme) point of ( ; ( 0 ), . ( 0 ), 0 ) is that the derivatives of ( ; ( 0 ), . ( 0 ), 0 ) must be equal to zero [3,4,13]. This implies that ( ; ( 0 ), . ( 0 ), 0 ) = 0 (6.6) We now proceed to solve for equation (6.6). Observe that ( ; ( 0 ), . ( 0 ), 0 ) = ⎣ ⎢ ⎢ ⎢ ⎡ � 1 0 −1 � 1 0 −2 −1 2 � 1 0 − � 2 − 2 0 −2 1 0 � 1 0 − 2 � −1 2 � 1 0 − � 2 ⎦ ⎥ ⎥ ⎥ ⎤ = 1 0 −2 −1 2 � 1 0 − � 2 �� 1 0 −1 � − 1 0 1 0 � 1 0 − 2 �� By equation (6.6) it follows that 1 0 −2 −1 2 � 1 0 − � 2 �� 1 0 −1 � − 1 0 1 0 � 1 0 − 2 �� = 0. Since 1 0 −2 −1 2 � 1 0 − � 2 > 0 ∀ ∈ + , we must have that � 1 0 −1 � − 1 0 1 0 � 1 0 − 2 � = 0 By simplifying the above equation we have 2 (1 − 0 ) − 2 0 + 1 0 = 0 On the Generalized Power Transformation of Left Truncated Normal Distribution 1 Global Journal of Science Frontier Research Volume XXI Issue IV Year 2021 23 ( F ) © 2021 Global Journals Version I 3. U. A. Osisiogu, (1998). An Introduction to Real Analysis, Bestsort Educational Book. Nigeria, 1998. R ef