Managerial economics guide 5

CAPM

Equilibrium assumption

- Open market

- All the risky assets refer to all the tradeable stocks available to all

- There is a risk-free asset (for borrowing and/or lending in unlimited quantities) with

interest rate r f

- All information is available to all such as covariances, variances, mean rates of return of stocks

and so on.

- Everyone is a risk-averse rational investor who uses the same nancial engineering

mean-variance portfolio theory from Markowitz.

→ Everyone has a portfolio on the same efficient frontier, and hence has a portfolio that is a

mixture of the risk-free asset and a unique efficient fund F (of risky assets). In other words,

everyone sets up the same optimization problem, does the same calculation, gets the same

answer and chooses a portfolio accordingly.

→This efficient fund used by all is called the market portfolio and is denoted by M. The fact

that it is the same for all leads us to conclude that it should be computable without using all

the optimization methods from Markowitz: The market has already reached an equilibrium so

that the weight for any asset in the market portfolio is given by its capital value (total worth

of its shares) divided by the total capital value of the whole market (all assets together).

CAPM on gretl

Time series data google_for_CAPM

β

The most important parameter is . This parameter measures the systematic risk of the portfolio.

Using the google and Nasdaq data, we will perform a simple regression with the returns of one specific

stock as the dependent variable, in this case we have Google’s returns over the constant and the market

returns

What we will use is the OLS,

Gretl: File > Open data > google_for_CAPM.csv

Model > Ordinary least square > Dependent variable: google return , independent: Nasdaq

return

We have 52 close price but only 51 observations

β = 1

.06419

Here we have , you will find this value under coefficient, this is also the beta of google

i

return, it implies that Google’s stock is slightly riskier than the overall market (beta>1)

Analysis > confidence intervals for coefficients

95% confidence interval means that 95% of the interval will contain the true parameter

The confidence interval for beta goes from 0.69 to 1.43, this means that the estimation for beta is not

precise because 0.69 means the asset’s aggregate risk is much smaller than that of the market and 1.4

means the asset’s aggregate risk is much higher than that of the market. The larger extreme of the

interval also exaggerate the variance of the model and increase the overall riskiness of the market.

The standard error, square root of the variance of the estimates. We have low sd for the constant and

large sd for beta.

Test of significance of the coefficient

Can we reject the null hypothesis that beta=0?

If you reject the H0 that beta=0 means that you accept the estimated coefficient and that variable has

some effects and can somehow explain the dependent variable and vice versa. Otherwise, you can

eliminate that variable from your model. (β−β )

H 0

t-ratio represent the statistic test we need, which is calculated by . this is the discrepancy

Standard error

measure. If the threshold level is greater than the t-ratio I reject the null hypothesis otherwise, I’ll accept

it.

We don’t see a threshold value here but we have the p-value. The p-value is the probability of getting a

α

)

number greater than the t-ratio, we then compare this to our significant level ( , if p-value is greater

α

than , we accept the null hypotheses, otherwise we can reject it.

We can see that the Nasdaq return’s p-value is smaller than its t-ratio, so we reject the null hypothesis

and accept Nasdaq return as a significant variable whose coefficient will never be 0. Of course, in real

life this number won’t be always be 1.06 because the two extreme of the confidence interval are very

different as explained above.

We have the mean of the dependent variable which represent the average return of Google’s stock

The sum squared error represents the difference between real returns of Google’s stock and their

estimated value. For the model to be a good fit, we need to minimize this value and as we see here it is

quite small.

R-squared of 0.41 means 41% of real values is captured by the model or 59% of the real values is

captured by the model.

F test is the test of co significant excluding the constant value but here we only have 1 independent

variable so the result won’t be different from that of the t-test, we are rejecting the null hypothesis

Standard error of the regression gives you an idea of the SD of the disturbances

If the disturbances has the same variance, our estimation is not precise anymore thanks to

heterokedascity. So to test this problem we can use White’s test

Test > Heterokedascity > White’s test

We can see that the dependent variable is the residuals so the question here is whether the variance of

the shocks depends on either of the listed independent variable. So, if the coefficients can jointly be 0

(our null hypothesis), then the variance of the shocks doesn’t depend on these variable and there is no

heterokedascity.

We can see that the p-value is high so we should accept the null hypothesis,

If we had heterokedascity, we would have received the robust parameter in the final model

if we are interested in the normality of the residual we can run the test Test> Normality of the residual

this test compare the %observation in our variable and the %observation that would be there if it were

a normal distribution

the p value is inside the bracket (0.10) and therefore is greater than our significant value, so we accept

the null hypothesis, the variable is normal

so our model is

Google return= 0.00097+1.064*Nasdaq return

Cross section data

Woodridge data on the determinants of wage,

We can see there are dummy variables, which only have two values which are represented by 0 and 1.

For example, non white=1; if the person is white, the variable takes value 0 otherwise, the variable take

value 1

These data are undated so we have a cross-section data

Question: Is wage affected by the parents’ education?

Model

lwage= meduc + feduc

Our dependent variable is log of wage

Our first independent variable is mother education and the second is father education

Sub-text:

Are the coefficients significantly different from 0? Looking at the p value, constant have a p value of 0

and t-ratio 0f 103, which lies far on the right of the threshold level and therefore we reject the null

hypothesis

Doing the same with other coefficients, we can see that they are all significantly different from 0

With this, we can read the final model as

l wage = 6

.39 + 0

.02meduc + 0

.02f educ

Another way to read this

3 stars mean that the estimate is significant at 1% statistical level)(extremely), which is extremely high- 2

stars mean their significance level is 5% and 3star is 10%,

How does this model perform overall?

F test has a value of 24 and a small p-value (6.42e-11) so we reject the null hypothesis and the model

holds

R squared is really small, only 6% of real values of lwage is captured by this model. There’s a huge

amount of unexplained variability. If we didn’t have lwage but wage as a dependent variable, what

would the coefficient stand for? It would be the derivative of wage in respect to meduc and feduc while

the constant value will be the wage at 0

The coefficients tell us how much the dependent variable increase/decrease ( in unit) when the

independent variable increase/decrease by 1 unit.

Id we have different type of regression, log of wage, log of feduc… it would represent the elasticity, so

the percentage change in the dependent variable for an unit change in the independent variable

The variables are all strongly significant

R squared is 0.25, which is not bad given that the cross section data has a lot of variability

F test, the coefficients is 44 and the pvalue is 1.16*10^-54, so you reject the null hypothesis

Wage is increasing in experience since the coefficient is .014,

Model

L

wage = 5

.39550 + 0.014 e

xper + 0.012 t

enure + 0.2 (

1if married|0 if not married)⏟

* * * you will get higher wage if youre marr

A dummy variable doesn’t have a numerical value, just assigned value

If you increase your exp by 1 year, your wage increase by 0.014 %. If you rmarried, your wage

increase by 0.2

If we don’t have a variable proxy for your skill, it maybe that the wage is hig not because your

education value is high but because you are more skilled. So if you want to know the return of

an education on your future wage, this is a bad way to do it

Lets use IQ as a proxy for skill, modify model ad see what happen