1 Estimating risk factors for IBM - using data 95-06

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1 1 Estimating risk factors for IBM - using data Basic estimation of asset pricing models, using IBM returns data Market model r IBM = a + br m + ɛ CAPM Fama French 1.1 Using octave/matlab er IBM = a + ber m + ɛ er IBM = a + ber m + b h HML + b s SMB + ɛ Exercise 1. You want to find the risk of the stock IBM. To do so you think the market model r it = α i + β i r mt + e it is a reasonable description of the risk return relationship. To estimate the parameters α i and β i you need a history of stock returns and index returns. Collect monthly returns for IBM and a broad based US stock market index, for example the S&P 500. Take data for 1995:1 to 2006:12. Estimate the model. What is the R 2 in your estimation? Use Matlab/Octave to implement the estimations. Solution to Exercise 1. The data is read into a couple of series Plotting one against the other 1

2 Running the regression, get >> r=ibm(:,2); >> rm=sp500(:,2); >> X=[ones(144,1) rm]; >> b=x\r b = We estimate the parameters as â = ˆb = Next, plotting the observations and comparing it to the actual regrssion. >> plot(rm,r,"*",rm,x*b) 2

3 Check for any obvious problems by calculating the residuals and plotting them against r m: >> plot(rm,e,"*"); 3

4 Calculate the R 2 >> SSR=e *e SSR = >> TSS=(r-mean(r)) *(r-mean(r)) TSS = >> R2=1-SSR/TSS R2 = The R 2 of the regression is Exercise 2. You want to measure the risk on your equity position in IBM. You want to apply the CAPM. The CAPM can be written as E[r it ] r ft = β i (E[r mt ] r ft We rewrite this in excess return form, defining er it = r it r ft and er mt = r mt r ft. The CAPM thus imposes E[er it ] = β it E[er mt ] Consider now the regression er it = α i + β i er mt + ε it (1) Estimate this regression using historical data on the returns on IBM and a stock market index. Use monthly returns. Take data for 1995:1 to 2006:12. Discuss the fit of the model using the R 2 of the regression What restriction on the parameters of (3) does the CAPM impose? Test this restriction. 4

5 Solution to Exercise 2. Now, first get data on monthly returns of IBM and a stock market, using the S&P500. Then, gather relatively short term interest rates. I have used a series of 3 month treasury bill rates from the fed. All data from 1995:1 to 2006:12. This data is input in matrices ibm, sp500 and treas, with the first column holding the date, and the next holding the returns. The following is the first observations for the two series: >> treas(1,1) ans = >> treas(1,2) ans = >> ibm(1,1) ans = >> ibm(1,2) ans = Note that the interest rates are given in annualized percentages, need to transform a bit here. Also need to be careful about timing: Interest rates are given at their observation data, and valid for the following period. Therefore necessary to lag a bit carefully before doing excess returns. >> ri=ibm(2:144,2); >> rf=treas(1:143,2)./1200; >> rm=sp500(2:144,2); >> er_i=ri-rf; >> er_m=rm-rf; Now ready to estimate >> b=[ones(143,1) er_m]\er_i b = We thus estimate â = and ˆb = Let us plot the observations and the fitted line 5

6 Calculate the R 2 : >> e=er_i-x*b; >> SSR=e *e SSR = >> TSS=(r-mean(r)) *(r-mean(r)) TSS = >> R2=1-SSR/TSS R2 = Now, what restrictions do the CAPM impose on the parameters? Well, that in this regression a = 0. Set up a hypotesis test: H 0 : a = a 0 = 0 H A : a 0 Need some estimates of parameter variance, and then construct z = â a 0 σ(â) = â σ(â) >> S2=e *e/143 S2 = >> V=S2*inv(X *X) V = e e e e-02 >> sqrt(v(1,1)) ans = >> sqrt(v(2,2)) ans = >> z=b(1)/sqrt(v(1,1)) z =

