HW # Saisical Financial Modeling ( P Theodossiou) 1 The following are annual reurns for US finance socks (F) and he S&P500 socks index (M) Year Reurn Finance Socks Reurn S&P500 Year Reurn Finance Socks Reurn S&P500 1975 00 37 1990 08 31 1976 84 39 1991 491 305 1977 05 7 199 33 76 1978 99 66 1993 106 101 1979 184 186 1994 35 13 1980 158 35 1995 541 376 1981 106 49 1996 35 30 198 97 16 1997 48 334 1983 07 6 1998 114 86 1984 193 63 1999 41 10 1985 197 317 000 57 91 1986 344 187 001 90 119 1987 150 53 00 146 1 1988 07 166 003 310 87 1989 85 317 004 96 109 Consider he single index model (SIM) for sock reurns, R R e F, F F M, F, which is equivalen o he sandard regression model Y X u where Y R, X R, u e, and Use he aached excel spreadshees o F, M, F, F F 1 Esimae and inerpre he inercep and slope of he above regression model Consruc he daa scaer diagram, graph he sample regression model and plo he residuals 3 Verify numerically ha Y Yˆ and uˆ 0 4 Verify numerically ha y yˆ uˆ and compue he coefficien of deerminaion or R 5 Compue he sample regression variance, sandard errors and -values of he inercep and slope 6 Tes he hypoheses ha he sample regression line inercep and slope are differen from zero
Repea he above exercise using he daa below P/E raios and TL/TA of a random sample of 30 S&P500 companies OBS P/E TL/TA OBS P/E TL/TA 1 30 310 16 148 70 16 663 17 155 394 3 16 663 18 114 399 4 188 459 19 138 735 5 1 575 0 114 399 6 188 44 1 94 54 7 117 505 131 57 8 00 656 3 77 604 9 180 551 4 110 699 10 144 346 5 00 656 11 15 51 6 13 434 1 173 508 7 89 53 13 151 461 8 151 461 14 60 710 9 1 575 15 115 453 30 97 54 Model o be esimaed is: PE TLTAu
HW # - SOLUTION - SINGLE INDEX MODEL (SIM) 1 Esimae and inerpre he inercep and slope of he above model The single index model (SIM) for any sock or a porfolio of socks is specified as: R R e F, F F M, F, where R F, and R M, are he annual reurns for a porfolio of financial socks and he S&P 500 sock marke index, α F and β F are he inercep and slope (bea) of he model and e F, is an error erm wih he usual regression sochasic properies The S&P500 sock index is used as a proxy for he reurn of a well diversified porfolio of he financial asses in he US For simpliciy of noaion, he SIM model is expressed in he sandard regression form as Y X u where Y = R F,, X = R M,, u = e F,, α = α F, and β = β F The parameers α = α F and β = β F are esimaed using,, x y X Y TXY ˆ x X TX 3,97631 0567 7,54955 ˆ Y ˆ X 135 0567 149 56333 The accompanying excel spreadshee provides wha s needed for he compuaion The esimaed sample regression equaion for R F, is Rˆ 563330567R F, M, is an esimaor of he condiional expecaion E R R R F, M, M, and provides condiional forecass of R F for given values of R M, The inercep of he equaion gives a forecas for he reurn of financial socks when he marke reurn is zero, ie, R F = 56333 for R M = 0 3
Scaer Plo and Regression of he Financial Socks and he S&P 500 sock Marke Reurns R F, =56333 + 056695 R M, RF, 60 50 40 30 0 10 0-30 -0-10 -10 0 10 0 30 40 50-0 R M, -30 The slope gives he rade-off beween he reurn R F and he marke reurn, ie, ˆ R R 0567 F The laer implies ha an 1% increase in he marke reurn will be followed by a 0567% increase in he financial socks reurn Consruc he scaer diagrams for he daa and regression model and he residuals The figure above provides he scaer plo of R M, vs R F, and he plo for he sample regression line Rˆ 563330567R F, M, M, which passes hrough he daa poins Noe ha here roughly as many daa poins above as here are below he regression line The verical disance beween he daa poins and he sample regression line represen visually he residuals u Y Yˆ These are ploed on he scaer diagram ˆ below: 4
