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Testing for the 'January Effect' under the CAPM framework
 

Testing for the 'January Effect' under the CAPM framework

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In this project, we used the Capital Assets Pricing Model (CAPM) to test for the ‘January effect’ - a calendar‐related market anomaly in the financial market where financial security prices ...

In this project, we used the Capital Assets Pricing Model (CAPM) to test for the ‘January effect’ - a calendar‐related market anomaly in the financial market where financial security prices increase in the month of January.

Please refer to "Chapter 2 – The Capital Asset Pricing Model: An Application of Bivariate Regression Analysis" of the book "The Practice of Econometrics" by Ernst R. Bernd for the test data, background and problem statement.

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    Testing for the 'January Effect' under the CAPM framework Testing for the 'January Effect' under the CAPM framework Document Transcript

    • Financial Markets & Financial Institutions    Prof. Dr. Bernhard Schipp  Group 6: Alice Nušlová, Rocio Cataño, Lov Loothra, Nikhil Garg      EXERCISE 2: LEAST SQUARES ESTIMATES OF ß  The choice of a ‘risky’ and a ‘safe’ industry is one which cannot be made without some reasonable  level of doubt. However, after an internal debate in the group and referring to a few online sources,  we chose Airlines as the highly risky industry with DELTA as the corresponding company and, on the  other  hand,  we  choose  Foods  as  the  relatively  ‘safe’  industry  with  GENMIL  as  the  corresponding  company.  We  also  decided  to  work  with  the  latter  half  of  the  data  and  hence  chose  the  60  observations corresponding to January 1983 to December 1987.    (a) In order to estimate the α and ß parameters by the ordinary least squares method, we ran the  following regression on both the chosen companies:    rp – rf = α + ß (rm − rf )   Here, rp is the return of the company, rm is the overall return of the market and rf is the risk free rate  of return.  The following is the summary of the regression analysis:    Industry Company Intercept (α) Slope (ß) LSE SE p-val LSE SE t-stat p-val Airlines DELTA -0.00429631 0.0112209 0.7032 0.639525 0.189369 3.37714 0.0013 Foods GENMIL 0.0123792 0.00799058 0.1268 0.566974 0.134852 4.20442 0.0001   From the above result, we observe that the estimated value of ß for DELTA of the Airlines industry is  larger  than  GENMIL  of  the  Foods  industry.  This  signifies  that  the  DELTA  is,  in  fact,  riskier  than  GENMIL which is in‐line with prior intuition.    (b) We made a time sequence plot of DELTA airlines using the 60 observations from January 1983 to  December 1987.     Time Sequence Plot     ‐0.3 ‐0.2 ‐0.1 0 0.1 0.2 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 Predicted Rm ‐ Rf DELTA‐Rf
    • From  the  unusual  residual  data,  the  value  corresponding  to  observation  7  (July  1983)  had  a  studentized residual value of ‐3.07, a residual of ‐0.246406 and a corresponding value of ‐0.26614  (the least value).  Upon inspecting the historical data, the value can be accounted for by the July 17,  1983 hijacking of Delta 722 Miami‐Tampa 727.    Furthermore,  if  we  observe  the  value  corresponding  to  observation  58,  it  is  nearly  as  low  as  observation  7.  However,  it  is  not  an  outlier!  This  is  because  that  value  corresponds  to  the  Black  Monday Crash of October 19, 1987 which is accounted for in the market variable as well.    (c) We now test for the following hypothesis:    H0: α = 0 Ha: α ≠ 0   From  the  regression  analysis  summary  in  part  (a),  we  can  observe  that  the  p‐values  for  all  observations are greater than 0.05. Hence, at the 95% confidence level, we cannot reject the null  hypothesis that H0: α = 0. α does not play a significant role in CAPM and is only used to fit the  regression.    (d) We now prepare the 95% confidence intervals for DELTA and GENMIL. From the statistics we  obtained in part (a), we have the following:    DELTA: 0.639525 ± 0.378738 => 0.260787 to 1.018263 GENMIL: 0.566974 ± 0.269704 => 0.29727 to 0.836678 We now test for the following hypothesis:    H0: ß = 1 Ha: ß ≠ 1 Now, in this case, the t‐value is given by: t-stat = (ß – 1) / SE (ß); where SE is the Standard Error.  Hence, the t‐test value for DELTA is –1.903558 and for GENMIL it is –3.211120. The corresponding p‐ values for two‐tailed and one‐tailed tests are given below:    p-value (two-tailed t-test) p-value (one-tailed t-test) DELTA 0.0618 0.0309 GENMIL 0.0021 0.00105   For GENMIL we can reject H0 but the same is not possible for DELTA. This is because the p‐value for  the two‐tailed t‐test is greater than 0.05. There are no surprises here.     (e) The proportion of stock that is non‐diversifiable is given by the R2  statistic from the regression  analysis we did in part (a). For DELTA, the R2  statistic is 16.4326%. Hence the proportion of stock that  is  non‐diversifiable  for  DELTA  is  16.4326%  and  the  proportion  that  is  diversifiable  is  83.5674%.  Similarly, for GENMIL, the R2  statistic is 23.3586% which determines the proportion of stock that is  non‐diversifiable.  Thus  the  proportion  that  is  diversifiable  is  76.6414%.  In  both  the  cases,  the  proportion which is non‐diversifiable is less than a typical stock.    The results are not surprising. GENMIL and DELTA are both less diversified in the market, so even  though these stocks do move with the market, they have a large proportion of “company‐specific”  risk.   
    • (f) From our sample data, we have the following observations:    DELTA: ß = 0.639525 & R2 = 16.4326% GENMIL: ß = 0.566974 & R2 = 23.3586% Hence in our sample data, a large value of ß does not necessarily correspond with a higher R2  value.  Given that R2 is defined as R2 = ß2 σm 2 /(ß2 σm 2 + σε 2 ),  the  relationship  would  be  exact  if  firms  have  roughly similar firm‐specific risk. However, as observed above, this is not the case for our sample.      EXERCISE 8: IS JANUARY DIFFERENT?    There  is  some  tentative  evidence  which  supports  the  notion  that  stock  returns  in  the  month  of  January are relatively higher. This is referred to as the “January Effect”, a calendar‐related market  anomaly in the financial market where financial security prices increase in the month of January. This  effect was first observed by investment banker Sidney B. Wachtel.    This creates an opportunity for investors to buy stock for lower prices before January and sell them  after their value increases. This effect is curious because even if we consider investors selling losing  stocks in December, the expectation of higher January returns should shift supply‐demand curves  and equilibrate returns.    We will try to empirically and statistically test this hypothesis in our report.    (a) If the “January premium” affected the overall market return rm and the risk‐free return rf by the  same amount jm, the market risk premium, MRP, would be:   MRP = r'm – r'f = (rm + jm) – (rf + jm)  = rm – rf + jm – jm    = rm – rf    Therefore, the market risk premium would not be affected by the January premium.    We will now try and ascertain whether the “January is different” hypothesis be tested within the  CAPM framework of Eq. (2.17). The estimable equation relating the total risk premium of security, j,  to the market risk premium and to the stochastic disturbance term is:     rj – rf = αj + ßj (rm – rf) + εj    We conclude, therefore, that the hypothesis cannot be tested within the CAPM framework, because  the independent variable of the regression, that is, the market risk premium, would be unchanged.    Furthermore,  we  believe  that  it  would  not  be  more  reasonable  to  assume  that  the  “January  is  different” hypothesis referred only to risky assets because if January is actually different, the returns  of all stocks (including the risk‐free returns) should differ and not just risky assets.    (b) If r'm = rm + jm and the risk‐free assets return is unaffected, then the market risk premium, MRP,  would be:    MRP = r'm – rf = rm + jm – rf    Therefore, the market risk premium would be affected. 
