Questions

Q1

What is the simple linear regression equation for the best-fitting straight line which predicts the rate of return for the stock from the rate of return of the market?

The parameter representing the \(y\)-intercept obtained from running the simple linear regression of ACME stock returns on market returns is \(0.00416\), while the slope is \(0.73659\).

\[ \begin{equation*} R_{stock} = \alpha + \beta_{stock} \cdot (R_{Market}) + u \end{equation*} \]

Substituting these values give the following ways of describing the linear regression equation:

\[ \begin{equation*} R_{stock} = 0.00416 + 0.73659 \cdot (R_{Market}) + u \end{equation*} \]

\[ \begin{equation*} \widehat{R_{stock}} = 0.00416 + 0.73659 \cdot (R_{Market}) \end{equation*} \]

Q2

Can you be fairly certain that return on the market provides any predictability of the return on ACME in the population? (make sure to state the Null and Alternative Hypotheses)

In order to test if market returns provide any predictability of the return on ACME, we make the following hypotheses, i.e the null hypothesis and alternate hypothesis:

\[ \begin{equation*} H_0:\beta_{stock} = 0 \end{equation*}\] \[\begin{equation*} H_a:\beta_{stock} \ne 0 \end{equation*} \] We can say that the market provides any predictability of the return on ACME in the population if the \(p\)-value obtained from the \(F\)-value is significant. The \(p\)-value (\(<0.0001\)) obtained from the \(F\)-value (\(48.56\)) indicates that there is statistical significance at a \(95\%\) confidence level. Therefore, we can reject \(H_0\) with reasonable confidence. We are reasonably confident that at least one of the independent variables offers some predictability of the dependent variable in the population.

Q3

What is your best estimate for the Return on ACME assuming the return on the Market in a certain period is 2%?

Assuming the return on the market in a certain period is \(2\% \Longleftrightarrow R_{Market}=0.02\), we plug this into the regression equation and get our best estimate.

\[ \begin{equation*} \widehat{R_{stock}} = 0.00416 + 0.73659 \cdot (R_{Market}) \end{equation*} \] \[ \begin{equation*} \widehat{R_{stock}} = 0.00416 + 0.73659 \cdot (0.02) \end{equation*} \] \[ \begin{equation*} \widehat{R_{stock}} = 0.0188918 \end{equation*} \] \[ \begin{equation*} \widehat{R_{stock}} =1.88918\% \end{equation*} \]

Thus, the best estimate we have for a stock given a \(2\%\) market return is \(1.88918\%\).

The brown star above shows the point \((2, 1.88918)\), which means the expected return of the stock is \(1.89\%\) when the weekly market return is \(2\%\).

Q4

What proportion of the total variation in ACME’s returns is predictable from the market’s return (what is the coefficient of determination)?

The coefficient of determination can be found through the \(R^2\) value derived from the SAS output. In this case, the \(R^2\) is equal to \(0.4927\) meaning that \(49.27\%\) of the proportion of the total variation in ACME’s returns is predictable from the market’s return in the sample. The adjusted-\(R^2\), which is \(48.25\%\) represents the proportion of the total variation in ACME’s returns that is predictable from the market’s return in the population.

Q5

Between what values do you expect ACME’s beta to fall 90% of the time? (calculate the confidence interval for the slope with an alpha of 10%)

Through the regression output with an additional condition in the SAS code, we found that a \(90\%\) confidence interval has a value of \(\alpha = 0.10\). The SAS output gives a confidence interval of \([0.55944, 0.91375]\), meaning we are \(90\%\) confident that the true population parameter representing ACME’s beta is between \(0.55944\) and \(0.91375\). Another way to compute the confidence interval is by substituting the appropriate parameters into the formula. The expression of the confidence interval is given by:

\[ [\widehat{\beta}_x -t_{\alpha/2}\cdot s_{\beta x}, \widehat{\beta}_x +t_{\alpha/2}\cdot s_{\beta x}] \]

\[[0.73659 - t_{\alpha/2} \cdot 0.10571,0.73659 + t_{\alpha/2} \cdot 0.10571]\]

The \(t\)-value given the alpha and degrees of freedom can be computed in a few ways. The first way is to use a table. We know \(n=52\) and \(p=2\), representing the number of observations and number of parameters, respectively, meaning the degrees of freedom is \(50\). Using the table from here, \(t_{0.05}=1.676\). Substituting this into the prior expression gives:

\[[0.73659 - 1.676 \cdot 0.10571,0.73659 + 1.676 \cdot 0.10571]\] \[[0.55944, 0.91375]\]

Alternatively, using R, we can obtain the \(t\)-value using the fact that \(1-\frac{\alpha}{2}=0.95\) and \(df=n-p=52-2=50\) to find the \(t\)-value to be \(1.676\) using the below function.

qt(0.95, 50)
## [1] 1.675905

Q6

What is the value of the standard deviation of the residuals from this model?

Using the proc means function in SAS, we obtained the standard deviation of the residuals to be \(0.0141073\) or \(1.41073\%\). Furthermore, we verified these results in R by running:

#fix this
sd(merged$reg_residuals)
## [1] 0.01350979

SAS

Code

Importing the datasets

proc import datafile= “/home/u63973439/Financial Modeling and Econometrics - FINA 6271/Assignments/Case 2/case2data.xlsx” dbms=xlsx out=ACME replace; Sheet=“ACME”;

proc import datafile= “/home/u63973439/Financial Modeling and Econometrics - FINA 6271/Assignments/Case 2/case2data.xlsx” dbms=xlsx out=MKT replace; Sheet=“MKT”;

Merging the datasets

data merged; merge ACME MKT; by week;

Lagging the columns and finding the weekly returns of the stock and market

data merged; set merged; input = week; lacme = lag1(ACME); lmkt = lag1(MKT); mktreturns = (MKT - lmkt)/lmkt; acmereturns = (ACME - lacme)/lacme;

Running the regression and getting the residuals

proc reg data = merged; model acmereturns = mktreturns/clb alpha = 0.1; output out=merged r=residuals;

Retrieving information about the standard deviation of residuals

proc means data=merged; var residuals;

Printing the datasets

proc print data=merged; run;

Output

\[ \begin{equation*} \mathrm{\underline{Regression \; Information}:} \end{equation*} \]

\[ \begin{equation*} \mathrm{\underline{Residual \; Information}:} \end{equation*} \]