Solution Set 2 Foundations of Finance1Problem Set 2 SolutionI. Buying Stock on Margin: BKM, Chapter 3, Question 17 (for part b. assume that intereston the broker’s loan does not accrue until the end of the year).Part a.A. One Answer.1. You have $5000 today.Assets LiabilitiesCash 5000 Net Worth 5000Total Asset 5000 Total Liab & Net W. 50002. You borrow $5000 today.Assets LiabilitiesCash 10000 Loan 5000Net Worth 5000Total Asset 10000 Total Liab & Net W. 100003. You invest the $10000 cash in 400 shares of Telecom. Percent Margin =5000/10000 = 50%. Assets LiabilitiesTelecom (400@$25) 10000 Loan 5000Net Worth 5000Total Asset 10000 Total Liab & Net W. 100004. At the end of the period, the value of Telecom shares grows 10% to$27.50 while the amount outstanding on the loan grows 8%. PercentMargin = {11000-5400}/11000 = 50.909%.Solution Set 2 Foundations of Finance2Assets LiabilitiesTelecom(400@$27.5) 11000 Loan (5000 x 1.08) 5400Net Worth 5600Total Asset 11000 Total Liab & Net W. 110005. At the end of the period, close out the position. Assets LiabilitiesCash 5600 Net Worth 5600Total Asset 5600 Total Liab & Net W. 56006. Your return is:{Net Worth at End - Net Worth at Start}/Net Worth at Start = {5600 -5000}/5000 = 12%.B. Second Answer.At start, $10000 is invested in Telecom by buying 400 shares.Profit = Value of 400 Telecom shares at end - Loan Outstanding at end - Net Worth at Start= 400 x($25 x 1.1) - ($5000 x 1.08) -$5000= $11000 - $5400 - $5000 = $600.Return = Profit/{Net Worth at Start} = $600/$5000 = 12%. Part b. Know Margin = (Value of Stock Position - Loan Outstanding)/ Value of Stock Position.So0.30 = (400 x Price of Telecom - $5000)/(400 x Price of Telecom) 0.30 (400 x Price of Telecom) = (400 x Price of Telecom - $5000)280 x Price of Telecom = $5000 Price of Telecom = $17.85.II. Short-selling when the Stock Pays a Dividend: BKM, Chapter 3, Question 20 (assume theinterest rate on any required margin for the short position is zero). Assume that there is the margin requirement is 50%. For short sales, Margin = Net Worth in Broker’s Account / Value of Stock. So need to have a net balance in broker’s account of 0.5 x (100 x $14) =$700. Soneed to deposit $700 into the broker’s account on 1/1.A. One Answer.1. On 1/1, you borrow 100 shares of Zenith at $14.00 per share. The loan isdenominated in shares of IBM not in dollars.Solution Set 2 Foundations of Finance3Assets Liabilities100 sh @ 14 1400 Loan (100 sh @ 14) 1400Cash 700 Net Worth 700Total Asset 2100 Total Liab & Net W. 21002. You sell the borrowed shares at $14 per share on 1/1.Assets LiabilitiesCash 2100 Loan (100 sh @ 14) 1400less Commission(100 sh @ 0.5)-50 Net Worth 650Total Asset 2050 Total Liab & Net W. 20503. On the 3/1, Zenith pays a $2 dividend per share which you have to pay.