    • Furthermore, if the CAPM model were true and the α and ß parameters were constant, then the  expected portfolio return would be:     r'p = rf + α + ß (r'm – rf)  => r'p = rf + α + ß (rm + jm – rf)   => r'p = rf + α + ß (rm – rf) + ß jm  Now since, rf + α + ß (rm – rf) = rp, our equation becomes:    r'p ≡ rp + ß jm    Now, re‐writing the CAPM regression equation using the right‐hand sides of the above expressions,  we get:    r'p – rf = α + ß (r'm – rf ) + ε  => rp + ß jm – rf = α + ß (rm + jm – rf) + ε    = α + ß (rm – rf ) +ß jm + ε    Now, considering the term ß jm to be unobservable, we subtract it from both the sides to get:    rp – rf = α + ß (rm – rf) + ε    Comparing  the  above  equation  with  Eq.  2.17,  we  observe  that  the  equation  has  reduced  to  the  original CAPM equation sans the January premium. With this we therefore conclude that we cannot  estimate the “January premium” within the CAPM framework under these assumptions as well.    (c)  From  the  conclusions  drawn  from  parts  (a)  and  (b),  we  now  abandon  the  CAPM  model  and  examine  an  alternative  method  of  testing  the  “January  is  different”  hypothesis.  We  decided  to  choose the following industries and their corresponding companies: Computers (IBM and DATGEN),  Foods (GERBER and GENMIL) and Banks (CONTIL and CITCRP).     For each of these companies we ran the following regression:    rp = α + ß (DUMJ)  Here, rp is the return of the company, and DUMJ is a dummy variable, which takes the value of unity  if the month is January and zero for all other months. The following is the summary of the regression  analysis:    Industry Company Intercept Slope (DUMJ) LSE SE p-val LSE SE t-stat p-val Computers IBM 0.00817273 0.005633 0.1495 0.017327 0.019512 0.888016 0.3763 DATGEN 0.00405455 0.012163 0.7395 0.041146 0.042133 0.976555 0.3308 Foods GERBER 0.0157636 0.008398 0.063 0.007636 0.029093 0.262482 0.7934 GENMIL 0.0170909 0.006225 0.007 -0.00609 0.021566 -0.28244 0.7781 Banks CONTIL -0.0064818 0.014327 0.6518 0.064582 0.04963 1.30127 0.1957 CITCRP 0.0118455 0.007753 0.1292 0.000155 0.026857 0.005754 0.9954  
    • The  above  table  gives  the  values  of  the  estimated  regression  parameters  (LSE  =  Least  Squares  Estimate,  which  is  the  coefficient),  the  standard  error  (SE)  of  the  estimate,  the  corresponding  p‐ values for the slope and the intercept and the t‐statistic for the slope.    Now  we  test  the  null  hypothesis  that  the  coefficient  on  the  DUMJ  variable  is  zero  against  the  alternative hypothesis that it is not zero. We therefore have:    H0: ß = 0  Ha: ß ≠ 0   If  we  use  a  5%  significance  level  or  a  95%  confidence  interval,  we  cannot  reject  H0  that  the  coefficient on the DUMJ variable is zero for any of the chosen companies. For each of the chosen  companies, the p‐value is larger than 0.05. Furthermore, the critical value of the t‐distribution for a  two‐sided test with 95% confidence interval is 1.98 for 120 observations (since we’re working on 120  months  in  total).  Therefore,  we  can  make  an  equivalent  observation  that  the  t‐statistic  for  the  DUMJ variable is less than 1.98 for all companies.    Based on these results and the reasonable 5% significance level, we conclude that January is not  different.    (e) We now move on to yet another way of examining the “January is different hypothesis”. In this  part of the exercise, we use the risk premiums of the various companies we chose in part (c) (rp – rf),  the  market  premium  (rm – rf)  and  the  DUMJ  variable  which  is  unity  whenever  the  observation  corresponds to the month of January and zero otherwise. We then ran the following regression for  every company:     rp – rf = α + ß1(DUMJ) + ß2(rm – rf) + ε    By  doing  this,  we  have  restricted  the  slope  coefficients  to  be  the  same  for  all  months  but  have  allowed the intercept term  for January to be different from the common  intercept for the other  months. The following is the summary of the regression analysis:    Industry Company Intercept DUMJ Marker Risk Premium LSE p-val LSE t-stat p-val LSE p-val Computers IBM -0.001173 0.8093 0.008424 0.501443 0.617 0.454218 0.0000 DATGEN -0.0084713 0.4102 0.02148 0.605076 0.5463 1.02418 0.0000 Foods GERBER 0.00545394 0.462 -0.00453 -0.17696 0.8598 0.626992 0.0000 GENMIL 0.00875192 0.148 -0.01159 -0.5566 0.5789 0.273791 0.0015 Banks CONTIL -0.0172848 0.207 0.050747 1.07557 0.2843 0.715408 0.0003 CITCRP 0.00129034 0.8415 -0.01284 -0.57597 0.5657 0.670978 0.0000   Now, we will use the above results to test for the null hypothesis that “January is different”. Now, in  this case, the said hypothesis can be considered to be equivalent to rejecting the null hypothesis that  ß1 is zero. Therefore, we have:    H0: ß1 = 0 Ha: ß1 ≠ 0   If we use a 5% significance level and check the p‐values of the DUMJ variable, we observe that we  cannot reject the null hypothesis that ß1 is zero for all the chosen companies. This is because for 
    • each of the observations, the p‐value is larger than 0.05 and, equivalently, the t‐statistic is smaller  than  1.98 (which,  as discussed  in  part (c)  above,  is the critical  value of the t‐distribution for  120  observations  at  95%  confidence  level).  We  therefore  conclude  that  the  intercept  in  the  CAPM  regression is the same for January and the remaining 11 months of the year. As can be observed, the  common intercept for the remaining months of the year is not significantly different from zero in  every regression.    We now use the above results to test the null hypothesis that “January is better”. In this situation,  the said hypothesis corresponds to a one‐sided test for the DUMJ variable. Therefore, we have:    H0: ß1 = 0  Ha: ß1 > 0    In this test, we use the t‐statistic of the ß1 parameter and compare each of the observations with  1.658, which is the critical value for 120 observations from the t‐distribution at the 95% confidence  level for a one‐sided test. We can observe from our table of results that the t‐statistic is less than  1.658 in all the cases.    We therefore conclude that the “January is better” hypothesis is false for all our chosen companies.    (f) From the above analysis, we can conclude that January is not different.    We assumed in part (a) that the “January premium” affected the returns of both the risk‐free and  the risky assets. In part (b), we assumed that the premium affected only the risky assets returns.  However,  as  observed  in  both  parts  (a)  and  (b),  if  a  “January  premium”  does  exist,  it  cannot  be  tested for within the CAPM framework.     We investigated an alternative methodology of testing the “January is different” hypothesis in part  (c)  by  using  6  companies  from  3  different  industries  (viz.  Computers,  Foods  and  Banks).  By  introducing a dummy variable for January (DUMJ) and running the regression: rp = α + ß (DUMJ),  we rejected the hypothesis that “January is different” at 5% significance level for every company.     In  part  (e)  we  investigated  yet  another  way  of  analyzing  the  given  hypothesis  by  allowing  for  a  difference only in the intercept term within the CAPM framework. By running the regression: rp – rf = α + ß1(DUMJ) + ß2(rm – rf) + ε and analyzing the subsequent results, we concluded that at a 95%  confidence  interval,  the  intercept  does  not  change  significantly  in  January  for  all  the  chosen  companies.     Hence, based on the results in each part of the given exercise we are in a position to conclude that  the returns and the risk‐premiums are not significantly different in January as compared to the other  months of the year, i.e., January is not different.                     
