Investment Science Chapter 3 Sols
Investment Science Chapter 3
Dr. James A. Tzitzouris
3.1 Use A= 1− rP
1 (1+r)n
with r = 7/12 = 0.58%, P = $25, 000, and n = 7 × 12 = 84, to obtain A = $377.32. 3.2 Observe that since the net present value of X is P , the cash flow stream arrived at by cycling X is equivalent to one obtained by receiving payment of P every n + 1 periods (since k = 0, . . . , n). Let d = 1/(1 + r). Then
∞
P∞ = P k=0 (dn+1 )k .
Solving explicitly for the geometric series, we have that P∞ = Denoting the annual worth by A, we must have A= rP , 1 − dn P . 1 − dn+1
so that solving for P as a function of P∞ and substituting the result into the equation for A, we arrive at 1 − dn+1 A=r P∞ . 1 − dn 1
Investment Science Chapter 4 Solutions to Suggested Problems
Dr. James A. Tzitzouris
4.1 (One forward rate) f1,2 = (1 + s2 )2 1.0692 −1= − 1 = 7.5% (1 + s1 ) 1.063
4.2 (Spot Update) Use f1,k . Hence, for example, f1,k . All values are f1,2 f1,3 f1,4 f1,5 f1,6 5.60 5.90 6.07 6.25 6.32 (1.061)6 = 1.05
1/5
(1 + sk )k = 1 + s1
1/(k−1)
−1
− 1 = 6.32%
1
4.3 (Construction of a zero) Use a combination of the two bonds: let x be the number of 9% bonds, and y teh number of 7% bonds. Select x and y to satisfy 9x + 7y = 0, x + y = 1. The first equation makes the net coupon zero. The second makes the face value equal to 100. These equations give x = −3.5, and y = 4.5, respectively. The price is P = −3.5 × 101.00 + 4.5 × 93.20 = 65.90. 4.5 (Instantaneous rates) (a) es(t2 )t2 = es(t1 )t1 eft1 ,t2 (t2 −t1 ) =⇒ ft1 ,t2 = (b) r(t) = limt→t1 (c) We have d(ln x(t)) = r(t)dt, = s(t)dt + s (t)dt, = d[s(t)t]. Hence, ln x(t) = ln x(0) + s(t)t, and finally that x(t) = x(0)es(t)t . This is in agreement with the invariance property of expectation dynamics. Investing continuously give the same result as investing in a bond that matures at time t. 4.6 (Discount conversion) s(t)t−s(t1 )t1 t−t1
s(t2 )t2 −s(t1 )t1 t2 −t1
=
d[s(t)t] dt
= s(t) + s (t)
2
The
Dr. James A. Tzitzouris
3.1 Use A= 1− rP
1 (1+r)n
with r = 7/12 = 0.58%, P = $25, 000, and n = 7 × 12 = 84, to obtain A = $377.32. 3.2 Observe that since the net present value of X is P , the cash flow stream arrived at by cycling X is equivalent to one obtained by receiving payment of P every n + 1 periods (since k = 0, . . . , n). Let d = 1/(1 + r). Then
∞
P∞ = P k=0 (dn+1 )k .
Solving explicitly for the geometric series, we have that P∞ = Denoting the annual worth by A, we must have A= rP , 1 − dn P . 1 − dn+1
so that solving for P as a function of P∞ and substituting the result into the equation for A, we arrive at 1 − dn+1 A=r P∞ . 1 − dn 1
Investment Science Chapter 4 Solutions to Suggested Problems
Dr. James A. Tzitzouris
4.1 (One forward rate) f1,2 = (1 + s2 )2 1.0692 −1= − 1 = 7.5% (1 + s1 ) 1.063
4.2 (Spot Update) Use f1,k . Hence, for example, f1,k . All values are f1,2 f1,3 f1,4 f1,5 f1,6 5.60 5.90 6.07 6.25 6.32 (1.061)6 = 1.05
1/5
(1 + sk )k = 1 + s1
1/(k−1)
−1
− 1 = 6.32%
1
4.3 (Construction of a zero) Use a combination of the two bonds: let x be the number of 9% bonds, and y teh number of 7% bonds. Select x and y to satisfy 9x + 7y = 0, x + y = 1. The first equation makes the net coupon zero. The second makes the face value equal to 100. These equations give x = −3.5, and y = 4.5, respectively. The price is P = −3.5 × 101.00 + 4.5 × 93.20 = 65.90. 4.5 (Instantaneous rates) (a) es(t2 )t2 = es(t1 )t1 eft1 ,t2 (t2 −t1 ) =⇒ ft1 ,t2 = (b) r(t) = limt→t1 (c) We have d(ln x(t)) = r(t)dt, = s(t)dt + s (t)dt, = d[s(t)t]. Hence, ln x(t) = ln x(0) + s(t)t, and finally that x(t) = x(0)es(t)t . This is in agreement with the invariance property of expectation dynamics. Investing continuously give the same result as investing in a bond that matures at time t. 4.6 (Discount conversion) s(t)t−s(t1 )t1 t−t1
s(t2 )t2 −s(t1 )t1 t2 −t1
=
d[s(t)t] dt
= s(t) + s (t)
2
The