Question

Exercise 4 Leta(c)-c1/2 and let c2 > cı > 0 be given. Let: π1c1+12c2. where π2 = 1-T1. (i) Sketch the function u and indicate in your sketch the points (C1, u(a), (c, u(c)), and (c2,u(c2)). (ii) Draw the line that connects the two points (ci, u(cı)) and (c2, u(c2)) and represent that line algebraically. Hint: Find the slope and intercept in terms of the two points, (c1, u(c) and (c,,u (сг)).] (iii) Use that algebraic result to show that the point (c-Tu(A) + Co.ulC222 π2u(c2)) is on that line.

Answer 1

u(c) = squareroot c rightarrow Concave funcn. 0 < c_1 < c_2, C^bar = pi_1 c_1 + pi_2 c_2 C^bar is convex combination of c_1 & c_2, (a) point A rightarrow [c_1, u(c_1)], B rightarrow [c_2, u(c_2)] c rightarrow [c^bar, u(c^bar)] b) Now line AB rightarrow Algebraic form Y - Y_1 = -(Y_2 - Y_1/X_2 - X_1) (X - X_1) Y - u(c_1) = + [u(c_2) - u(c_1)/c_2 - c_1] (x - c_1) So u = u(c_1) + (u(c_2) + u(c_1)/c_2 - c_1) (c - c_1) Ans \r\n a) Now point (c^bar, pi_1 u_1 + pi_2 u_2) lie on line, put c = c^bar = pi_1 c_1 + pi_2 c_2 then u = u_1 + (u_2 - u_1/c_2 - c_1) (pi_1 c_1 + pi_2 c_2 - c_1) = u_1 + (u_2 - u_1/c_2 - c_1) ((1 - pi_2) c_1 + pi_2 c_2 - c_1) = u_1 + (u_2 - u_1/c_2 - c_1) (c_1 - c_1 pi_2 + pi_2 c_2 - c_1) = u_1 + (u_2 - u_1/c_2 - c_1) (pi_2 (c_2 - c_1) = u_1 + pi_2 u_2 - u_1 pi_2