(Assume that Zenith’s price is still $14 per share.)Assets LiabilitiesCash 2050 Loan (100 sh @ 14) 1400less Dividend (100 sh @ 2)-200 Net Worth 450Total Asset 1850 Total Liab & Net W. 18504. On the 4/1, Zenith’s price is $9 per share.. Assets LiabilitiesCash 1850 Loan (100 sh @ 9) 900Net Worth 950Total Asset 1850 Total Liab & Net W. 1850Solution Set 2 Foundations of Finance45. To close out the position on 4/1, take the following steps:(1) purchase 100 shares of Zenith stock.Assets LiabilitiesCash 950 Loan (100 sh @ 9) 900less Commission(100 sh @ 0.5)-50100 sh @ 9 900 Net Worth 900Total Asset 1800 Total Liab & Net W. 1800(2) repay the stock loan.Assets LiabilitiesCash 900 Net Worth 900Total Asset 900 Total Liab & Net W. 900Profit = Net Worth (4/1) - Net Worth (1/1) = $900 - $700 = $200.B. Second Answer.Profit = Proceeds Sale of 100 sh at $14 on 1/1 - Cost of Purchasing 100 sh at $9 on 4/1 - Payment of Dividend of $2 per share on 100 sh - 2 x Commission on 100 sh at $0.50 = $1400 - $900 - $200 -2x $50 = $200.III. Expected Return, Return Standard Deviation, Covariance and Portfolios:State Probability Asset A Asset B Riskless AssetBoom 0.25 24% 14% 7%Normal Growth 0.5 18% 9% 7%Recession 0.25 2% 5% 7%A. What is the expected return on each asset?E[RA] = 0.25 x 24% + 0.5 x 18% + 0.25 x 2% = 15.5%.E[RB] = 0.25 x 14% + 0.5 x 9% + 0.25 x 5% = 9.25%.E[Rf] = Rf = 0.25 x 7% + 0.5 x 7% + 0.25 x 7% = 7%.B. What is the standard deviation of return on each asset?Solution Set 2 Foundations of Finance5First, calculate varianceσ[RA]2 = 0.25 x (24x24) + 0.5 x (18x18) + 0.25 x (2x2) - (15.5x15.5) = 307 - 240.25 = 66.75.σ[RB]2 = 0.25 x (14x14) + 0.5 x (9x9) + 0.25 x (5x5) - (9.25x9.25) = 95.75 - 85.5625 = 10.1875.σ[Rf]2 = 0.25 x (7x7) + 0.5 x (7x7) + 0.25 x (7x7) - (7x7) = 49 - 49 = 0.Then calculate standard deviationσ[RA] = 8.1701%.σ[RA] = 3.1918%.σ[Rf] = 0%.C. What is the correlation and covariance between the returns on1. assets A and B?Covariance:σ[RA,RB] = 0.25 x (24x14) + 0.5 x (18x9) + 0.25 x (2x5) - (15.5x9.25) = 167.5 - 143.375 =24.125.Correlation:ρ[RA,RB]= σ[RA,RB] / {σ[RA] σ[RB]}= 24.125/{8.1701 x 3.1918} = 0.92512. asset A and the riskless asset?Covariance:σ[RA,Rf] = 0.25 x (24x7) + 0.5 x (18x7) + 0.25 x (2x7) - (15.5x7) = 108.5 - 108.5 = 0.Correlation:ρ[RA,Rf]= σ[RA,Rf] / {σ[RA] σ[Rf]} = 0/0 which is not well defined.3. asset B and the riskless asset?Covariance:σ[RB,Rf] = 0.Correlation:ρ[RB,Rf]= σ[RB,Rf] / {σ[RB] σ[Rf]} = 0/0 which is not well defined.Solution Set 2 Foundations of Finance6IV. Using Dividend Yield Information: Suppose the following data is to be used by Ms Q (arisk-averse investor) to form a portfolio that consists of the small firm fund and T-bills.E[RSmall(t)] = 1.369% σ[RSmall(t)] = 8.779%E[DP(start t)] = 4.446% σ[DP(start t)] = 1.513%σ[DP(start t),RSmall(t)] = 1.967where DP(start t) is the dividend yield on the S&P 500 known at the start of month t. RSmall(t) is the return on the small firm fund in month t.A. What is the intercept and slope coefficients from a regression of RSmall(t)(dependent variable) on DP(start t)?Slope: = 1.967/(1.513x1.513) = 0.859φSmall,DP'σ[RSmall(t), DP(start t)]σ[DP(start t)]2.Intercept: = 1.369 - 0.859 x 4.446 = -2.451.µSmall,DP' E[RSmall(t)] & φSmall,DPE[DP(start t)]B. What is the standard deviation of the residual from the regression of RSmall(t) onDP(start t)?Defining eSmall,DP(t) to be the residual from the regression of RSmall(t) on DP(start t), then σ[RSmall(t)]2' φ2Small,DPσ[DP(start t)]2% σ[eSmall,DP(t)]2.So σ[eSmall,DP(t)]2' σ[RSmall(t)]2& φ2Small,DPσ[DP(start t)]2' 8.7792& 0.8592× 1.5132' 75.381and σ[eSmall,DP(t)] = 8.682. C.