Finance-Theory · Unit 18 · Video 8 · Interactive Practice

Testing the CAPM: Is Alpha Zero? Biogen vs Motorola

IKey Formulas

Regression	Form	What the CAPM predicts
$R_i - R_f = \alpha_i + \beta_i(R_m - R_f) + \epsilon$	Excess returns	$\alpha_i = 0$
$R_i = \alpha_i + \beta_i R_m + \epsilon$	Total returns	$\alpha_i = R_f(1 - \beta_i)$
$\alpha = \text{intercept} - R_f(1 - \beta)$	Alpha from a total-return fit	The gap between estimate and prediction
$\alpha / \text{SE}(\alpha)$	Standard errors from zero	About $2$ is the conventional bar

Key Insight: Every term must share one period. In a monthly regression the risk-free rate is the annual rate divided by $12$ , and $\text{E}[\epsilon] = 0$ is what lets the expected returns collapse back to the pricing equation.

IIVisualization 1 — Where the CAPM Puts the Intercept

In total returns every CAPM line pivots through $(R_f, R_f)$ , so the predicted intercept is $R_f(1 - \beta)$ .

Beta (β) 0.80 Regression form

IIIVisualization 2 — Stock A: Alpha in Standard Errors

Stock A, monthly: $\beta = 0.8$ , intercept $0.9\%$ (SE $0.6\%$ ), $R_f = 3.6\%$ a year. Is its alpha distinguishable from zero?

Risk-free rate entered as

💡 Economic size and statistical precision are separate questions: Biogen's alpha of about $21\%$ a year was just as inconclusive.

IVQuiz Questions

Problem 1 · The Predicted Intercept

Given: Stock B's monthly total returns are regressed on the market's monthly total return, $R_i = \alpha_i + \beta_i R_m + \epsilon$ , giving $\beta = 1.25$ . The risk-free rate is $3.6\%$ a year. Find the intercept the CAPM predicts, per month.

✅ Correct! With

\beta \gt 1

the line is steeper than

\beta = 1

, so it crosses the vertical axis below zero:

0.3\% \times (1 - 1.25) = -0.075\%

❌ Check the sign. The prediction is

R_f(1 - \beta)

, not

R_f(\beta - 1)

. With

\beta = 1.25

the factor

1 - \beta

is negative.

❌ Not quite. A zero intercept is the prediction for the excess-return regression. This one is in total returns, where the CAPM predicts

R_f(1 - \beta)

❌ Units mismatch. That uses the annual

3.6\%

. The regression is monthly, so the risk-free rate must be

3.6\% / 12 = 0.3\%

per month.

Show solution

Expand the pricing relation in total returns: $R_i = R_f + \beta_i R_m - \beta_i R_f + \epsilon$ , so the CAPM intercept is $R_f(1 - \beta_i)$ .

The regression is monthly, so convert the rate first:

R_f = \frac{3.6\%}{12} = 0.3\% \text{ per month}

R_f(1 - \beta) = 0.3\% \times (1 - 1.25) = 0.3\% \times (-0.25) = -0.075\% \text{ per month}

Problem 2 · The Annual-Rate Trap

Given: Stock B as in Problem 1 ( $\beta = 1.25$ ), with an estimated intercept of $-0.1\%$ per month (standard error $0.5\%$ ). An analyst plugs the annual $3.6\%$ straight into $R_f(1 - \beta)$ . Find the predicted intercept she uses, and the correct alpha.

The predicted intercept she uses

The correct alpha

✅ Correct! Her prediction of

-0.9\%

turns a near-zero alpha into

-0.1\% - (-0.9\%) = +0.8\%

per month, an apparent

1.6

standard errors. With matching units, alpha is only

-0.025\%

per month.

❌ Not quite. She computes

3.6\% \times (1 - 1.25)

, keeping the annual figure.

❌ Check the factor. That is

3.6\% \times (-1.25)

, multiplying by

-\beta

. The prediction uses

1 - \beta = -0.25

3.6\% \times (-0.25) = -0.9\%

❌ Check the sign. With

\beta = 1.25

the factor

1 - \beta

is negative, so her (wrong-unit) prediction is negative too:

-0.9\%

❌ That is the correct prediction, not hers. She skipped the division by

12

❌ Not quite. Alpha is the estimated intercept minus the monthly prediction

R_f(1 - \beta) = -0.075\%

❌ Close, but… In a total-return regression the intercept is not alpha by itself. Subtract the CAPM prediction

-0.075\%

first.

