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CAPM Betas The Capital Asset Pricing Model (“CAPM”)

[ Rs- Rf ] = b0 + b1 [ RM - Rf ] + e

• People commonly refer to the b0 in this model as the stock’s “alpha” and the b1 is simply called “beta.”

• “R” stands for return. The subscript “s” indicates that the model is for stock “s” (e.g., Coca Cola or Microsoft). The subscript “f” stands for the risk-free security (e.g., a 30-day Treasury bill). The subscript “M” stands for the stock market.

• [ Rs- Rf ] is the risk premium of a stock; [ RM - Rf ] is the risk premium of the market.

• The “e” designates the error term. • If you work with daily data, all your returns should be daily returns and you

are calculating a daily beta of the stock. If you work with monthly data, all your returns should be monthly returns and you are calculating the stock’s monthly beta. Etc.

• If you want to calculate the CAPM beta for a mutual fund or an investment portfolio, just use the returns from the mutual fund or portfolio instead of the returns of an individual stock.

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CAPM Betas (continued) From earlier slide, the Capital Asset Pricing Model (“CAPM”) is:

[ Rs- Rf ] = b0 + b1 [ RM - Rf ] + e

To estimate b0 ( “alpha”) and the b1 (CAPM “beta”), you estimate the model (i.e., perform the OLS regression):

y = b0 + b1 x + e

Note that [ Rs- Rf ] is “y” and [ RM - Rf ] is “x”. Don’t reverse them!

You’ll need to perform all the steps.• Step 1: Collect the Raw Data.• Step 2: Calculate descriptive statistics.• Step 3: Plot the data.• Step 4: Specify the model.• Step 5: Estimate the model (i.e., perform the regression). • Step 6: Determine whether the model is good in a statistical

sense.

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Can outdoor temperature be estimated using

cricket chirps? Step 1 – Collect the Raw Data:

Go outside some evening with a thermometer. Count the number of cricket chirps you hear for 15 seconds. Repeat on various evenings when the temperature is different.

Temperature Chirps57 1860 2064 2165 2368 2771 3074 3477 39

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Can outdoor temperature be estimated using

cricket chirps? Step 2 – Calculate descriptive statistics:

Can use Excel’s formulas for mean and sample standard deviation.

Temperature Chirps57 1860 2064 2165 2368 2771 3074 3477 39

Mean (Excel f x = AVERAGE) 67.0 26.5

Std. Dev. (Excel f x = STDEV) 6.845 7.387

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Can outdoor temperature be estimated using

cricket chirps? Step 3 – Plot the data:

Can use Excel’s XY (Scatter) chart type. Temperature is “y”. Chirps is “x”.

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Can outdoor temperature be estimated using

cricket chirps? Step 4 – Specify the model:

y = b0 + b1 x + e

• Temperature is “y”. • Chirps is “x”.• “e” is “error”, which is the difference between our model’s prediction of

temperature and the temperature we actually observed.

• We’ll estimate b0 and b1 using ordinary least squares (“OLS”) regression.

• If our model’s any good, once we’ve estimated b0 and b1 , we can use the following formula to estimate temperature just by counting cricket chirps:

Temperature = b0 + (b1 )( # of chirps) + 0

• Note: our model assumes that “y” and “x” have a linear relationship --- i.e., our model is the equation of a line with intercept b0 and

slope b1 .

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How will we determine the best line? We’ll use Excel to perform an ordinary least squares (OLS) regression.

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Can outdoor temperature be estimated using

cricket chirps? Step 5 – Estimate the model (i.e., perform the regression):

y = b0 + b1 x + e

• We’ll need Excel’s “Analysis ToolPak” Add-In. You may already have added it. If not, here’s how you do it:– For Excel 1997-2003: On the tool bar, click “Tools”, “Add-Ins”, “Analysis

ToolPak”. – For Excel 2007: On the upper left-hand corner “MS” logo, click “Excel

Options”, “Add-Ins”, “Analysis ToolPak”.

• If you’ve added the “Analysis ToolPak” Add-In., you’ll be able to:– For Excel 1997-2003: On the tool bar, click “Tools”, “Data analysis”,

“Regression”. – For Excel 2007: On the tool bar, click “Data”, “Data analysis”,

“Regression”.

>>>>

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Can outdoor temperature be estimated using

cricket chirps?

