The iformatio required by the mea-variace approach is substatial whe the umber of assets is large; there are mea values, variaces, ad )/2 covariaces - a total of 2 + )/2 parameters. Sigle-factor model: Suppose that there are assets, idexed by i, with rates of retur r i, i =, 2,...,. Suppose the that there is a sigle factor f, a radom quatity, ad assume that the rates of retur ad the factor are related by r i = a i + b i f + e i, ) for i =,...,. I Equatio ), a i ad b i are costats ad e i is a radom quatity that represets the error of the model. b i is termed the factor loadig that measures the sesitivity of the retur to the factor. For the error we assume Ee i ] = 0 i, because ay o-zero mea could be trasferred to a i. Moreover, we assume that the errors are ucorrelated with f ad with reach other; that is, Cove i, e j ] = Ee i ē i )e j ē j )] = Ee i e j ] = 0 i j Covf, e i ] = Ef f)e i ē i )] = Ef f)e i ] = 0 i. The variaces of the error terms, deoted by σ 2 e i, are assumed kow. If we agree to use a sigle-factor model, the stadard parameters for mea-variace aalysis ca be determied as r i = a i + b i f σ 2 i = b 2 i σ 2 f + σ2 e i σ ij = b i b j σ 2 f, i j b i = Covr i, f] σf 2. The oly parameters to be estimated for this model are a i 's, b i 's, σe 2 i 's, ad f ad σf 2 's - a total of just 3 + 2 parameters. Suppose that there are assets with rates of retur govered by the sigle-factor model ), ad suppose that a portfolio of these assets is costructed with weights w i, with w i =. The the rate of retur r of the portfolio is r = w i r i = w i a i + w i b i f + w i e i, ad deotig a = w ia i, b = w ib i ad e = w ie i, we ca write the above formula as r = a + bf + e. Similarly to the sigle asset case, a ad b are costats ad e is a radom variable. Because Ee i e j ] = 0, the variace of e is ) ] σe 2 = Ee 2 ] = E w i e i w j e j = E wi 2 e 2 i = wi 2 σe 2 i, 2) ad because Covf, e i ] = 0 i Covf, e] = 0, the overall variace of the portfolio is σ 2 = Vara + bf + e] = b 2 Varf] + 2bCovf, e] + Vare] = b 2 σ 2 f + σ2 e. 3)
The CAPM ca be iterpreted as a special case of a sigle-factor model. Suppose we model the excess rates of retur of stocks r i r f where r f is the risk-free rate) with a sigle-factor model, with the factor beig the excess rate of retur of the market r M r f. The the factor model becomes r i r f = α i + β i r M r f ) + e i. Arbitrage Pricig Theory APT) is a alterative theory of asset pricig that ca be derived from the factor model framework. Istead of the strog equilibrium assumptio of the CAPM theory, APT assumes that i) whe returs are certai, ivestors prefer greater retur to lesser retur, ii) the uiverse of assets beig cosidered is large, ad iii) there are o arbitrage opportuities. Specically, APT assumes that all asset rates of retur satisfy a sigle-factor model with o error term, ad hece the ucertaity with a retur is due oly to the ucertaity i the factor f. For assets i ad j, we write r i = a i + b i f r j = a j + b j f. 4) We the select w so that the coeciet of f i this equatio is zero; that is, r = wa i + b i f) + w)a j + b j f) = wa i + w)a j + wb i + w)b j ) f w = }{{} b j b i =0 If there is a risk-free asset r f, the retur of this portfolio must have this same rate, ad eve if there is o explicit risk-free asset, all portfolios costructed this way must have the same rate of retur - otherwise there would be a arbitrage opportuity. We deote this rate by λ 0 ad d that r = wa i + w)a j = λ 0 b ja i b ia j = λ 0 b j b i b j b i λ 0 b j b i ) = a i b j a j b i a j λ 0 b j = a i λ 0 b i = c. The last equatio holds for all i ad j because we could select ay assets i ad j that satisfy 4)), which is why it must be a costat c. We use this iformatio to write a simple formula for the expected rate of retur of asset i as follows: r i = a i + b i f = λ0 + b i c + b i f = λ0 + b i λ, where λ = c+ f is a costat same for all assets i). This statemet ca be geeralized for multiple factors: r i = a i + r i = λ 0 + m b ij f j m b ij λ j, where the value λ j is the price of risk associated with the factor f j, ofte called the factor price. b j
. L8.) A simple portfolio) Someoe who believes that the collectio of all stocks satises a sigle-factor model with the market portfolio servig as the factor gives you iformatio o three stocks which make up a portfolio. See Table.) I additio, you kow that the market portfolio has a expected rate of retur of 2% ad a stadard deviatio of 8%. The risk-free rate is 5%. a) What is the portfolio's expected rate of retur? b) Assumig the factor model is accurate, what is the stadard deviatio of this rate of retur? Table : Simple Portfolio. Stock Beta Stadard deviatio of radom error term Weight i portfolio A.0 7.0% 20% B 0.80 2.3% 50% C.00.0% 30% Solutio: r f = 5% r M = 2% σ M = 8% β i σ ei w i A.0 7.0% 20% B 0.80 2.3% 50% C.00.0% 30% a) The beta of the portfolio is a weighted combiatio of the idividual betas: β p = w A β A + w B β B + w C β C = 0.92 Hece, applyig the CAPM to the portfolio we d r p = r f + β p r M r f ) = 0.05 + 0.920.2 0.05) = 0.44 =.44%. b) Usig formula 2) The variace of the error terms of the portfolio is σe 2 = wi 2 σe 2 i = 0.2 2 0.07 2 + 0.5 2 0.023 2 + 0.3 2 0.0 2 = 0.00033725 The, usig formula 3), the overall variace of the portfolio is σ 2 = b 2 σf 2 + σ2 e = βpσ 2 f 2 + σ2 e = 0.92 2 0.8 2 + 0.00033725 = 0.02776, ad the stadard deviatio of the portfolio is σ = σ 2 = 0.67 = 6.7%.
