econ 321 midterm prep

7/27/2019 ECON 321 Midterm Prep

1/59

ECON 321 Midterm


2/59

Summation Operator

xi= x1+ x2+ + xn

For any constant c, c = nc

For any constants a and b, (axi+

byi )=

a xi+

b yii=1

n

i=1

n

i=1

n


3/59

Statistics

If x~N(, 2), then

Expectation:

If c is a constant, then E(c) = c

If {a1, a2, , an} are constants and {x1, x2, , xn}

are random variables, then

z= x ms ~N(0,1)

E(a1x1+a2x2+...+anxn)= a1E(x1)+a2E(x2)+...+anE(xn)


4/59

Statistics (Continued)

Variance and Covariance:

If c is a constant, then Var(c) = 0

Cov(x,y) =E[(x m x )(y m y )]=E{[x E(x)][y E(y)]}

=E(xy) E(x)E(y)

=E[(x m x )y]=E[x(y m y )]


5/59


If x and y are independent, then Cov(x,y) = 0

because E(xy) = E(x)E(y) when they are

independent

However, zero covariance between x and y

does not necessarily imply that x and y are

independent


6/59


Try the following. If you can prove this, then

you are all set for properties of expectation,

variance, and covariance!

Var(ax+by)=a2Var(x)+b2Var(y)+2abCov(x,y)


7/59


Correlation coefficient:

If x and y are independent, then x,y= 0, but zero

correlation itself does not necessarily imply

independence

r [ 1,1]


8/59

Ordinary Least Squares

Least squares refers to the minimizing the sum

of squared residuals

Trivia: why not

Cannot be minimized because solution can be

either positive or negative

min ui2

i=1

n

min uii=1

n

?


9/59

Ordinary Least Squares (Continued)

Estimators:

First order conditions or method of moments

Hint: the point is always on the OLS

regression line (SRF line)

(x,y)

y= 0+ 1x 0=y 1x


10/59

Statistical Properties of OLS

An estimator is unbiasedif its expected value

(or mean of its sampling distribution) equals

the population value

S i i l P i f OLS


11/59


(Continued)

Gauss-Markov assumptions:

iidindependently (i.e., each random variable is

mutually independent of every other random variable)

and identically distributed (i.e., each random variable

has the same probability distribution as every other

random variable)

ui ~iid(0,s u2 )E(u |x) =E(u)

= 0

S i i l P i f OLS


12/59


(Continued)

E(u|x) = E(u) = 0 is also referred to as the zeroconditional mean assumption; the expectationof u is zero given any values of x

Cov(ui, uj) = 0 for all i != j; also referred to asno serial correlation (if you choose to go on totime series, especially financial econometrics)

for all i = 1, 2, , n Also referred to as the constant

variance/homoskedasticity assumption

Var(ui )=s

u

2


13/59

Unbiasedness of 1

1= (xi x )yi

i=1

n

(xi x )2

i=1

n

E( 1) =E(xi x )yi

i=1

n

(xi x )2

i=1

n

=E

(xi x )( 0+i=1

n

1xi+ui )

(xi x )2

i=1

n


14/59

Unbiasedness of 1(Continued)

=E 0 (xi x )+

i=1

n

1 xi (xi x )+i=1

n

ui (xi x )i=1

n

(xi x )2

i=1

n

=E 1+ui (xi x )

i=1

n

(xi x )2

i=1

n

=E( 1)+Eui (xi x )

i=1

n

(xi x )2

i=1

n


15/59

Unbiasedness of 1(Continued)

= 1 +E[ui (xi x )

i=1

n

]

(xi x )2

i=1

n

= 1

The numerator is essentially

Cov(x,u) (E(u) = 0), which is

assumed to be zero per the

zero conditional mean

assumption (E(u|x) = E(u) =

0)


16/59

Unbiasedness of 0

0=y 1x

= 0+ 1x+u 1x= 0+ ( 1 1)x+uE( 0 ) =E[ 0+ ( 1 1)x+u]=E( 0 )+E[( 1 1)x ]+E(u)= 0+E[( 1 1)x ]=

