Financial Econometrics Exercise Sheet 4 Solutions PDF

FIN 3618 Financial Econometrics – Exercise sheet 4: A brief overview of the classical linear regression model – 1 Self-Assessment 1. [TRUE/FALSE] The x variable on the right hand side of a linear regression is sometimes called an explanatory variable. TRUE. 2. [TRUE/...

FIN 3618 Financial Econometrics – Exercise sheet 4: A brief overview of the classical linear regression model – 1 Self-Assessment 1. [TRUE/FALSE] The x variable on the right hand side of a linear regression is sometimes called an explanatory variable. TRUE. 2. [TRUE/FALSE] The y variable on the left hand side of a linear regression is sometimes called a regressor. FALSE. 3. [TRUE/FALSE] When computing the intercept and slope parameters with OLS, we mini- mize the squared horizontal distances between the model’s predictions and the values of the dependent variable. FALSE. 4. [TRUE/FALSE] OLS selects parameter values that minimize the sum of the squared residu- als. TRUE. 5. [TRUE/FALSE] The sample regression function contains a disturbance term. FALSE. 6. [TRUE/FALSE] A model that is non-linear and cannot be transformed into a linear model also cannot be estimated using OLS. TRUE. 7. [TRUE/FALSE] The CLRM assumes that the variance of errors is zero. FALSE. 8. [TRUE/FALSE] The CLRM assumes that the errors are normally distributed. TRUE. 9. [TRUE/FALSE] If the CLRM assumptions hold, then the OLS estimators are BLUE. TRUE. 10. [TRUE/FALSE] Consistency is a weaker condition than unbiasedness. TRUE. 11. [TRUE/FALSE] The standard error of the slope parameter is the square root of its variance. TRUE. 1 12. [TRUE/FALSE] The precision of a parameter estimate can be evaluated using its standard error. TRUE. 13. [TRUE/FALSE] Whenever we perform a hypothesis test, we will specify null and alternative hypotheses. TRUE. 14. [TRUE/FALSE] A hypothesis test may be either one-sided or two-sided. TRUE. 15. [TRUE/FALSE] The test statistic for a parameter is computed by subtracting its hypothe- sized value under the null and then dividing by its standard error. TRUE. 16. [TRUE/FALSE] The test of significance and confidence interval approaches may yield differ- ent findings with respect to the null hypothesis. FALSE. 17. [TRUE/FALSE] If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. TRUE. 18. [TRUE/FALSE] A Type II Error involves rejecting the null when it is true. FALSE. 19. [TRUE/FALSE] A t-test tells us whether a parameter value is statistically significantly dif- ferent from zero. TRUE. 2 Exercises 1. The following x- and y- values are given t x y 1 8.5 18 2 7.8 20 3 7.5 20 4 6.2 25 5 6.5 29 6 6.0 31 7 6.5 33 8 4.6 37 9 4.0 43 10 3.3 44 2 (a) Calculate the arithmetic sample mean and the sample variance of x and y. Arithmetic sample means of x and y 1 x̄ = (8.5 +... + 3.3) = 6.09 10 1 ȳ = (18 +... + 44) = 30 10 Sample variances of x and y 1 σx2 = (8.5 − 6.09)2 +... + (3.3 − 6.09)2 ≈ 2.8277   10 − 1 1 σy2 (18 − 30)2 +... + (44 − 30)2 ≈ 88.2222   = 10 − 1 (b) Compute the sample covariance and the sample correlation between x and y. Sample covariance 1 σx,y = [(8.5 − 6.09) (18 − 30) +... + (3.3 − 6.09) (44 − 30)] ≈ −15.1778 10 − 1 Sample correlation −15.1778 ρx,y = √ √ ≈ −0.9610 2.8277 · 88.2222 3 (c) Determine the sample regression function. Sample regression function: ŷt = α̂ + β̂ xt Determine β̂ P10 t=1 xt · yt − 10 · x̄ · ȳ 1690.4 − 10 · 6.09 · 30 β̂ = P10 2 = ≈ −5.3676 t=1 xt − 10 · x̄ 2 396.33 − 10 · 6.092 Determine α̂ α̂ = ȳ − β̂ x̄ = 30 − (−5.3676) · 6.09 ≈ 62.689 Sample regression function: ŷt = 62.689 − 5.3676 · xt 4 (d) In some econometrics books you can find the following formula σx,y β̂ = σx2 which looks slightly different than the one you find on the lecture slides. i. Use your results from exercise part (a) and check whether the above formula gives