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2. Consider the single-period market model $\mathcal{M}=\left(B, S^{1}, S^{2}\r...
Aug 5, 2024
Solution by Steps
step 1
Identify the given data: The interest rate is r=110r = \frac{1}{10}, initial stock prices are S01=1S_0^1 = 1 and S02=2S_0^2 = 2, and the stock prices at time t=1t = 1 are given in the table
step 2
Calculate the discounted stock prices at time t=1t = 1: S~11=S111+r={12101.1,11.1,7101.1}={1211,1011,711} \tilde{S}_1^1 = \frac{S_1^1}{1 + r} = \left\{ \frac{\frac{12}{10}}{1.1}, \frac{1}{1.1}, \frac{\frac{7}{10}}{1.1} \right\} = \left\{ \frac{12}{11}, \frac{10}{11}, \frac{7}{11} \right\} S~12=S121+r={11.1,5101.1,16101.1}={1011,511,1611} \tilde{S}_1^2 = \frac{S_1^2}{1 + r} = \left\{ \frac{1}{1.1}, \frac{\frac{5}{10}}{1.1}, \frac{\frac{16}{10}}{1.1} \right\} = \left\{ \frac{10}{11}, \frac{5}{11}, \frac{16}{11} \right\}
step 3
Check for arbitrage opportunity: An arbitrage opportunity exists if there is a portfolio (x,y)(x, y) such that the initial cost is zero and the final value is non-negative in all states and positive in at least one state
step 4
Set up the system of inequalities for the portfolio (x,y)(x, y): x⋅1211+y⋅1011≥0 x \cdot \frac{12}{11} + y \cdot \frac{10}{11} \geq 0 x⋅1011+y⋅511≥0 x \cdot \frac{10}{11} + y \cdot \frac{5}{11} \geq 0 x⋅711+y⋅1611≥0 x \cdot \frac{7}{11} + y \cdot \frac{16}{11} \geq 0 x⋅1+y⋅2=0 x \cdot 1 + y \cdot 2 = 0
step 5
Solve the system of inequalities: From the equation x+2y=0x + 2y = 0, we get x=−2yx = -2y. Substitute x=−2yx = -2y into the inequalities: −2y⋅1211+y⋅1011≥0  ⟹  −2411y+1011y≥0  ⟹  −1411y≥0  ⟹  y≤0 -2y \cdot \frac{12}{11} + y \cdot \frac{10}{11} \geq 0 \implies -\frac{24}{11}y + \frac{10}{11}y \geq 0 \implies -\frac{14}{11}y \geq 0 \implies y \leq 0 −2y⋅1011+y⋅511≥0  ⟹  −2011y+511y≥0  ⟹  −1511y≥0  ⟹  y≤0 -2y \cdot \frac{10}{11} + y \cdot \frac{5}{11} \geq 0 \implies -\frac{20}{11}y + \frac{5}{11}y \geq 0 \implies -\frac{15}{11}y \geq 0 \implies y \leq 0 −2y⋅711+y⋅1611≥0  ⟹  −1411y+1611y≥0  ⟹  211y≥0  ⟹  y≥0 -2y \cdot \frac{7}{11} + y \cdot \frac{16}{11} \geq 0 \implies -\frac{14}{11}y + \frac{16}{11}y \geq 0 \implies \frac{2}{11}y \geq 0 \implies y \geq 0 Since y≤0y \leq 0 and y≥0y \geq 0, we conclude y=0y = 0. Thus, x=0x = 0
step 6
Conclusion: Since x=0x = 0 and y=0y = 0 is the only solution, there is no arbitrage opportunity in this market model
Answer
There is no arbitrage opportunity in this market model.
Key Concept
Arbitrage Opportunity
Explanation
An arbitrage opportunity exists if a portfolio can be constructed with zero initial cost and non-negative final value in all states, with a positive value in at least one state. In this case, no such portfolio exists.