    • APPENDIX    Programming Package used: STATGRAPHICS Centurion XVI, Microsoft Excel 2010    Sample regression analysis summary for Exercise 2, Part (a):  Regression: rp – rf = α + ß (rm − rf ) for Delta:   Simple Regression - delta_last_60-Rf vs. Rm-Rf Dependent variable: delta_last_60-Rf Independent variable: Rm-Rf Linear model: Y = a + b*X Coefficients Least Squares Standard T Parameter Estimate Error Statistic P-Value Intercept -0.00429631 0.0112209 -0.382883 0.7032 Slope 0.639525 0.189369 3.37714 0.0013 Analysis of Variance Source Sum of Squares Df Mean Square F-Ratio P-Value Model 0.0859235 1 0.0859235 11.41 0.0013 Residual 0.43696 58 0.00753379 Total (Corr.) 0.522883 59 Correlation Coefficient = 0.405372 R-squared = 16.4326 percent R-squared (adjusted for d.f.) = 14.9918 percent Standard Error of Est. = 0.0867974 Mean absolute error = 0.0681243 Durbin-Watson statistic = 2.16048 (P=0.7273) Lag 1 residual autocorrelation = -0.0871333 The StatAdvisor The output shows the results of fitting a linear model to describe the relationship between delta_last_60-Rf and Rm-Rf. The equation of the fitted model is delta_last_60-Rf = -0.00429631 + 0.639525*Rm-Rf Since the P-value in the ANOVA table is less than 0.05, there is a statistically significant relationship between delta_last_60-Rf and Rm-Rf at the 95.0% confidence level. The R-Squared statistic indicates that the model as fitted explains 16.4326% of the variability in delta_last_60-Rf. The correlation coefficient equals 0.405372, indicating a relatively weak relationship between the variables. The standard error of the estimate shows the standard deviation of the residuals to be 0.0867974. This value can be used to construct prediction limits for new observations by selecting the Forecasts option from the text menu. The mean absolute error (MAE) of 0.0681243 is the average value of the residuals. The Durbin- Watson (DW) statistic tests the residuals to determine if there is any significant correlation based on the order in which they occur in your data file. Since the P-value is greater than 0.05, there is no indication of serial autocorrelation in the residuals at the 95.0% confidence level.
    • Sample regression analysis summary for Exercise 8, Part (c):  Regression: rp = α + ß (DUMJ) for Continental Illinois:   Simple Regression - contil vs. DUMJ Dependent variable: contil Independent variable: DUMJ Linear model: Y = a + b*X Coefficients Least Squares Standard T Parameter Estimate Error Statistic P-Value Intercept -0.00648182 0.0143269 -0.452422 0.6518 Slope 0.0645818 0.04963 1.30127 0.1957 Analysis of Variance Source Sum of Squares Df Mean Square F-Ratio P-Value Model 0.0382324 1 0.0382324 1.69 0.1957 Residual 2.66429 118 0.0225787 Total (Corr.) 2.70252 119 Correlation Coefficient = 0.118941 R-squared = 1.4147 percent R-squared (adjusted for d.f.) = 0.579227 percent Standard Error of Est. = 0.150262 Mean absolute error = 0.0875947 Durbin-Watson statistic = 1.93443 (P=0.3605) Lag 1 residual autocorrelation = 0.0256169 The StatAdvisor The output shows the results of fitting a linear model to describe the relationship between contil and DUMJ. The equation of the fitted model is contil = -0.00648182 + 0.0645818*DUMJ Since the P-value in the ANOVA table is greater or equal to 0.05, there is not a statistically significant relationship between contil and DUMJ at the 95.0% or higher confidence level. The R-Squared statistic indicates that the model as fitted explains 1.4147% of the variability in contil. The correlation coefficient equals 0.118941, indicating a relatively weak relationship between the Plot of Fitted Model delta_last_60-Rf = -0.00429631 + 0.639525*Rm-Rf -0.27 -0.17 -0.07 0.03 0.13 0.23 Rm-Rf -0.27 -0.17 -0.07 0.03 0.13 0.23 delta_last_60-Rf
