 # Financial Mathematics Problems

## Transcribed Text

BOOK REVIEW PROBLEMS: [Solution from Ross on next page for Example 9.3d] Exercise 9.10 Use the approximation to E [U (W )] given by (9 .5) to determine the optima! amounts to invest in each security in Examp le 9.3a when using the utility function U( x) = 1 - e- .oo5x . Compare your results with those obtained in that example . Exercise 9.13 Find the solution of Example 9.3d. Example 9.3d Let us reconsider Example 9.3b, this time using the utility function U(x ) = ./x. Then, with a 1 = a and a2 = 1 - a we have A = 1 + .15a + .18(1 - a) , B = .04a 2 + .0625(1 - a )2 - 2(.02)a (l - a ), and we must choose the value of a that maximizes The solution can be obtained by setting the derivative equal to zero and then solving this equation numerically. O Example 9.3b Suppose you are thinking about investing your fortune of 100 in two securities whose rates of retum have the following expected values and standard deviations: r 1 = .15, v1 = .20; r 2 = .18, v2 = .25 . lf the correlation between the rates of return is p = - .4, find the optima! portfolio when employing the utility function U(x ) = 1 - e-.OOsx_ Solution. If w1 = y and w2 = 100 - y, then from Equation (9.2) we obtain E[W ] = 100 + .15y + .18(100 - y)= 118 - .03y. Also, since c(l, 2) = pv1v2 = - .02, Equation (9.3) gives Var(W) = y2 (.04) + (100 - y)2(.0625) - 2y(l 00 - y)( .02) = .1425y2 - 16.5y + 625. We should therefore choose y to maximize 118 - .03y - .005( .1425y2 - 16.5y + 625)/ 2 or, equivalently, to maximize .01125y - .0007125y2/2. Simple calculus shows that this will be maximized when = · 01125 = 15.789. y .0007125 That is, the maximal expected utility of the end-of-period wealth is obtained by investing 15.789 in investment 1 and 84.211 in investment 2. Substituting the value y = 15.789 into the previous equations gives E[W] = 117.526 and Var(W) = 400 .006, with the maximal expected utility being 1 - exp{- .005(117.526 + .005 (400.006) / 2)} = .44 16. This can be contrasted with the expected utility of .3904 obtaine d when all 100 is invested in security 1 or the expected utility of .4413 when all 100 is invested in security 2. □ Exercise 11.2 Find the optima! strategy and the maximal retum in Example 11. 2a when you have 8 to invest. Use the method of Example 11.2a. Example 11.2a Suppose that three investrnent proje cts with the following return fun ctions are available: lOx f ¡(X) = -- , x = O, 1, . .. , l+ x fz(x) = ,Jx, x = O, 1, ... , h(x) = 10(1 - e- x), x = O, 1, ... , and that we want to maximize our return when we have 5 to invest. Now, lOx V¡(X) = f¡ (X) = -- , y ¡(X) = X. l + x Because V2(x) = max {fz(y) + V1(x - y)} = max { ✓Y + --- , l0(x - y)} O::,y::,x 0::,y::,x 1 + X - y wesee that V2(1) = max {l0/ 2, l} = 5, Y2(l) = O, V2(2) = max {20/ 3, 1 + 5, h } = 20/ 3, Y2(2) = O, V2(3) = max {30/ 4 , 1 + 20 / 3, v2 + 5, -vJ} = 23/3, y2(3) = 1, V2(4) = max {40/ 5, 1 + 30 /4, h + 20/3, ✓3 + 5, ✓4 } = 8.5, Y2(4) = 1, V2(5) = max {50/ 6, 1 + 8, h + 7.5, J3 + 20/3, ✓4 + 5, ✓5 } = 9, Y2(5) = l. Continuing, we have that V3(x) = max {h(y) + V2(x - y)} = max {10(1- e- Y) + V2(x - y)} . O::,y::,x O::,y::,x Using that 1 - e- 1 = .632, l - e- 2 = .865, 1 - e- 3 = .950, 1 - e -4 = .982, l - e- 5 = .993 , we obtain V3(5) = max{9, 6.32 + 8.5 , 8.65 + 23/3, 9.50 + 20/3, 9.82 + 5, 9.93) = 16.32, Thus , the maximal sum of returns from investing 5 is 16.32; the optima) amount to invest in project 3 is YJ (5) = 2; the optima! amount to invest in project 2 is y2 (3) = l; and the optimal amount to invest in project 1 1 is y1(2) = 2. o Exercise 11.3 Use the method of Exam ple 11.2b to solve the preceding exercise . Example 11.2b Let us reconsider Example 11.2a, where we have 5 to invest among three projects whos e return functions are lOx f ¡(X) = 1 + x ' fz( x ) = .Jx, h( x) = 10(1 - e- x) . Let x ;(j ) denote the optimal amount to invest in proj ect i when we have a total of j to invest. Because max{j 1(1), h (l), h (l)} = max{5, 1, 6.32} = 6.32, we see that Since max{f¡(x;([) + 1) - f ;(x;(l))} = max{5, 1, 8.65 - 6.32} = 5, 1 we have Because max{f ;(x ;(2) + 1) - f¡(x ;(2))} = max{20/ 3 - 5, 1, 8.65 - 6.32} i = 2.33, it follows that Since max{f ;(x;(3) + 1) - f¡( x;(3))} = max{20/ 3 - 5, 1, 9.50 - 8.65} i = 1.67, we obtain X¡(4) = 2, X2(4) = O, X3(4) = 2. Finally, max {f;(x;( 4) + 1) - f¡(x;(4))} = max{30/4 - 20/ 3, 1, 9.50 - 8.65} i = 1, giving that x 1(5) = 2, x 2(S) = 1, x 3(5) = 2. The maximal retum is thus 6.32 + 5 + 2.33 + 1.67 + 1 = 16.32. O The following algorithm can be used to solve the problem when m is to be invested among n projects, each of which has a concave retum function. Toe quantity k will represent the current amount to be invested, and x; will represent the optima! amount to invest in project i when a total of k is to be invested. Algorithm (1) Set k = O and x; = O, i = 1, .. . , n. (2) m; = f ;(x; + 1) - f¡(x;), i = 1, ... , n. (3) k = k + l. (4) Let J be such that m1 = max ; m¡. (5) lf J = j , then Xj ➔ X¡+ 1, mj ➔ f¡(x¡ + 1) - f¡(x¡) . (6) lf k < m, go to step (3). Step (5) means that if the value of J is j, then (a) the value of x¡ should be increased by 1 and (b) the value of m¡ should be reset to equal the difference off¡ evaluated at 1 plus the new value of x¡ and f¡ evaluated at the new value of x¡ . Remark: When g(x ) is defined for ali x in an interval, then g is concave if g' (t ) is a decreasing function of t (that is, if g" (t) ::: O). Hence, for g concave ·+1 . !' g' (s)d s ::: [ ' g' (s)ds 1 , - 1 yielding that g (i + 1) - g(i) ::: g(i) - g(i - 1) wbich we used as the definition of concavity for g defined on the intege rs. Exercise 11.6 Continue withE xampl e 11.2c and find the optimal strategy when you have 25 to invest. Example 11.2c Suppose you have 25 to invest among three projects whose cost and retum values are as follows. Then V(x ) = O, x ~ 4, Project 1 2 3 Cost per share 5 9 15 V(x ) = 7, i(x ) = 1, x = 5, 6, 7, 8, Retum per share 7 12 22 V(9) = max{7 + V(4) , 12 + V(O)} = 12, i(9) = 2, V(x ) = max{7 + V(x - 5), 12 + V(x - 9)} = 14, i (x ) = 1, X = 10, 11, 12, 13, V(14) = max{7 + V(9), 12 + V(5)} = 19, i (x) = 1 or2 , V(15) = max{7 + V(lO), 12 + V(6), 22 + V(O)} = 22, i(15) = 3, V(16) = max{7 + V(ll ), 12 + V(7), 22 + V(l)} = 22, i(16) = 3, V(17) = max{7 + V(12), 12 + V(8), 22 + V(2)} = 22, i(17) = 3, V(18) = max{7 + V(13), 12 + V(9), 22 + V(3)} = 24, i(18) = 2, and so on. Thus, for instance, with 18 it is optima} to first purchase one share of project i(18) = 2 and then purchase one share of project i (9) = 2. Thati s, with 18 it is optimal to purchase two shares ofproject2 for a total return of 24. O

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