7 If the parameters are normally distributed, z is normally distributed with a standard normal, will not reject the null at any reasonable size of the test. The typical output from running this regression will be Coeff se t value pvalue(norm) pvalue (t) Constant er m R Adj R F Pval F 0 Exercise 3. In a series of papers Eugene Fama and Ken French has introduced an alternative asset pricing model to the CAPM, the three factor model, where expected returns follows: E[r it ] r ft = β 1 (E[r mt ] r ft + b 2 SMB t + b 3 HML t Defining er it = r it r ft and er mt = r mt r ft, we consider the regression er it = α i + b 1 er mt + b 2 SMB t + b 3 HML t + ε it (2) Estimate this regression using historical data on the returns on IBM. The data on a market index and the other factors can be found on Ken French s homepage. Use monthly returns. Take data for 1995:1 to 2005:12. Discuss the fit of the model using the R 2 of the regression. Which of the tree factors seem important for pricing of IBM? Use Matlab/Octave to implement the estimations. Solution to Exercise 3. The data is read into matrices IBM, RMRF, HML and SMB. Take a look at the first few data: >> IBM(1:3,1) ans = >> IBM(1:3,2) ans = >> RMRF(1:3,1) ans = >> RMRF(1:3,2) ans = The dates are the same, which is good, but the excess market return from Ken French is in percentage terms. Pull the observations, at the same time dividing the French data by 100. >> ibm=ibm(:,2); >> rmrf=rmrf(:,2)/100; >> smb=smb(:,2)/100; >> hml=hml(:,2)/100; >> size(ibm) 7

8 ans = >> n=132; Now are set for doing the regression >> y=ibm; >> X=[ones(n,1) rmrf smb hml]; >> bhat=x\y bhat = r ibm = a + b 1er mt + b 2SMB t + b 3HML t The regression estimates thus Variable coeff Constant a Market er m b SMB b HML b Do the usual diagnostics, plot residuals against explanatory variables >> e=y-x*bhat; >> plot(rmrf,e,"*") >> plot(smb,e,"*") >> plot(hml,e,"*") 8

9 9

10 Calculate the R 2 of the regression. >> SSR=e *e SSR = >> TSS=(y-mean(y)) *(y-mean(y)) TSS = >> R2 = 1-SSR/TSS R2 = Find R 2 = 38.6%. To say something about what are the important factors relevant for the pricing of IBM, need an estimate of the standard deviations of the various parameters. First find the variance covariance matrix of the parameters. >> s2=e *e/n s2 = >> V=s2*inv(X *X) V = e e e e e e e e e e e e e e e e-02 The estimated standard deviations of the four parameters are then calculated as the square root of the diagonal terms. >> Stdevs = sqrt(diag(v)) Stdevs =

11 To make probability statements, normalize by the standard errors bhat = >> Stdevs Stdevs = >> tvalues= bhat./stdevs tvalues = The t-values are the basis for most inference, and they are typically printed out by standard econometric packages. Another typical way of reporting this is by finding the probability values >> pvalues_n = 2*(1-normcdf(abs(tvalues))) pvalues_n = e e e e-02 >> pvalues_t=2*(1-tcdf(abs(tvalues),n-4)) pvalues_t = e e e e-02 Values typically reported following such an analysis: Coeff st err t val p val(normal) p val(t-dist) Const RMRF SMB HML R % Adjusted R n 132 F pval f e-13 Exercise 4. You consider pricing of IBM stock using the Fama French three factor model. er ibm,t = a + b 1 er m,t + b 2 SMB t + b 3 HML t + e t Test whether the thee explanatory factors er m,t, SMB t and HML t has any explanatory power, i.e, test the hypothesis b 1 = 0, b 2 = 0 and b 3 = 0 jointly. Use monthly returns data on IBM and data for the three factors from Ken French s homepage. Take data for 1995:1 to 2005:12. Solution to Exercise 4. This is often termed the F test, and is the result which is always printed in regression outputs, It is calculated as It has an F distribution F (r, n k). F = (n k) r R 2 1 R 2 11