Plo of Regression Residuals Residual u = R F, - (56333 + 056695 R M,) u 40 30 0 10 0-30 -0-10 -10 0 10 0 30 40 50-0 -30-40 -50 R M, 3 Verify numerically ha Y Yˆ and uˆ 0 I is clear from he excel spreadshee Y ˆ Y = 4048 and uˆ 0 4 Verify numerically ha y yˆ uˆ Calculae he coefficien of deerminaion or R The resuls in he excel spreadshee verify ha yˆ uˆ,0943 + 9,4799 = 11,574 y The model s R-square value is R y ˆ y 1 uˆ y 1 ESS TSS 9, 47985 1 01809 11,57416 or 1809% The above number indicaes ha he oal variaion of financial sock reurns explained by he regression model is 1809% 5
5 Compue he sample regression variance, sandard errors and -values of he inercep and slope The sample regression variance is and s uˆ ˆ 9,47985 338566 T k 30 s 338566 18400 The sandard errors of he esimaes for he inercep and slope are X SE ˆ s T x 14, 307 307,54955 18400 4613 SE ˆ 1 s x 1 18400 01177 7,754955 Their respecive -values are and ˆ ˆ SE ˆ ˆ ˆ SE ˆ 56333 1 4613 0567 49 01177 6 Tes he hypoheses ha he inercep and slope of he regression model are differen from zero Because he ˆ 1 < c,8,5% = 048, he null hypohesis, ha he inercep is saisically equal o zero, is acceped a he 5% level of significance Noe ha (df) ˆ follows he disribuion wih 8 degrees of freedom Because he ˆ 49 > c,8,5% = 048, he alernaive hypohesis, ha he slope is saisically differen from zero, is acceped a he 5% level of significance Noe ha of freedom (df) ˆ follows he disribuion wih 8 degrees 6
PRICE/EARNINGS (P/E) RATIO RELATIVE TO THE TOTAL LIABILITIES RATIO MODEL 1 Esimae and inerpre he inercep and slope of he above model The price o earnings (P/E ) raio model P E TL TA u where TL/TA is oal liabiliies o oal asses raio for a randomly seleced firms included in he S&P 500 sock marke index The model posulaes a linear relaionship beween he P/E raio and he financial leverage of a firm Is sandard regression formulaion is Y X u where Y = P/E and X = TL/TA Using he resuls, included in excel spreadshee, we find ha,, x y ˆ x 4941 01009 4,89506 ˆ Y ˆ X 137 01009 514 189091 Thus, he esimaed P/E model (ie, sample regression model) is PE 189091 01009 TLTA This equaion gives he forecased P/E raio of a firm in relaion o is TL/TA raio (ie, given is TL/TA raio) For example, a firm wih 50% leverage is expeced o have a P/E raio of PE 18909101009 50% 1890915045 134591 The inercep of he equaion, which is equal o 189091, gives he forecased value of a firm s P/E wih zero TL/TA In oher words, a non-levered firm (ie, zero deb firm) is expeced o have a P/E raio of 189091 The slope of he sample regression line ˆ PE 01009 TL TA gives he rade-off beween TL/TA raio and he P/E raio I implies ha an increase in he TL/TA of a firm by 10 unis (eg, from 40% o 50%) is expeced o resul in 1009 unis reducion in is P/E raio 7
Scaer Plo and Regression of he P/E and he TL/TA Raios for a Sample S&P 500 Socks P/E = 189091-01009 TL/TA P/E 5 0 15 10 5 0 0 0 40 60 80 TL/TA Residuals, u = P/E - (189091-01009 TL/TA ) u 10 8 6 4 0 - -4-6 -8 0 0 40 60 80 TL/TA Consruc he scaer diagrams for he daa and regression line he residuals The op figure provides he scaer plo of TL/TA and P/E and a graphical illusraion of he sample regression line The sample regression line has a negaive slope indicaing ha, on average, firms wih higher leverage 8