0

If 1isunbiased, then

the second

term is equal tozero


17/59

Gauss-Markov Theorem

Under the Gauss-Markov assumptions, the OLSestimator(s) is/are the BLUE(s)

Best

Linear*Unbiased*

Estimator

The theorem only applies to comparisonsbetween unbiased estimators


18/59

Measures of Goodness of Fit

Total sum of squares (SST/TSS):

Total sample variation (spread) in thedependent variable about its sample average

Explained sum of squares (SSE/ESS):

Total sample variation of the fitted values in a(multiple) regression model

(yi y )2

i=1

n

(

yi )

( yi y)2

i=1

n

M f G d f Fit


19/59

Measures of Goodness of Fit(Continued)

Sum of squared residuals/residual sum of

squares (SSR):

The sum of the squared OLS residuals across all

observations; variation of residuals in the sample

In my own words, goal of least squaresminimize unobserved SSR to maximize the

explanatory power of the independent variable(s)

ui2

i=1

n

M f G d f Fit


20/59


SST = SSE + SSR

Coefficient of determination (R-squared):

The proportion of the total sample variation in the

dependent variable that is explained by the

independent variable(s)

R2 =

SSE

SST

=1SSR

SST

M f G d f Fit


21/59


100*R2is the percentage of the sample variationin the dependent variable that is explained by theindependent variable(s)

Note that R-squared only measures the linearrelationship between dependent andindependent variables

R2can be easily inflated by increasing the size of

the sample (and the number of independentvariables [other issues will also arise, as you willsee later in the course (?)])


22/59

Units of Measurement

Change in the unit of measurement of theindependent variable:

Now, lets change x for , where the latter is x/10,then

Previously, for every increase in x, the fitted valueincreases by 1.5

Now, for every increase in , the fitted value increasesby 15, but each independent variable has beenchanged to be 1/10 of their original value, so therelationship remains the same

yi= 2+1.5xi

xiyi= 2+15xi

xi


23/59

Units of Measurement (Continued)

This can be proved very simply (your professorpresented a more formal proof, but this is forintuition):

The converse is also true (i.e., change x to ,valued at 10xdivide by 10)

yi= 2+15xi

= 2+10(1.5)(xi

10)

= 2+1.5xi

xi 1


24/59


Change in the unit of measurement of the

dependent variable:

Now, lets change to , where the latter is 1/10

the value, then

Do we still have the same relationship? Of course!

The converse applies also in this case

yi= 2+1.5xi

yi yi

yi= 0.2+0.15xi


25/59


Change in unit of measurement in both

variables:

Now, we will combine the two changes ( and )

yi= 2+1.5xi

xi yiyi

10=

2

10+10

1

10

(1.5)

xi

10

yi=0.2+10 110

(1.5)xi

yi=0.2+1.5xi


26/59

Statistical Inference

Terms:

Null hypothesisone takes this hypothesis as

true and requires the data to provide substantial

evidence that suggests otherwise H0

Alternative hypothesishypothesis against

which the null hypothesis is tested

H1/Ha


27/59

Statistical Inference (Continued)

Never accept the null hypothesisonly do

not reject

Type I errorrejection of H0when it is true

Type II errorfailure to reject H0when it is

false

Significance levelprobability of type I error in

hypothesis testing

alpha


28/59


One-sided alternative one-sided test

H0: < 0,H1: > 0

H0: > 0, H1: < 0

Two-sided alternative two-sided test

H0: = 0,H1: != 0


29/59


t-statistic for 0and 1:

Why n2?