12 Variable coeff serr t-val p-val(t) Constant er m SMB HML R F Adj R pval F DW 2.07 Exercise 5. You consider pricing of IBM stock. You have two alternative models in mind. One is a single factor mode er ibm,t = α + βer m,t + e t where the excess return of IBM (er ibm,t only depends on one factor, the excess return on a market portfolio er m,t. The other model you have in mind is Fama And French es three factor model er ibm,t = a + b 1 er m,t + b 2 SMB t + b 3 HML t + e t Using monthly returns data on IBM and data for the three factors from Ken French s homepage, estimate the two models. Take data for 1995:1 to 2005:12. Compare the R 2 of the two models. You wonder if the first model is good enough, and you want to formally test whether the one factor model is a sufficient description of IBM s returns. Do this as a hypothesis test, where the null hypothesis is that the first model is enough. That is, you want to test H 0 : b 2 = 0 and b 3 = 0. (Hint: This is a joint test, which can not be solved by looking at the coefficients b 1 and b 2 in isolation.) Solution to Exercise 5. Let us first do the estimation. The one factor model The three factor model Coefficient estimate t-value p-value(t) constant α e-01 market β e-14 R AdjR F pval F e-14 Coefficient estimate t-value p-value(t) constant a e-01 market b e-09 SMB e-02 HML e-02 R AdjR F pval F e-13 First, the R 2 of the second model is better (0.387 vs 0.362), even the adjusted R 2 is better(0.372 vs 0.357), which indicates the first model improves on the second. We now wonder whether the first model is sufficient. Some indication can be had from looking at the p-values of the two parameters SMB and HML, of which only the first is significant. However, this is not a formal test of the joint hypothesis of b 2 = b 3 = 0. This can be tested using a likelihood ratio type test, in this setting typically called an F test, We will be calculating ( 1 r SSE ( b) (ˆb)) SSE F = 1 SSE (ˆb) n k 12

13 where b is the restricted estimate and ˆb the unrestricted, r the number of restrictions, k the number of parameters, and n the number of observations. Here we start with an unrestricted model of four parameters (ˆb = [a, b 1, b 2, b 3]. This is to be compared to a restricted model with two parameters ( b = [a, b 1]. Here r = 2 and k = 4. Do the calculations >> y=ibm; >> X=[ones(n,1) rmrf smb hml]; >> bhat = ols(y,x) bhat = >> SSR_unrestr = (y-x*bhat) *(y-x*bhat) SSR_unrestr = >> X=[ones(n,1) rmrf]; >> btilde = ols(y,x) btilde = >> SSR_restr = (y-x*btilde) *(y-x*btilde) SSR_restr = >> r=2 r = 2 >> k=4 k = 4 >> F= ( (1/r)*(SSR_restr-SSR_unrestr))/((1/(n-k))*SSR_unrestr) F = >> pval_f_f = 1-fcdf(F,r,n-k) pval_f_f = At the 5% level we do not reject the null, and accept the one factor model as a sufficient explanation. 13

14 1.2 Using R Doing exactly the same exercises with R as the tool. Exercise 6. Consider estimation of the risk of the stock IBM using the market model r it = α i + β i r mt + e it Collect monthly returns for IBM and a broad based US stock market index, for example the S&P 500. Take data for 1995:1 to 2006:12. (Available on course homepage) Estimate the model using OLS. Use R to perform the estimation Solution to Exercise 6. The following R code does this rets <- read.table("../data_set/ibm_sp500.txt",header=true,sep=";") n <- dim(rets)[1] rm <- rets[,2] ibm<- rets[,3] mod <- lm(ibm rm) summary(mod) The above code results in estimates of OLS: lm(formula = ibm rm) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) rm <2e-16 *** Residual standard error: on 142 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 1 and 142 DF, p-value: < 2.2e-16 Exercise 7. You want to apply the CAPM. The CAPM can be written as E[r it ] r ft = β i (E[r mt ] r ft We rewrite this in excess return form, defining er it = r it r ft and er mt = r mt r ft. The CAPM thus imposes E[er it ] = β it E[er mt ] Consider now the regression er it = α i + β i er mt + ε it (3) Estimate this regression using historical data on the returns on IBM and a stock market index. Use monthly returns. Take data for 1995:1 to 2006:12. Discuss the fit of the model using the R 2 of the regression What restriction on the parameters of (3) does the CAPM impose? Test this restriction. Use R to implement the estimations. Solution to Exercise 7. The data is read in. Here the input file contains data aligned on observation date. But some thought necessary here. The interest rate is observed looking forward, while returns are looking backward from observation date. Also note that the risk free rate is in annualzed percentages, which need correcting 14