raios have lower P/E raios The second figure provides a plo of he regression s residuals u Y Yˆ 3 Verify numerically ha Y Yˆ and uˆ 0 ˆ I is clear from he excel spreadshee Y ˆ Y = 4117 and uˆ 0 4 Verify numerically ha y yˆ uˆ Calculae he coefficien of deerminaion or R The resuls in he excel spreadshee verify ha yˆ uˆ 499 + 3968 = 4467 y The model s R-square value is R y ˆ y ESS TSS 499 01117 4467 or 1117% The above number indicaes ha he oal variaion of he P/E raios is explained by he regression model is 01117 or 1117% 5 Compue he sample regression variance, sandard errors and -values of he inercep and slope The sample regression variance is s uˆ ˆ 3968 1417 T k 30 and s 1417 37646 The sandard errors of he esimaes for he inercep and slope are X SE ˆ s T x 84,07164 304,89506 37646 8484 SE ˆ 1 s x 1 33676 005381 4,89506 Their respecive -values are ˆ ˆ SE ˆ 189091 664 8484 9
and ˆ ˆ SE ˆ 01009 188 005381 6 Tes he hypoheses ha he inercep and slope of he regression model are differen from zero Because he ˆ 664 > c,8,5% = 048, he alernaive hypohesis, ha he inercep is saisically significan, is acceped a he 5% level of significance Noe ha of freedom (df) ˆ follows he disribuion wih 8 degrees Because he ˆ 188 < c,8,5% = 048, he null hypohesis, ha he slope is saisically equal o zero, is acceped a he 5% level of significance Noe ha (df) ˆ follows he disribuion wih 8 degrees of freedom 10
MATHEMATICAL RESULTS For he mahemaical proof of he above resul, we use he already esablished resuls (a) X TX and Y TY (his resul follows easily from he definiion of he sample mean) (b) Yˆ ˆ ˆ X (sample regression line) (c) u ˆ ˆ Y Y or Y ˆ ˆ Y u (he residual is he difference beween Y and Yˆ ) (d) ˆ Y ˆ X or Y aˆ ˆ X (OLS esimaor for regression inercep) (e) ˆ xyx (OLS esimaor for regression slope) (d) Y Yˆ or Y Yˆ (sum of he acual values is equal o he sum of he prediced values) (e) Y Yˆ uˆ (he residual is he difference beween he acual and prediced value of Y ) 1 Mahemaical proof ha Y Yˆ and uˆ 0 The sum of all he sample regression values Yˆ ˆ ˆ X (ie, for = 1,,, T ) is ˆ ˆ ˆ ˆ Yˆ ˆ X ˆ X T ˆ T X T ˆ X Furher, subsiue Y aˆ ˆ X above Y TY Y ˆ I follows easily from he above ha Y ˆ T Y T or Y Y ˆ These resuls imply ha he sum (average) of he observed values of Y is equal o he sum (average) of he prediced values ˆ Also, i implies ha he sample regression line passes hrough he poin X, Y Y The sum of he regression residuals is 11
uˆ Y Yˆ Y Yˆ 0, because Y Yˆ Tha is, on average he sample regression errors (residuals) are zero Mahemaical proof of he equaliy y yˆ uˆ I follows easily from (b), (c) and (d) ha ˆ ˆ y Yˆ Yˆ Yˆ Y ˆ X ˆ X ˆ ˆ X X ˆ x This resuls shows ha he relaionship beween y and x is y ˆ ˆ x Tha is, he regression slope of variables expressed as deviaions from heir respecive sample means remains he same as ha of he original daa Nex subrac Y from Y Yˆ uˆ and y Y Y Yˆ Y uˆ y yˆ uˆ Square boh he lef- and righ-hand side of he equaliy y yˆ uˆ yˆ uˆ yu ˆ ˆ The sum he above equaliy across all sample daa gives because yu ˆˆ 0 y yˆ uˆ yu ˆ ˆ yˆ uˆ To prove his resul, subsiue ino he above equaion uˆ y yˆ and hen y ˆ x, o ge ˆ 1
yu ˆˆ yˆ y yˆ yy ˆ yˆ Noe ha The resul ˆ xy ˆ x ˆ xy ˆ x 0 ˆ ˆ ˆ ˆ x ˆ x xy x x xy yu ˆˆ ˆ xuˆ 0, implies ha cov yˆ, uˆ ˆ cov x, uˆ 0 In oher words, i implies ha he forecased value of y canno be improved furher using uˆ 13