2. L8.2) APT factors) Two stock are believed to satisfy the two-factor model r = a + 2f + f 2 r 2 = a 2 + 3f + 4f 2. I additio, there is a risk-free asset with a rate of retur r f = 0%. It is kow that r = 5% ad r 2 = 20%. What are the values of λ 0, λ ad λ 2 for this model? Solutio: Whe the rates of retur of the stocks deped o two factors as the expected rate of retur ay asset i is r i = a i + b i f + b i2 f 2, r i = λ 0 + b i λ + b i2 λ 2. Because there is a risk-free asset, λ 0 = r f = 0%. Because we kow the expected rates of retur r i of stocks ad 2, we ca solve λ ad λ 2 from the equatio system 5% = 0% + 2λ + λ 2 Solvig this yields 20% = 0% + 3λ + 4λ 2. λ = 2% λ 2 = %. Hece the prices of risks of factors ad 2 are 2% ad %, respectively.
3. L8.4) Variace estimate) Let r i, for i =, 2,...,, be idepedet samples of a retur r of mea r ad variace σ 2. Dee the estimates ˆ r = r i Show that Es 2 ] = σ 2. s 2 = r i ˆ r) 2. Solutio: We show that s 2 is a ubiased estimate of the variace. First we write the formula of the average ˆ r i the formula of the sample variace. The we add ad subtract the true expected rate of retur r iside the squared term. These modicatios yield ] 2 2 Es 2 ] = E r i ˆ r) 2 = E r i r j = E r i r) r j r). Extractig r i from the ier summatio ad movig the factor / ) outside the expected value yields E ) 2 r i r) r j r). We the expad the square iside the summatio to get E ) 2 r i r) 2 2 ) 2 r i r) r j r) + r j r). Movig the expected value iside the summatio ad expadig the term /) 2 r j r)] yields 2 E ) r i r) 2] 2 ) E r i r)r j r)] + 2 E r j r)r k r)]. k i The samples were assumed idepedet ad hece have zero covariace, that is, E r i r)r j r)] = σ ij = 0, i j. Note also that E r i r) 2] = σ 2. These modicatios give the above formula as ) ] 2 σ 2 + )σ2 = σ 2 ) ] 2 + ). 2 2 This ca be simplied ito σ 2 2 + 2 + ] 2 = σ 2 ) = σ 2 = σ2 Thus, we have show that Es 2 ] = σ 2. Note that if the true expected value r was kow, the deomiator of the variace estimate would be istead of, because E /) r i r) 2] = /) E r i r) 2] = /)σ 2 = σ 2.
4. L8.7) Clever, but o cigar) Kalle Virtae gured out a clever way to get 24 samples of mothly returs i just over oe year istead of oly 2 samples; he takes overlappig samples; that is, the rst sample covers Ja to Feb, ad the secod sample covers Ja 5 to Feb 5, ad so forth. He gures that the error i his estimate of r, the mea mothly retur, will be reduced by this method. Aalyse Kalle's idea. How does the variace of his estimate compare with that of the usual method of usig 2 o-overlappig mothly returs? Solutio: First divide the year ito half-moth itervals ad idex these time poits by i. Let r i be the retur over the i-th full moth but some will start midway through the moth). We let r m ad σ 2 deote the mothly expected retur ad variace of that retur. Now let ρ i be the retur over the i-th half-moth period. Assume that these returs are ucorrelated. The retur over ay mothly period is a sum of two half-moth returs; that is, the mothly retur r i is r i = ρ i + ρ i+. It is easy to see that r i = ρ i + ρ i+ r m = 2 ρ ρ = r m /2 σ 2 = Varρ i + ρ i+ ] = Varρ i ] + Varρ i+ ] = 2σ 2 ρ σ 2 ρ = σ 2 /2, where ρ is the expected half-mothly retur ad σ 2 ρ is the variace of that retur. The covariace of two mothly returs r i ad r j is the Covr i, r j ] = Covρ i + ρ i+, ρ j + ρ j+ ] = Covρ i, ρ j ] + Covρ i, ρ j+ ] + Covρ i+, ρ j ] + Covρ i+, ρ j+ ] Varr i ] = σ 2, if i = j Covr i, r j ] = Varρ i ] = σ 2 /2, if i j = 0, otherwise. Now for Kalle's scheme we form the estimate We eed to evaluate Var ˆ r ] ] 24 = Var r i = 24 24 24 2 = 24 2 Varr ] + Covr, r 2 ] + 24 23 ˆ r = 24 Cov r i, r j ] i=2 24 r i. Covr i, r i ] + Varr i ] + Covr i, r i+ ]) + Covr 23, r 24 ] + Varr 24 ] ] Hece we have Var ˆ r ] = 24 2 24σ 2 + 2 23σ 2 /2 ] = 24 2 24σ 2 + 24σ 2 σ 2] = 2 24 24 2 σ 2 24 2 σ2 = 2 σ2 24 2 σ2.
For twelve o-overlappig moths of data we would have ] 2 Varˆ r] = Var r i = 2 2 2 2 σ2 = 2 σ2. The dierece σ 2 /24 2 0.007σ 2 is caused by the loger estimatio period for the half-moth case; the last sample is actually partly from the followig year, because sample r 24 covers Dec 5 to Ja 5. Hece the estimatio precisio caot be icreased by samplig with half-moth itervals; the gai of greater sample size is lost because of the correlatio of the overlappig samples.