Chapter012

Chapter012 - The Theory of Interest - Solutions Manual...

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The Theory of Interest - Solutions Manual 141 Chapter 12 1. () [] 1 1 1 1 1 EE 1 E 1 from independence 1. n t t n t t n an i i i = = ⎡⎤ + ⎢⎥ ⎣⎦ + =+ 2. 1 11 1 1 1 E 1 from independence nt s n ts s n t ni t ai i ia == = Π + =ΣΠ + =Σ + = 3. ( a ) Year 1: 8% given. Year 2: .5 .07 .09 .08, or 8%. Year 3: .25 .06 2 .08 .10 .08, or 8%. += ++ = ( b ) 22 2 2 2 Year 1: 0, no variance. Year 2: .5 .07 .08 .09 .08 .0001 .0001 .01. Year 3: .25 .06 .08 2 .08 .08 .10 .08 .0002 σ = =− + = + + = .0002 .01 2 ( c ) 1000 1.08 1.09 1.10 $1294.92. = ( d ) 1000 1.08 1.07 1.06 $1224.94. = ( e ) 3 1000 1.08 $1259.71. = ( f ) ( ) ( ) ( ) ( )( ) [ ] .25 1000 1.08 1.09 1.10 1.08 1.09 1.08 1.08 1.07 1.08 1.08 1.07 1.06 .25 1294.92 1271.38 1248.05 1224.94 $1259.82 + + + =
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The Theory of Interest Chapter 12 142 ( g ) () 22 2 .25 1294.92 1259.82 1271.38 1259.82 1248.05 1259.82 1224.94 1259.82 2720.79 2720.79 52.16. σ =− + +− = == 4. ( a ) .09 1 .07 .09 .07 1 11 E1 .09 .07 1 1 ln 1 .925952. .09 .07 Then set 1 .925952 and solve .07997. t id t t t ii ⎡⎤ += ⎣⎦ −+ =+ = = ( b ) We have () ( ) 3 1 3 1.07997 .79390. a (c) .09 2 2 .07 .09 .07 1 .09 .07 1 .857412. .09 .07 1 Then set 1 .857412 and solve .16630. t t t t kk + =⋅ = ⎢⎥ = ( d ) Applying formula (12.10), we have [ ] 36 1 Var 3 .857412 .925952 .0000549 a =−= and the standard deviation is .0000549 .00735. = 5. ( b ) Applying formula (12.11), we have 3 3 3 .07997 E 2.5772. i aaa = ( d ) Applying formula (12.14), we have 2 21 2 33 3 3 2 2 Var .857412 .925952 2 .857412 2.2229 2.5772 2.5772 .857412 .925952 .857412 .925952 .005444 aa a ki i mm m a + =−− −− + = and the standard deviation is .005444 .0735. =
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The Theory of Interest Chapter 12 143 6. The random variable () 2 /2 t i will be normal with 3% μ = and .25%. σ = ( a ) Applying formula (12.1), we have [ ] 4 E 100 4 100 1.03 112.55. a == Applying formula (12.3), we have {} 4 8 22 4 4 28 Var 100 10,000 1 2 1 10,000 1 2 .03 .03 .0025 1.03 119.828 ai i s i =+ + + + ⎡⎤ ⎣⎦ + + = and the standard deviation is 119.828 10.95 = . ( b ) Applying formula (12.5), we have 44 . 0 3 E 100 100 430.91. ss ±± Applying formula (12.8), we have 1 2 2 1.03 1 2 .03 .03 .0025 1.0634 s s m m = + + = and [] ( )( ) 2 4 1.0634 1.03 2 1.0634 Var 100 10,000 4.67549 4.3091 4.3091 1.0634 1.03 1.0634 1.03 944.929 s + =− ⎢⎥ −− = and the standard deviation is 944.929 30.74. = 7. ( a ) 11 EE1 1 . nn n i s ++ = ⎡⎤ ⎡ ⎣⎦ ⎣ ( b ) Var Var 1 Var n s = ⎡ ⎤ ⎣ ⎦ . ( c ) EE 1 1 n i aa a = + . ( d ) Var Var 1 Var n a = . 8. We know that 1 i + is lognormal with .06 = and 2 .01. = From the solution to Example 12.3(1), we have .067159 i = and then ( ) ( ) 2 2 2 .06 .01 .01 .13 .01 1 .011445. se e e e ee μσ = =
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The Theory of Interest Chapter 12 144 We then apply formula (12.4 a ) to obtain () [] ( ) 2 22 5 21 0 Var 1 2 1 1 2 .067159 .067159 .011445 1.067159 .09821 n n an i i s i =+ + + −+ ⎡⎤ =+ + + ⎣⎦ = and the standard deviation .09821 .3134 == agreeing with the other approach.
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This note was uploaded on 11/18/2010 for the course MATH 320 taught by Professor Dr.k during the Spring '10 term at Nevada.

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Chapter012 - The Theory of Interest - Solutions Manual...

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