15 > rets <- read.table("../data_set/ibm_sp500_95_06.txt", header=true,sep=";") > n <- dim(rets)[1] > rf <- rets[1:n-1,4]/1200 > rm <- rets[2:n,2] > ibm <- rets[2:n,3] Calculating execess returns er IBM,t = r IBM,t r ft er m,t = r m,t r ft > er_ibm <- ibm-rf > er_m <- rm-rf Plotting one against the other Running the regression er ibm = a + ber m + e > mod=lm(er_ibm er_m) Showing the resulting estimates > summary(mod) Call: lm(formula = er_ibm er_m) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) er_m <2e-16 *** Residual standard error: on 141 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 1 and 141 DF, p-value: < 2.2e-16 15

16 We estimate the parameters as â = ˆb = Next, plotting the observations and comparing it to the actual regrssion. > plot(er_m,er_ibm,main="regressing IBM on SP500") > lines (er_m,mod$fitted.values) Check for any obvious problems by calculating the residuals and plotting them against r m: > plot(er_m,mod$residuals,main="residuals against SP500") Now, what restrictions do the CAPM impose on the parameters? Well, that in this regression a = 0. Set up a hypotesis test: H 0 : a = a 0 = 0 H A : a 0 Here we can use the p-value in the R output: The intercept, estimated as , has a p-value of We can thus not reject the null of a zero intercept. Exercise 8. 16

17 You want to measure the risk of an equity position on IBM. In a series of papers Eugene Fama and Ken French has introduced an alternative asset pricing model to the CAPM, the three factor model, where expected returns follows: E[r it ] r ft = β 1 (E[r mt ] r ft + b 2 SMB t + b 3 HML t Defining er it = r it r ft and er mt = r mt r ft, we consider the regression er it = α i + b 1 er mt + b 2 SMB t + b 3 HML t + ε it (4) You have access to the time series of these variables on the course homepage. The data is for the period 1995:1 to 2005:12. Estimate this regression using historical data on the returns on IBM. Discuss the fit of the model using the R 2 of the regression. Which of the tree factors seem important for pricing of IBM? Use R to implement the estimations. Solution to Exercise 8. Reading the data: > rets <- read.table("../data_set/ibm_ff_95_05.txt",header=true,sep=";") > eribm <- rets$eribm > RMRF <- rets$rmrf/100 > HML <- rets$hml/100 > SMB <- rets$smb/100 Doing the analysis: > mod <- lm( eribm RMRF + HML + SMB) Call: lm(formula = eribm RMRF + HML + SMB) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) RMRF e-09 *** HML SMB * --- Signif. codes: 0 âăÿ***âăź âăÿ**âăź 0.01 âăÿ*âăź 0.05 âăÿ.âăź 0.1 âăÿ âăź 1 Residual standard error: on 128 degrees of freedom Multiple R-squared: 0.387,Adjusted R-squared: F-statistic: on 3 and 128 DF, p-value: 1.433e-13 Do the usual diagnostics, plot residuals against explanatory variables > plot (mod$residuals,rmrf) > plot (mod$residuals,hml) > plot (mod$residuals,smb) 17

18 Estimate Std. Error t value Pr(> t ) (Intercept) RMRF HML SMB

19 Observe the R 2 of the regression. R 2 = 38.7%. To say something about what are the important factors relevant for the pricing of IBM, use the estimates of the standard deviations of the various parameters. The t-values are the basis for most inference, and they are typically printed out by standard econometric packages. Another typical way of reporting this is by finding the probability values In terms of the results find that RMRF very significant. SMB significant at the 5% level, with a p value of 4.1%, while HML not significant, p value of 7.92%. Exercise 9. You consider pricing of IBM stock using the Fama French three factor model. er ibm,t = a + b 1 er m,t + b 2 SMB t + b 3 HML t + e t Test whether the thee explanatory factors er m,t, SMB t and HML t has any explanatory power, i.e, test the hypothesis b 1 = 0, b 2 = 0 and b 3 = 0 jointly. Use monthly returns data on IBM and data for the three factors from the course homepage. The data is for 1995:1 to 2005:12. Use R to implement the estimations. Solution to Exercise 9. This is often termed the F test, and is the result which is always printed in regression outputs, It is calculated as F = (n k) r It has an F distribution F (r, n k). This is the printout of the regression estimation in R. > mod2 <- lm( eribm RMRF + HML + SMB ) > summary(mod2) Call: lm(formula = eribm RMRF + HML + SMB) Residuals: Min 1Q Median 3Q Max R 2 1 R 2 Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) RMRF e-09 *** HML SMB * --- Signif. codes: 0 âăÿ***âăź âăÿ**âăź 0.01 âăÿ*âăź 0.05 âăÿ.âăź 0.1 âăÿ âăź 1 19

