Solutions to Extra Problems 5a

Solutions to Extra Problems 5a - 1. R1 = D (1:03) ; R2 = D...

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ACTSC 231 Solutions to Extra Problems 5a 1. R 1 = D (1 : 03) ; R 2 = D (1 : 03) 2 ; R 3 = D (1 : 03) 3 ; ::: If the next dividend was 1 year from now the PV would be D (1 : 03) i : 03 Since the dividend is actually in 6 months time the PV is P ( i ) = D (1 : 03)(1+ i ) 1 = 2 i : 03 d ( i ) = d ln P ( i ) = d ln D (1 : 03)(1+ i ) 1 = 2 i : 03 ± = d ² 1 2 ln (1 + i ) + ln ( D (1 : 03)) ln ( i : 03) ³ = n d i 1+ i d i i : 03 o But 1 + i = e = ) i = e 1 = ) d i = d ´ e 1 µ = e = 1 + i = ) d ( i ) = d i i : 03 = h 1 2 1+ i 1+ i 1+ i i : 03 i = 2 : 03+ i 2( i : 03) and d (8%) = 2 : 03+ : 08 2( : 08 : 03) = 21 : 1 2. a) Mary should put all her money into 2-year zero coupon bonds. I.e. A 2 = 1210 and P A ( i ) = 1210 v 2 (note: P A (10%) = 1000) . This is called absolute matching or dedication. We have P A ( i ) = P L ( i ) for all i . b)
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Solutions to Extra Problems 5a - 1. R1 = D (1:03) ; R2 = D...

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