❌ That is the analyst's number. It sets the annual

3.6\%

against a monthly intercept.

❌ Check the subtraction. Subtracting a negative prediction adds:

-0.1\% - (-0.075\%) = -0.1\% + 0.075\%

Show solution

Her prediction (annual rate, wrong):

3.6\% \times (1 - 1.25) = 3.6\% \times (-0.25) = -0.9\%

She would report $-0.1\% - (-0.9\%) = 0.8\%$ per month, which is $0.8 / 0.5 = 1.6$ standard errors, a spurious signal created by the units.

Correct (monthly rate): $R_f = 3.6\% / 12 = 0.3\%$ , so the prediction is $0.3\% \times (-0.25) = -0.075\%$ and

\alpha = -0.1\% - (-0.075\%) = -0.025\% \text{ per month}

Problem 3 · Does Stock B Reject the CAPM?

Given: Stock B's correct alpha is $-0.025\%$ per month, and its estimated intercept has a standard error of $0.5\%$ . Find how many standard errors alpha lies from zero, and the verdict.

Standard errors from zero

Verdict

✅ Correct! An estimate a twentieth of a standard error from zero is exactly what

\alpha = 0

plus noise looks like.

❌ Not quite. Divide the estimate by its standard error:

\alpha / \text{SE}

❌ Close, but… That divides the raw intercept

-0.1\%

. The test is on alpha, the intercept minus the CAPM prediction.

❌ Inverted. The count is estimate over standard error,

-0.025 / 0.5

, not the other way round.

❌ That is the units-mismatch alpha

0.8\%

divided by

0.5\%

. Use the correct alpha.

❌ Not quite. An estimate is never exactly zero, even when the true alpha is. What matters is its distance from zero in standard errors: here

0.05

, against a bar of about

2

❌ Not quite. In total returns the CAPM does not predict a zero intercept; it predicts

R_f(1 - \beta) = -0.075\%

. The test is on alpha,

-0.025\%

, only

0.05

standard errors from zero.

❌ Not needed. The test uses only the regression output and the risk-free rate: alpha

= -0.025\%

0.05

standard errors from zero, far below about

2

❌ Not quite. An estimate is never exactly zero; the question is whether it is far from zero relative to its standard error, with about

2

as the bar.

Show solution

\frac{\alpha}{\text{SE}} = \frac{-0.025\%}{0.5\%} = -0.05

Alpha is $0.05$ standard errors below zero, nowhere near the conventional bar of about $2$ . Stock B is consistent with the CAPM, just as Motorola was.

Problem 4 · Same Alpha, Sharper Estimate

Given: Stock A's alpha is $0.84\%$ per month (intercept $0.9\%$ , $\beta = 0.8$ , $R_f = 0.3\%$ per month). Suppose a longer sample gives the same estimates but the intercept's standard error falls from $0.6\%$ to $0.35\%$ . Find the new number of standard errors and the verdict.

Standard errors from zero

Verdict

✅ Correct! Precision, not size, moved the verdict: the same

0.84\%

is now

2.4

standard errors from zero.

❌ Not quite. Divide the unchanged alpha by the new standard error.

❌ That is the old count,

0.84 / 0.6

. Use the new standard error

0.35\%

❌ Close, but… That divides the intercept

0.9\%

. Alpha is

0.9\% - 0.06\% = 0.84\%

❌ Inverted. The count is estimate over standard error, not standard error over estimate.

❌ Not quite. The estimate is unchanged, but the verdict depends on it relative to its standard error:

0.84 / 0.35 = 2.4

now clears the bar of about

2

❌ Not quite. Economic size alone decides nothing: the same

\approx 10\%

a year sat at

1.4

standard errors and could not be rejected. The rejection comes from

2.4

standard errors.

❌ Not quite. The standard error does not change alpha; it stays

0.84\%

. A smaller standard error makes the same alpha more standard errors from zero:

2.4

, past the bar of about

2

❌ Not quite. The verdict depends on alpha measured in standard errors. The economic size was the same at

1.4

standard errors, when the test could not reject.

Show solution

Alpha is unchanged: $0.9\% - 0.3\% \times (1 - 0.8) = 0.9\% - 0.06\% = 0.84\%$ per month.

\frac{0.84\%}{0.35\%} = 2.4

At $2.4$ standard errors the estimate clears the conventional bar of about $2$ , so $\alpha = 0$ is rejected for stock A. Annualised, $0.84\% \times 12 \approx 10.1\%$ a year: the same economic size that, at $1.4$ standard errors, could not be told apart from zero.

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