Solution by Steps
step 1
Identify the given data: The interest rate is r=110r = \frac{1}{10}, initial stock prices are S01=1S_0^1 = 1 and S02=2S_0^2 = 2, and the stock prices at time t=1t = 1 are given in the table
step 2
Calculate the discounted stock prices at time t=1t = 1: S~11=S111+r={12101.1,11.1,7101.1}={1211,1011,711} \tilde{S}_1^1 = \frac{S_1^1}{1 + r} = \left\{ \frac{\frac{12}{10}}{1.1}, \frac{1}{1.1}, \frac{\frac{7}{10}}{1.1} \right\} = \left\{ \frac{12}{11}, \frac{10}{11}, \frac{7}{11} \right\} S~12=S121+r={11.1,5101.1,16101.1}={1011,511,1611} \tilde{S}_1^2 = \frac{S_1^2}{1 + r} = \left\{ \frac{1}{1.1}, \frac{\frac{5}{10}}{1.1}, \frac{\frac{16}{10}}{1.1} \right\} = \left\{ \frac{10}{11}, \frac{5}{11}, \frac{16}{11} \right\}
step 3
Check for arbitrage opportunity: An arbitrage opportunity exists if there is a portfolio (x,y)(x, y) such that xS01+yS02=0xS_0^1 + yS_0^2 = 0 and xS~11(ωi)+yS~12(ωi)≥0x\tilde{S}_1^1(\omega_i) + y\tilde{S}_1^2(\omega_i) \geq 0 for all ii with at least one strict inequality
step 4
Set up the system of equations for no initial cost: x⋅1+y⋅2=0  ⟹  x+2y=0  ⟹  x=−2y x \cdot 1 + y \cdot 2 = 0 \implies x + 2y = 0 \implies x = -2y Substitute x=−2yx = -2y into the inequalities: −2y⋅1211+y⋅1011≥0  ⟹  −24y+10y≥0  ⟹  −14y≥0  ⟹  y≤0 -2y \cdot \frac{12}{11} + y \cdot \frac{10}{11} \geq 0 \implies -24y + 10y \geq 0 \implies -14y \geq 0 \implies y \leq 0 −2y⋅1011+y⋅511≥0  ⟹  −20y+5y≥0  ⟹  −15y≥0  ⟹  y≤0 -2y \cdot \frac{10}{11} + y \cdot \frac{5}{11} \geq 0 \implies -20y + 5y \geq 0 \implies -15y \geq 0 \implies y \leq 0 −2y⋅711+y⋅1611≥0  ⟹  −14y+16y≥0  ⟹  2y≥0  ⟹  y≥0 -2y \cdot \frac{7}{11} + y \cdot \frac{16}{11} \geq 0 \implies -14y + 16y \geq 0 \implies 2y \geq 0 \implies y \geq 0 From the inequalities, y≤0y \leq 0 and y≥0y \geq 0 imply y=0y = 0. Therefore, x=0x = 0
step 5
Conclusion: Since x=0x = 0 and y=0y = 0, there is no portfolio that satisfies the conditions for arbitrage
Answer
There is no arbitrage opportunity in this market model.
Key Concept
Fundamental Theorem of Asset Pricing (FTAP)
Explanation
The FTAP states that a market model admits no arbitrage if and only if there exists a risk-neutral measure. By checking the conditions for arbitrage, we determined that no such opportunity exists in this model.
Solution by Steps
step 1
Identify the common elements in arrays AA and BB. The common elements are {a,b}\{a, b\}
step 2
Find all pairs of indices (i,j)(i, j) such that A[i]=B[j]A[i] = B[j]. The pairs are: (1,5)(1, 5), (2,1)(2, 1), (3,5)(3, 5), (4,3)(4, 3), (4,4)(4, 4), (6,3)(6, 3), (6,4)(6, 4)
step 3
From these pairs, extract three common subsequences of length 2: (2,1),(4,3)(2, 1), (4, 3); (2,1),(6,3)(2, 1), (6, 3); (3,5),(4,3)(3, 5), (4, 3)
step 4
Identify the longest common subsequences. The longest common subsequences are: (2,1),(4,3)(2, 1), (4, 3); (2,1),(6,3)(2, 1), (6, 3); (3,5),(4,3)(3, 5), (4, 3)
Answer
The three common subsequences of length 2 are (2,1),(4,3)(2, 1), (4, 3); (2,1),(6,3)(2, 1), (6, 3); (3,5),(4,3)(3, 5), (4, 3). The longest common subsequences are the same as the common subsequences of length 2.
Key Concept
Common subsequences
Explanation
A common subsequence is a sequence that appears in both arrays in the same order. The longest common subsequence is the longest sequence that can be found in both arrays.
