MATH362 Fundamentals of Mathematical Finance

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1 MATH362 Fundmentls of Mthemticl Finnce Solution to Homework Three Fll, 2007 Course Instructor: Prof. Y.K. Kwok. If outcome j occurs, then the gin is given by G j = g ij α i, + d where α i = i + d i We then hve i= nd g ij = G j = g jj α j i= { di if j = i if j i. i= i j α i = + g jj )α j = + d j )α j = i= i= i= α i α i + d i =, j =,2,,m. i= + d i + d i Therefore, the betting gme will lwys yield gin of exctly. 2. Suppose we hold α,α 2,α 3 ) units of the three securities, with problem, we cn set i= 3 α i,α i 0. In this 3 α i = since the rndom returns re greter thn one under ll sttes i= of world. Using log utility criterion, the growth fctor is m = E[ln R] = 2 ln4α + 2α α α 2 )) + 2 ln2α + 4α α α 2 )) = 2 ln3 + α α 2 ) + 2 ln3 α + α 2 ). i=

2 Applying the first order condition m α = 2 m α 2 = α α 2 2 ) 3 + α α Possible optiml strtegies re 2 = 0 3 α + α 2 3 α + α 2 = 3 + α α 2 α = α 2 = 0 3 α + α α α 2 = 3 α + α 2 α = α 2 2 ) 0 nd 3 3 ). 3 In fct, for ny portfolio choice with α 2 = α,α 3 = 2α,α 0, the return is 4α + 2α 2 + 3α 3 = 4α + 2α + 3 2α ) = 3 or 2α + 4α 2 + 3α 3 = 2α + 4α α ) = 3. Therefore, the optiml strtegies lwys yield return of 3 for ll α. 3. ) By log-utility criterion, we hve ) m = E[ln R] = p j ln α j r j + α i. j= i= The first order condition gives so tht m α k = p j ) r k + p k = 0 α j r j + α i α k r k + α i j= j k i= i= p k r k α k r k + α i j= p j α j r j + i= i= = 0, α i k =,,n. b) Assume i= r i =. Let α i = p i,i =,,n, so tht α 0, nd Hence, this is possible strtegy. Now substitute into prt ), we hve α i = i= p i =. i= p k r k p k r k + p i i= j= p j p j r j + = p i i= Therefore, this is indeed n optiml trding strtegy. 2 j= r j = 0.

3 c) Suppose < p i r i for ll i =,,n. Furthermore, suppose α n = 0 is prt of n optiml solution. Write α i = L. From the result in prt ), we put k = n nd α n = 0 so tht i= L = n j= p j α j r j + L. Substituting the result into the other cse k n, we obtin p k r k α k r k + L = p nr n L. Rerrnging the terms, we hve pk α k = L ). r k It is desirble to express L in terms of θ = so tht L = α k = L k= The solution α k cn be expressed s α k = k= k= pk r k. We observe r k ) L = + θ. p k r k = + θ) r k ) = L θ p k r k r k + r k θ). However, this solution is vlid provided tht 0 L. Tht is, + θ 0 θ + i) nd + θ θ, ii) where θ = i= r k. Since p k r k >,k n, together with Ineq i), we obtin α k > 0,k n. Ineq ii) gives the necessry condition for the existence of solution. 3

4 4. Recll tht clss of the power utility functions includes the logrithm utility since [ lim 0 + x ] = ln x. This clss of function hs the sme recursive property s the log utility; tht is, the structure is preserved from period to period. This is seen from E[UX K )] = E[R KR K R X 0 ) ] = E[R K R K R ]X 0 = E[R K ]E[R K ] E[R ]X 0 where the lst equlity follows from the fct tht the expected vlue of product of independent rndom vribles is equl to the product of the expected vlues. Hence to mximize E[UX K )] with fixed-proportions strtegy it is only necessry to mximize E[R X 0 ) ], so gin to mximize E[UX K )] one needs only to mximize E[UX )]. 5. Let x,y ),x 2,y 2 ),x 3,y 3 ) B Reflexivity: x = x nd y y so tht x,y ) x,y ) Comprbility: If x > x 2, then x,y ) x 2,y 2 ). If x 2 > x, then x 2,y 2 ) x,y ). If x = x 2, then if y y 2, then x,y ) x 2,y 2 ) if y 2 y, then x 2,y 2 ) x,y ). Trnsitivity: Given x,y ) x 2,y 2 ),x 2,y 2 ) x 3,y 3 ). If x = x 2 > x 3, with y y 2, then x > x 3 so tht x,y ) x 3,y 3 ). If x > x 2 = x 3, with y 2 y 3, then x > x 3 so tht x,y ) x 3,y 3 ). If x = x 2 = x 3, with y y 2 y 3, then x = x 3,y y 3, so tht x,y ) x 3,y 3 ). If x > x 2 > x 3, then x > x 3, so tht x,y ) x 3,y 3 ). 6. Suppose x,y ) x 2,y 2 ). Cse I: x > x 2 : αx,y ) + α)x 2,y 2 ) = + αx x 2 ), + αy y 2 )) 2: βx,y ) + β)x 2,y 2 ) = + βx x 2 ), + βy y 2 )). α > β + αx x 2 ) > + βx x 2 ) nd since x x 2 > 0, so α > β αx,y ) + α)x 2,y 2 ) βx,y ) + β)x 2,y 2 ). Cse II: x = x 2,y y 2 α > β + αx x 2 ) = = + βx x 2 ), but + αy y 2 ) + βy y 2 ) nd since y y 2 0, so 4