    • variables. The standard error of the estimate shows the standard deviation of the residuals to be 0.150262. This value can be used to construct prediction limits for new observations by selecting the Forecasts option from the text menu. The mean absolute error (MAE) of 0.0875947 is the average value of the residuals. The Durbin- Watson (DW) statistic tests the residuals to determine if there is any significant correlation based on the order in which they occur in your data file. Since the P-value is greater than 0.05, there is no indication of serial autocorrelation in the residuals at the 95.0% confidence level.   Sample regression analysis summary for Exercise 8, Part (e):  Regression: rp – rf = α + ß1(DUMJ) + ß2(rm – rf) + ε for Citicorp  Multiple Regression - citicrp-rkfree Dependent variable: citicrp-rkfree Independent variables: DUMJ market-rkfree Standard T Parameter Estimate Error Statistic P-Value CONSTANT 0.00129034 0.00643722 0.20045 0.8415 DUMJ -0.0128418 0.0222961 -0.575967 0.5657 market-rkfree 0.670978 0.0901989 7.43887 0.0000 Analysis of Variance Source Sum of Squares Df Mean Square F-Ratio P-Value Model 0.250693 2 0.125346 27.67 0.0000 Residual 0.530045 117 0.0045303 Total (Corr.) 0.780738 119 R-squared = 32.1097 percent R-squared (adjusted for d.f.) = 30.9492 percent Standard Error of Est. = 0.0673075 Mean absolute error = 0.0518742 Durbin-Watson statistic = 1.84745 (P=0.2028) Lag 1 residual autocorrelation = 0.0708349 Plot of Fitted Model contil = -0.00648182 + 0.0645818*DUMJ 0 0.2 0.4 0.6 0.8 1 DUMJ -0.6 -0.2 0.2 0.6 1 contil
    • The StatAdvisor The output shows the results of fitting a multiple linear regression model to describe the relationship between citicrp-rkfree and 2 independent variables. The equation of the fitted model is citicrp-rkfree = 0.00129034 - 0.0128418*DUMJ + 0.670978*market-rkfree Since the P-value in the ANOVA table is less than 0.05, there is a statistically significant relationship between the variables at the 95.0% confidence level. The R-Squared statistic indicates that the model as fitted explains 32.1097% of the variability in citicrp-rkfree. The adjusted R-squared statistic, which is more suitable for comparing models with different numbers of independent variables, is 30.9492%. The standard error of the estimate shows the standard deviation of the residuals to be 0.0673075. This value can be used to construct prediction limits for new observations by selecting the Reports option from the text menu. The mean absolute error (MAE) of 0.0518742 is the average value of the residuals. The Durbin-Watson (DW) statistic tests the residuals to determine if there is any significant correlation based on the order in which they occur in your data file. Since the P-value is greater than 0.05, there is no indication of serial autocorrelation in the residuals at the 95.0% confidence level. In determining whether the model can be simplified, notice that the highest P-value on the independent variables is 0.5657, belonging to DUMJ. Since the P-value is greater or equal to 0.05, that term is not statistically significant at the 95.0% or higher confidence level. Consequently, you should consider removing DUMJ from the model.     REFERENCES    [1] Berndt, "The Practice of Econometrics; Chapter 2 – The Capital Asset Pricing Model: An Application of Bivariate Regression Analysis” [2] Prof. Dr. Bernhard Schipp, Course Script: “Financial Markets and Financial Institutions (Essentials of Quantitative Finance)” Plot of citicrp-rkfree -0.29 -0.09 0.11 0.31 0.51 predicted -0.29 -0.09 0.11 0.31 0.51 observed