Solution by Steps
step 1
Define the problem: Given two arrays AA and BB, find the longest common subsequence (LCS)
step 2
Initialize a 2D array LL of size (n+1)×(m+1)(n+1) \times (m+1) where nn and mm are the lengths of AA and BB respectively
step 3
Set L[i][0]=0L[i][0] = 0 for all ii and L[0][j]=0L[0][j] = 0 for all jj
step 4
For each ii from 1 to nn and each jj from 1 to mm, do the following:
step 5
If A[i−1]=B[j−1]A[i-1] = B[j-1], then set L[i][j]=L[i−1][j−1]+1L[i][j] = L[i-1][j-1] + 1
step 6
Otherwise, set L[i][j]=max⁡(L[i−1][j],L[i][j−1])L[i][j] = \max(L[i-1][j], L[i][j-1])
step 7
The length of the LCS is L[n][m]L[n][m]
Answer
The polynomial-time algorithm for the LCS problem is based on dynamic programming. The length of the LCS is found in L[n][m]L[n][m].
Key Concept
Dynamic programming
Explanation
Dynamic programming is used to solve problems by breaking them down into simpler subproblems and storing the results of these subproblems to avoid redundant computations.
Solution by Steps
step 1
Initialize the 2D array LL for arrays A=[a,b,b,a,c,a]A = [a, b, b, a, c, a] and B=[b,d,a,a,b]B = [b, d, a, a, b]
step 2
Fill the array LL using the algorithm from part (b)
step 3
Trace back from L[n][m]L[n][m] to find the LCS
step 4
The LCS for the given example is [a,a,b][a, a, b]
Answer
The LCS for the arrays A=[a,b,b,a,c,a]A = [a, b, b, a, c, a] and B=[b,d,a,a,b]B = [b, d, a, a, b] is [a,a,b][a, a, b].
Key Concept
Traceback in dynamic programming
Explanation
Traceback is used to reconstruct the solution from the computed values in the dynamic programming table.
Solution by Steps
step 1
Define the problem: Given an undirected graph GG and a positive integer kk, determine if there is a strongly independent set of size kk
step 2
Show that the problem is in NP: A non-deterministic algorithm can guess a set SS of size kk and verify in polynomial time that SS is strongly independent
step 3
Reduce a known NP-complete problem to the STRONGLY-INDEPENDENT-SET problem
step 4
Use the 3-SAT problem for the reduction
step 5
Construct a graph GG such that there is a strongly independent set of size kk if and only if the 3-SAT instance is satisfiable
step 6
Show that the reduction can be done in polynomial time
step 7
Conclude that the STRONGLY-INDEPENDENT-SET problem is NP-complete
Answer
The STRONGLY-INDEPENDENT-SET problem is NP-complete.
Key Concept
NP-completeness
Explanation
A problem is NP-complete if it is in NP and every problem in NP can be reduced to it in polynomial time.
Solution by Steps
step 1
Calculate the initial market value of the portfolio. The market value of 10 shares of firm XX and 10 shares of firm YY is given by: Market Value=10×£10+10×£10=£200 \text{Market Value} = 10 \times £10 + 10 \times £10 = £200
step 2
Define the weights of the securities in the portfolio. Let wXw_X and wYw_Y be the weights of securities XX and YY respectively. Since the total market value is £200, we have: wX+wY=1 w_X + w_Y = 1
step 3