20 Residual standard error: on 128 degrees of freedom Multiple R-squared: 0.387,Adjusted R-squared: F-statistic: on 3 and 128 DF, p-value: 1.433e-13 The F statistic in this regression is exactly the test we want. It is a test of the null that all coefficients except the intercept is zero. The P-value of the F test is essentially zero. We reject the null of all coefficients being zero. Note that this F statistic is only usable when we test whether all the coefficients are jointly zero. It can not be used when testing whether only some of the coeffients are zero. Exercise 10. You consider pricing of IBM stock. You have two alternative models in mind. One is a single factor model er ibm,t = α + βer m,t + e t where the excess return of IBM (er ibm,t only depends on one factor, the excess return on a market portfolio er m,t. The other model you have in mind is Fama And French es three factor model er ibm,t = a + b 1 er m,t + b 2 SMB t + b 3 HML t + e t Using monthly returns data on IBM and data for the three factors from the course homepage, estimate the two models. Note that data is for 1995:1 to 2005:12. Compare the R 2 of the two models. You wonder if the first model is good enough, and you want to formally test whether the one factor model is a sufficient description of IBM s returns. Do this as a hypothesis test, where the null hypothesis is that the first model is enough. That is, you want to test H 0 : b 2 = 0 and b 3 = 0. (Hint: This is a joint test, which can not be solved by looking at the coefficients b 1 and b 2 in isolation.) Solution to Exercise 10. > library(xtable) > library(car) > rets <- read.table("../data_set/ibm_ff_95_05.txt",header=true,sep=";") > eribm <- rets$eribm > RMRF <- rets$rmrf/100 > HML <- rets$hml/100 > SMB <- rets$smb/100 > mod1 <- lm( eribm RMRF ) > summary(mod1) Call: lm(formula = eribm RMRF) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) RMRF e-14 *** --- Signif. codes: 0 âăÿ***âăź âăÿ**âăź 0.01 âăÿ*âăź 0.05 âăÿ.âăź 0.1 âăÿ âăź 1 Residual standard error: on 130 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: 74 on 1 and 130 DF, p-value: 2.18e-14 Estimate Std. Error t value Pr(> t ) (Intercept) RMRF

21 > mod2 <- lm( eribm RMRF + HML + SMB ) > summary(mod2) Call: lm(formula = eribm RMRF + HML + SMB) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) RMRF e-09 *** HML SMB * --- Signif. codes: 0 âăÿ***âăź âăÿ**âăź 0.01 âăÿ*âăź 0.05 âăÿ.âăź 0.1 âăÿ âăź 1 Residual standard error: on 128 degrees of freedom Multiple R-squared: 0.387,Adjusted R-squared: F-statistic: on 3 and 128 DF, p-value: 1.433e-13 Estimate Std. Error t value Pr(> t ) (Intercept) RMRF HML SMB Now the hypothesis test. Need to specify it in matrix form: > R = rbind(c(0,0,1,0),c(0,0,0,1)) > r=c(0,0) > R = rbind(c(0,0,1,0),c(0,0,0,1)) > r=c(0,0) > print(r) [,1] [,2] [,3] [,4] [1,] [2,] > print(r) [1] 0 0 And now finally performing the test: > linearhypothesis(mod2,r,r) Linear hypothesis test Hypothesis: HML = 0 SMB = 0 Model 1: restricted model Model 2: eribm RMRF + HML + SMB Res.Df RSS Df Sum of Sq F Pr(>F) Signif. codes: 0 âăÿ***âăź âăÿ**âăź 0.01 âăÿ*âăź 0.05 âăÿ.âăź 0.1 âăÿ âăź 1 At the 5% level we do not reject the null, and accept the one factor model as a sufficient explanation. 21