5 α > β αx,y ) + α)x 2,y 2 ) βx,y ) + β)x 2,y 2 ). 7. Let ux,y) = lnx + y), nd consider,0) nd 0,) B, we hve,0) 0,). But u,0) = ln = u0,). Hence, u cnnot be utility function representing the Dictionry Order. 8. Consider the HARA clss of utility functions Ux) = ) x + b ) ) ) = x + b ) ) ) Let = ),b 0 +. Then b) Let = 2. Then Ux) = x x s. Ux) = 2 x + b)2 = b2 2 + bx 2 2 x2 which is equivlent with V x) = bx 2 since they differ by only constnt. Now let 2 = c,b = /, V x) = x 2 cx2. 2 x2 c) Note tht Then + αx ) n e αx s n. Let b = n αx Ux) = + ) e αx s. ) ). = α). ) / r d) Let b 0 +, = c / ). Then Ux) = c / c) = cx. e) Tke c = from prt d. By Problem 7, Ux) is equivlent with x ln x s By setting Uc) = E[Ux)], where c is the certinty equivlent, we obtin U x)c x) U x) vrx) 2 5

6 so tht c x + U x) 2U x) vrx). 0. Recll tht W = W 0 + r f ) + r r f ) nd η = + From the previous result on d dw 0, we hve d dw 0 )W 0. η = + W 0 + r f )E[u W) r r f )] + E[u W) r r f ) 2 ] E[ u W) r r f ) 2 ] = + E[u W){W 0 + r f ) + r r f )} r r f )] E[ u W) r r f ) 2 ] = + E[u W) W r r f )] E[ u W) r r f ) 2 ] = + E[R R W)u W) r r f )] E[u W) r r f ) 2. ] Since u W) < 0 for concve utility function, we hve sign η ) = signe[r R W)u W) r r f )]). Suppose R R W) is n incresing function, then R R W) = R R W 0 + r f ) + r r f )) { RR W 0 + r f )) when r r f < R R W 0 + r f )) when r < r f. By the rule of conditionl probbility, we hve E[R R W)u W) r r f )] = E[R R W)u W) r r f ) r r f 0]Prob r r f 0) + E[R R W)u W) r r f ) r r f < 0]Prob r r f < 0). Consider E[R R W)u W) r r f ) r r f ) < 0], since u W) > 0 nd R R W) > 0, we hve so tht R R W) r r f ) > R R W 0 + r f )) r r f ) for r r f < 0 E[R R W)u W)r r f )] > R R W 0 + r f ))E[u W) r r f )] = 0 so tht η <. 6

7 . ) Outcomes %) F A F A F B F B F C F C b) F A > F B > F C, outcome; F A F B F C, outcome. X geo A) = = X geo B) = = X geo C) = = Hence, C is preferred to B nd A, nd B is preferred to A, ccording to geometric men. 2. Fx) domintes Gx) by the 3 th order stochstic dominnce TSD) if i) nd ii) Fy)dy dt Ft)dt Gt) dt Gy)dy dt x [,b] Now we show tht Fx) domintes Gx) iff ux)dfx) ux) dgx) c c ) for ll utility functions with u x) > 0,u x) < 0 nd u x) > 0 for ll x C, where C is the set of ll possible outcomes. Proof Let nd b be the smllest nd lrgest vlues F nd G cn tke on = ux)dfx) Gx)) = ux)[fx) Gx)] = u x) + b u x)[fx) Gx)]dx u x)[fx) Gx)]dx F) = G) = 0, Fb) = Gb) = ) x u x) [Fy) Gy)]dy x b [Fy) Gy)]dy dx 7

8 = u b) u x) [Fy) Gy)]dy + x u x) y t u x) [Fy) Gy)]dy dx = u x) x [Fy) Gy)]dy dt dx. [Fy) Gy)]dy dx [Fy) Gy)]dy dt b By i), we hve Here we ssume tht both u x) y [Fy) Gy)]dy dx 0. Fy)dy dt nd Gy)dy dt re continuous function of x, otherwise *) holds only t discontinuous point, since u x) < 0 nd u x) > 0. Also, by ii) nd u x) > 0, Hence TSD *). u b) [Fy) Gy)]dy 0. For the converse, we need to extend the clss of utility functions such tht u x) 0,u x) 0 nd u x) 0 with strict inequlity being held t lest for some intervl. Assume the contrry, suppose there is n intervl [c,d] over which Fy)dy dt > Gy) dy dt. ) Choose ux) such tht u x) = { α > 0, x [c,d] 0, otherwise. Then = d u x) α c [Fy) Gy)]dy dt dx [Fy) Gy)]dy dt dx < 0. We cn choose to be α sufficiently lrge nd pproprite vlues of u b) nd u b) so tht This leds to contrdiction. 8 ur)d[fx) Gx)] < 0.

9 Assume tht condition ii) of TSD breks down, tht is, Fx)dx > Gx) dx. Choose ux) so tht u b) = β > 0. Pick β to be lrge enough so tht to mke u b) [Fy) Gy)]dy 0 ux)d[fx) Gx)] < 0, contrdiction. Hence, TSD ). Combining both results, we conclude TSD ) for ll utility functions such tht u x) 0,u x) < 0 nd u x) > 0 with strict inequlity being hold t lest for some intervl. 9

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