Calculate the portfolio variance. The portfolio variance is given by: σp2=wX2σX2+wY2σY2+2wXwYσXσYρXY \sigma_p^2 = w_X^2 \sigma_X^2 + w_Y^2 \sigma_Y^2 + 2 w_X w_Y \sigma_X \sigma_Y \rho_{XY} where σX=20%\sigma_X = 20\%, σY=15%\sigma_Y = 15\%, and ρXY=−0.5\rho_{XY} = -0.5
step 4
Substitute the values into the portfolio variance formula: σp2=wX2(0.2)2+wY2(0.15)2+2wXwY(0.2)(0.15)(−0.5) \sigma_p^2 = w_X^2 (0.2)^2 + w_Y^2 (0.15)^2 + 2 w_X w_Y (0.2)(0.15)(-0.5) σp2=0.04wX2+0.0225wY2−0.03wXwY \sigma_p^2 = 0.04 w_X^2 + 0.0225 w_Y^2 - 0.03 w_X w_Y
step 5
Express wYw_Y in terms of wXw_X using the constraint wX+wY=1w_X + w_Y = 1: wY=1−wX w_Y = 1 - w_X Substitute wYw_Y into the portfolio variance formula: σp2=0.04wX2+0.0225(1−wX)2−0.03wX(1−wX) \sigma_p^2 = 0.04 w_X^2 + 0.0225 (1 - w_X)^2 - 0.03 w_X (1 - w_X)
step 6
Simplify the expression: σp2=0.04wX2+0.0225(1−2wX+wX2)−0.03wX+0.03wX2 \sigma_p^2 = 0.04 w_X^2 + 0.0225 (1 - 2w_X + w_X^2) - 0.03 w_X + 0.03 w_X^2 σp2=0.04wX2+0.0225−0.045wX+0.0225wX2−0.03wX+0.03wX2 \sigma_p^2 = 0.04 w_X^2 + 0.0225 - 0.045 w_X + 0.0225 w_X^2 - 0.03 w_X + 0.03 w_X^2 σp2=0.0925wX2−0.075wX+0.0225 \sigma_p^2 = 0.0925 w_X^2 - 0.075 w_X + 0.0225
step 7
Minimize the portfolio variance by taking the derivative with respect to wXw_X and setting it to zero: dσp2dwX=0.185wX−0.075=0 \frac{d\sigma_p^2}{dw_X} = 0.185 w_X - 0.075 = 0 wX=0.0750.185≈0.405 w_X = \frac{0.075}{0.185} \approx 0.405 wY=1−wX≈0.595 w_Y = 1 - w_X \approx 0.595
step 8
Calculate the number of shares to hold for each security. The total market value is £200, so: Value of X=0.405×£200=£81 \text{Value of } X = 0.405 \times £200 = £81 Value of Y=0.595×£200=£119 \text{Value of } Y = 0.595 \times £200 = £119 Since each share is £10: Number of shares of X=£81£10=8.1 \text{Number of shares of } X = \frac{£81}{£10} = 8.1 Number of shares of Y=£119£10=11.9 \text{Number of shares of } Y = \frac{£119}{£10} = 11.9
Answer
The portfolio should be rebalanced to hold approximately 8 shares of firm XX and 12 shares of firm YY.
Key Concept
Portfolio Rebalancing
Explanation
To minimize the portfolio risk, the weights of the securities are adjusted based on their variances and the correlation between them.
Solution by Steps
step 1
To find the market price of Bond A when the yield to maturity (YTM) changes, we use the present value formula for bonds. The formula is: P=∑t=1nC(1+r)t+F(1+r)n P = \sum_{t=1}^{n} \frac{C}{(1 + r)^t} + \frac{F}{(1 + r)^n} where P P is the price, C C is the annual coupon payment, r r is the yield to maturity, F F is the face value, and n n is the number of years to maturity
step 2
For Bond A with a face value of £1,000, a coupon rate of 2%, and a term to maturity of 3 years, the annual coupon payment C C is: C=0.02×1000=£20 C = 0.02 \times 1000 = £20 When the YTM is 10%, the price P P is: P=20(1+0.10)1+20(1+0.10)2+20(1+0.10)3+1000(1+0.10)3 P = \frac{20}{(1 + 0.10)^1} + \frac{20}{(1 + 0.10)^2} + \frac{20}{(1 + 0.10)^3} + \frac{1000}{(1 + 0.10)^3} P=201.10+201.21+201.331+10001.331 P = \frac{20}{1.10} + \frac{20}{1.21} + \frac{20}{1.331} + \frac{1000}{1.331} P=18.18+16.53+15.03+751.31 P = 18.18 + 16.53 + 15.03 + 751.31 P=£801.05 P = £801.05
step 3