you the same β̂ as the one you calculated in exercise part (c). −15.1778 β̂ = ≈ −5.3676 2.8277 5 ii. Rewrite the formula from above in such a way that you end up with the one for β̂ you find on the lecture slides. σx,y β̂ = σx2 T 1 P T −1 (xt − x̄) (yt − ȳ) t=1 = T 1 (xt − x̄)2 P T −1 t=1 T P (xt − x̄) (yt − ȳ) t=1 = T (xt − x̄)2 P t=1 T P xt yt − T x̄ ȳ t=1 = T x2t − T x̄2 P t=1 6 (e) A way to measure how well our regression model fits the actual data is to look at the R2 , defined as PT 2 (ŷt − ȳ)2 R = Pt=1 T 2. (1) t=1 (y t − ȳ) One possible interpretation is the following. What we are really interested in is to explain the variability of y about its mean value ȳ, known as the total sum of squares (TSS) T X TSS = (yt − ȳ)2. t=1 Interestingly, we can use the values ŷ determined in our regression model to split the distance between yt and ȳ into two parts T X T X 2 TSS = (yt − ŷ) + (ŷt − ȳ)2 t=1 t=1 The second term describes what we can explain using our regression, therefore we refer to this as the explained sum of squares (ESS). The first term, on the other hand, is the part PT we cannot explain PT using our model. This is the residual sum of squares (RSS), i.e., 2 2 2 t=1 (yt − ŷ) = t=1 ût. Based on this line of thought, the R can also be interpreted as ESS RSS R2 = or R2 = 1 − , resp. TSS TSS i. Calculate the explained sum of squares (ESS), the residual sum of squares (RSS), the total sum of squares (TSS), and the R2. T X ESS = (ŷt − ȳ)2 = 733.2144 t=1 XT RSS = (yt − ŷ)2 = 60.7861 t=1 TSS = ESS + RSS = 794.0006 ESS R2 = = 0.9234 TSS ii. Another possible interpretation of R2 is R2 = (ρx,y )2. (2) Rewrite Equation (1) in such a way that you end up with Equation (2). 7  PT 2 2 t=1 (ŷt − ȳ) R = PT 2 t=1 (yt − ȳ) PT 1 T −1 t=1 (ŷt − ȳ)2 = T 1 ȳ)2 P T −1 t=1 (yt − 2 σ = β̂ 2 x2 σy  2 2 σx,y σx = σx2 σy2 2 σx,y = σx2 σy2  2 σx,y = σx σy 2 = ρx,y 8 2. Explain, with the use of equations, the difference between the sample regression function and the population regression function. – PRF: true model – SRF: estimated from sample 9 3. Which of the following models can be estimated (following a suitable rearrangement if neces- sary) using ordinary last squares (OLS), where X, y, Z are variables and α, β, γ are parameters to be estimated? [Hint: The models need to be linear in the parameters.] (a) yt = α + β xt + ut Yes, we can use OLS since the model is the usual linear model we have been dealing with. (b) yt = eα xβt eut Yes, the model can be linearized by taking logs on both sides and by rearranging. (c) yt = α + β γ xt + ut Yes, we can estimate this model using OLS, but we would not be able to obtain the values of both β and γ. (d) ln(yt ) = α + β ln(xt ) + ut Yes, we can use OLS, since this model is linear in the logarithms. (e) yt = α + β xt zt + ut Yes, in fact we can still use OLS since it is linear in the parameters. as usual. 10 4. The capital asset pricing model (CAPM) can be written as E(Ri ) = Rf + βi [E(Rm ) − Rf ] (3) using the standard notation. The first step is to estimate the stock’s beta using the market model. The market model can be written as e e Ri,t = αi + βi Rm,t + ui,t (4) e e where Ri,t is the excess return for security i at time t, Rm,t is the excess return on a proxy for the market portfolio at time t, and ut is an i.i.d. random disturbance term. The coefficient beta in this case is also the CAPM beta security i. Suppose that you had estimated (4) and found that the estimated value of beta for a stock, β̂, was 1.147. The standard error associated with this coefficient, SE(β̂), is estimated to be 0.0548. A city analyst has told you that this security closely follows the market, but that it is no more risky, on average, than the market. This can be tested by the null hypothesis that the value of beta is one. The model is estimated over sixty-two daily