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Can outdoor temperature be estimated using

cricket chirps?

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Can outdoor temperature be estimated using

cricket chirps?

y = b0 + b1 x + e

= 42.997 + 0.906 x + e

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Best estimate: Temperature = 42.997 + (0.906)(Chirps) + e. In other words, the best line intercepts the y-axis at about 43, and it has a slope of about 0.9.

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For each of our temperature observations, we can determine how much our model’s prediction for temperature is in error.

“Predicted y” = b0 + b1 x + e

= 42.997 + 0.906 x + 0

Error (“e”) = “Predicted y” – “Actual y”.

For 1st observation, “Predicted y” = 42.997 + (0.906)(18) = 59.30.

Error = 59.30 – 57 = 2.30.

Temp. ChirpsPredicted

y Error (e)57 18 59.30 2.3060 20 61.11 1.1164 21 62.02 -1.9865 23 63.83 -1.1768 27 67.45 -0.5571 30 70.17 -0.8374 34 73.79 -0.2177 39 78.32 1.32

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For each of our temperature observations, we can determine how much our model’s prediction for temperature is in error.

“Predicted y” = b0 + b1 x + e

= 42.997 + 0.906 x + 0

Error (“e”) = “Predicted y” – “Actual y”.

For 1st observation, “Predicted y” = 42.997 + (0.906)(18) = 59.30.

Error = 59.30 – 57 = 2.30.

Ordinary least squares (OLS) finds the line that minimizes the sum of

the squared errors (e2).

Temp. ChirpsPredicted

y Error (e) e2

57 18 59.30 2.30 5.2960 20 61.11 1.11 1.2464 21 62.02 -1.98 3.9365 23 63.83 -1.17 1.3768 27 67.45 -0.55 0.3071 30 70.17 -0.83 0.6974 34 73.79 -0.21 0.0477 39 78.32 1.32 1.75

14.61 = sum of the e2

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It looks like our model is pretty good. We can tell how good it is in a statistical sense by examining the Adjusted R Square (R2) of the regression and the t-statistics and p-values of the coefficients.

Adjusted R2 = 94.8%.

This means our model has explained 94.8% of the variance of Temperature.

You could look up these T-stats in a statistical table to determine whether they were significantly different from 0.

It’s easier just to use the P-values.

The P-values are the probability of obtaining our coefficients (b0 , b1) if they were truly equal to 0.

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Step 6 – Determine whether the model is good in a statistical sense. The p-value of each of the coefficients is less than 5%. Therefore, we say each coefficient is significantly different from zero at the 5% level.The p-value of the F-stat tells us how good the model is overall. It also has a p-value less than 5%. Therefore, we say the regression is significant at the 5% level.

Parameter b 0 : "Intercept"

1) Coefficient estimate 42.9972) t-stat 19.673) p-value 1.12E-064) Is this coefficient significantly different

from zero at the 5% significance level? yes -- 0.000001 < .05

Parameter b 1 : "# of Chirps"

1) Coefficient estimate 0.90582) t-stat 11.353) p-value 2.81E-054) Is this coefficient significantly different

from zero at the 5% significance level? yes -- 0.000028 < .05

MEMO:1) Adjusted R-squared 0.9482) F-stat 128.733) p-value of the F-stat 2.81E-054) Is the regression significant

at the 5% level? yes -- 0.000028 < .05

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Suppose the p-value of the coefficient for CHIRPS had been > 5%. We would have concluded that it was NOT significantly different from zero at the 5% level.In that case, we would have to conclude that our best model for estimating temperature would be: Temperature = b0 . And b0 would just equal the average of our Temperatures (67 degrees). We would have to conclude that CHIRPS are of no help for determining outdoor temperature.Parameter b 0 : "Intercept"

1) Coefficient estimate 42.9972) t-stat 19.673) p-value 1.12E-064) Is this coefficient significantly different

from zero at the 5% significance level? yes -- 0.000001 < .05

Parameter b 1 : "# of Chirps"

1) Coefficient estimate 0.90582) t-stat 11.353) p-value 2.81E-054) Is this coefficient significantly different

from zero at the 5% significance level? yes -- 0.000028 < .05

MEMO:1) Adjusted R-squared 0.9482) F-stat 128.733) p-value of the F-stat 2.81E-054) Is the regression significant

at the 5% level? yes -- 0.000028 < .05