When the YTM changes to 5%, the price P P is: P=20(1+0.05)1+20(1+0.05)2+20(1+0.05)3+1000(1+0.05)3 P = \frac{20}{(1 + 0.05)^1} + \frac{20}{(1 + 0.05)^2} + \frac{20}{(1 + 0.05)^3} + \frac{1000}{(1 + 0.05)^3} P=201.05+201.1025+201.157625+10001.157625 P = \frac{20}{1.05} + \frac{20}{1.1025} + \frac{20}{1.157625} + \frac{1000}{1.157625} P=19.05+18.14+17.28+863.84 P = 19.05 + 18.14 + 17.28 + 863.84 P=£918.31 P = £918.31
step 4
When the YTM changes to 15%, the price P P is: P=20(1+0.15)1+20(1+0.15)2+20(1+0.15)3+1000(1+0.15)3 P = \frac{20}{(1 + 0.15)^1} + \frac{20}{(1 + 0.15)^2} + \frac{20}{(1 + 0.15)^3} + \frac{1000}{(1 + 0.15)^3} P=201.15+201.3225+201.520875+10001.520875 P = \frac{20}{1.15} + \frac{20}{1.3225} + \frac{20}{1.520875} + \frac{1000}{1.520875} P=17.39+15.12+13.15+657.52 P = 17.39 + 15.12 + 13.15 + 657.52 P=£703.18 P = £703.18
step 5
For Bond B with a face value of £1,000, a coupon rate of 5%, and a term to maturity of 4 years, the annual coupon payment C C is: C=0.05×1000=£50 C = 0.05 \times 1000 = £50 When the YTM is 10%, the price P P is: P=50(1+0.10)1+50(1+0.10)2+50(1+0.10)3+50(1+0.10)4+1000(1+0.10)4 P = \frac{50}{(1 + 0.10)^1} + \frac{50}{(1 + 0.10)^2} + \frac{50}{(1 + 0.10)^3} + \frac{50}{(1 + 0.10)^4} + \frac{1000}{(1 + 0.10)^4} P=501.10+501.21+501.331+501.4641+10001.4641 P = \frac{50}{1.10} + \frac{50}{1.21} + \frac{50}{1.331} + \frac{50}{1.4641} + \frac{1000}{1.4641} P=45.45+41.32+37.57+34.18+683.01 P = 45.45 + 41.32 + 37.57 + 34.18 + 683.01 P=£841.53 P = £841.53
step 6
When the YTM changes to 5%, the price P P is: P=50(1+0.05)1+50(1+0.05)2+50(1+0.05)3+50(1+0.05)4+1000(1+0.05)4 P = \frac{50}{(1 + 0.05)^1} + \frac{50}{(1 + 0.05)^2} + \frac{50}{(1 + 0.05)^3} + \frac{50}{(1 + 0.05)^4} + \frac{1000}{(1 + 0.05)^4} P=501.05+501.1025+501.157625+501.21550625+10001.21550625 P = \frac{50}{1.05} + \frac{50}{1.1025} + \frac{50}{1.157625} + \frac{50}{1.21550625} + \frac{1000}{1.21550625} P=47.62+45.35+43.19+41.13+822.70 P = 47.62 + 45.35 + 43.19 + 41.13 + 822.70 P=£999.99 P = £999.99
step 7
When the YTM changes to 15%, the price P P is: P=50(1+0.15)1+50(1+0.15)2+50(1+0.15)3+50(1+0.15)4+1000(1+0.15)4 P = \frac{50}{(1 + 0.15)^1} + \frac{50}{(1 + 0.15)^2} + \frac{50}{(1 + 0.15)^3} + \frac{50}{(1 + 0.15)^4} + \frac{1000}{(1 + 0.15)^4} P=501.15+501.3225+501.520875+501.74900625+10001.74900625 P = \frac{50}{1.15} + \frac{50}{1.3225} + \frac{50}{1.520875} + \frac{50}{1.74900625} + \frac{1000}{1.74900625} P=43.48+37.81+32.87+28.61+571.93 P = 43.48 + 37.81 + 32.87 + 28.61 + 571.93 P=£714.70 P = £714.70
Answer
The market prices of Bond A and Bond B when the YTM changes are as follows: - Bond A: £801.05 (10%), £918.31 (5%), £703.18 (15%) - Bond B: £841.53 (10%), £999.99 (5%), £714.70 (15%)
Key Concept
Present Value of Bonds
Explanation
The present value of a bond is calculated by discounting the future cash flows (coupon payments and face value) to the present using the yield to maturity. Changes in the yield to maturity affect the discount rate, thus changing the bond's market price.
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