observations. Test this hypothesis against a one-sided alternative that the security is more risky than the market, at the 5% level. Write down the null and alternative hypothesis. What do you conclude? Are the analyst’s claims empirically verified? The null hypothesis is that the true (but unknown) value of beta is equal to one, against a one sided alternative that it is greater than one: H0 : β = 1 and H1 : β > 1 The test statistic is given by β̂ − β ∗ 1.147 − 1 test stat = = = 2.682 SE(β̂ 0.0548 The critical t-value from the t-table is 1.671. The value of the test statistic is in the rejection region and hence we can reject the null hypothesis. 11 5. The analyst also tells you that shares in Chris Mining plc have no systematic risk, in other words that the returns on its shares are completely unrelated to movements in the market. The value of beta and its standard error are calculated to be 0.214 and 0.186, respectively. The model is estimated over thirty-eight quarterly observations. Write down the null and alternative hypothesis. Test this null hypothesis against a two-sided alternative. The null and alternative hypotheses are: H0 : β = 0 and H1 : β ̸= 0 The test statistic has the same format as before, and is given by β̂ − β ∗ 0.214 − 0 test stat = = = 1.150 SE(β) 0.186 The critical t-value is ±2.03 Since the test statistic is not within the rejection region, we do not reject the null hypothesis. 12 6. Form and interpret a 95% and 99% confidence interval for beta using the figures from the previous question. A confidence interval for beta is given by the formula   β̂ − tcrit SE(β̂) , β̂ + tcrit SE(β̂) Confidence intervals are 2-sided unless we are told otherwise, so we want to look up the values which put 2.5% in the upper tail and 0.5% in the upper tail for the 95% and 99% confidence intervals, respectively. The critical t-values are therefore ±2.03 and ±2.72 The confidence interval in each case is thus given by – (0.214 − 2.03 · 0.186 , 0.214 + 2.03 · 0.186) = (−0.164 , 0.592) for a 95% confidence interval – (0.214 − 2.72 · 0.186 , 0.214 + 2.72 · 0.186) = (−0.292 , 0.720) for a 99% confidence interval 13 Are hypotheses tested concerning the actual values of the coefficients (i.e., β) or their estimated values (i.e., β̂) and why? We test hypotheses about the actual coefficients, not the estimated values. 14 Using R: Download the Excel file “capm.xls” from the “data” directory in itslearning. It contains monthly time series of prices for the S&P500 index, Ford, General Electric, Microsoft, and Oracle from January 2002 to February 2018. Besides the file also contains the monthly time series of annualised US-Treasury bill (T-bill) yields for three months to maturity. Select one of the stock price series and estimate a CAPM beta for that stock. Test the null hypothesis that the true beta (slope) is one and also test the null hypothesis that the true alpha (intercept) is zero. What are your conclusions? R file: chp 4-3 capm.R Estimates, standard errors (first line in parentheses), and t-stats (second line in parentheses) Note that switching from stock prices to returns reduces the original sample size T from 194 to 193. You can’t compute a return at t = 0 because the information about the price at t − 1 is missing. Critical value tcrit = tT −2,α = t193−2, 5% = ±1.9725 Ford (RFord − rf )t = −0.9560 + 1.8898 (RM − rf )t (0.7931) (0.1916) (−1.2054) (9.8620) – H0 : β = 1 vs. H1 : β ̸= 1 (two-sided) 1.8898 − 1 test stat = = 4.6433 0.1916 → −1.9725 < 1.9725 < 4.6433 → reject H0 – H0 : α = 0 vs. H1 : α ̸= 0 (two-sided) −0.9560 − 0 test stat = = −1.2054 0.7931 → −1.9725 < −1.2054 < 1.9725 → do not reject H0 – both conclusions are confirmed by the p-values corresponding to the tests ∗ H0 : β = 1: p = 0.0000, i.e., reject H0 ∗ H0 : α = 0: p = 0.2295, i.e., do not reject H0 15