Degree of freedom (df)

# of observations# of estimated parameters (i.e.,estimators)

Decreases as you add more independent variables (later in thecourse) (i.e., intercept + 2 independent variablesn3)

t= 0 0

se( 0 ) ~tn 2t=

1 1se( 1) ~tn 2


30/59


Side note:

Var( 0

)=

s u2n 1 xi2i=1

n

(xi x )2

i=1

n


31/59


Hypothesis testing:

Lets work with an example (random data

generated in R)


32/59


Suppose that I want to test if 1is equal to-

0.37

Two-sided test

H0: 1= -0.37, H1: 1!= -0.37

t= 0.3671 ( 0.37)

0.1941

~t98

= 0.014940752


33/59


Critical valuethe value against which a test

statistic is compared to determined whether

to reject H0or not

One-sided testt > c or t < c, where c is the

critical value

Two-sided test|t|> c


34/59


How to find the critical value?

Determine your significance level (from earlier)

One-sided test

Two-sided test/2 Because you want 100*(1)% confidencethat you are not

wrongly rejecting H0middle region

Since this is a t-test, we will go to the t-table


35/59


Our df is 98 (100

observations2

estimators), so 100is the closest

approximation on

this table


36/59


Our df is 98 (100

observations2


approximation on

this table

Now, assume that

I want a

significance level

of 5% ( = 0.05).

However, this is a

two-sided test, so

I need /2 =0.025 for my tail

probability.

( )


37/59


Our df is 98 (100

observations2


approximation on

this table

Now, assume that

I want a

significance level

of 95%, = 0.05.

However, this is a

two-sided test, so

I need /2 =0.025 for my tail

probability.

l f ( d)


38/59


So my critical value is 1.984 per the table

Rejection rule: |t| > c|t| > 1.984

If this is true, you reject H0

Since my test statistic was 0.014940752, which isless than 1.984, I do not reject H0

Conclusionwith 95% confidence, I do not reject

the hypothesis that 1is equal to -0.37

l f ( d)


39/59


Caveat:

I do not know what kind of t-table you will use on

the midterm, so pay attention to the headings

Some tables have tail probabilities for both one- andtwo-sided tests

S i i l f (C i d)


40/59


What if you were given 2(i.e., it is known)?

Use the standard normal table

Back to our example

As you can see, the test statistic is now distributed

N(0,1)standard normal

t= 0.3671 ( 0.37)

0.1941~N(0,1)

= 0.014940752

S i i l I f (C i d)


41/59


Same significance level

95% two-sidedyou are now looking for 0.975 (1

/2) in the standard normal table

The value is 1.96

Standard normal critical values are easy to remember

0.9952.575

0.992.33

0.9751.96

0.951.645

0.91.28

S i i l I f (C i d)


42/59


Rejection rule: |t| > c|t| > 1.96

Again, this is not the case, so H0is not rejected

S i i l I f (C i d)


43/59


Overview: Identify the hypothesis test (one-/two-sided)

Determine the significance level (if not specified, isusually left as = 0.05)

Calculate the test statistic Determine the distribution of the test statistic (for OLS

estimators, is usually t distribution)

Find the critical value corresponding to the specified

significance level Determine the rejection rule

Compare test statistic with critical value Either reject or do not rejectH0

St ti ti l I f (C ti d)


44/59


p-value!

I believe this one confuses some people

Smallest significance level at which H0can be

rejected Or, the largest significance level at which H0cannot be

rejected



45/59


What does this mean?

For example, if you are comfortable with a 95%significance level (i.e., 5% chance that you rejectH

0when it should have been otherwise), but the

smallest significance level that the particular teststatistic can be rejected at is lower than 5%, itmeans the statistic is even more assuring thanwhat you initially set, so you can go ahead andreject H0knowing that the chance of wrongingdoing so is smaller than what you are comfortablewith (it sounds weird, but it is a good thing)



46/59


How to calculate the p-value:

Back to our example:

Need to scour the standard normal table for

0.014940752, or something close to it.

t=

0.3671 ( 0.37)

0.1941 ~N(0,1)

= 0.014940752



47/59


The value is somewhere between 0.5040 and0.50800.5060 is my guess



48/59


Since this is a two-sided test, we need to

calculate

If the rejection rule is t > c, then you need to

calculate P(T > t), or P(T < t)

See previous STAT courses or Wooldridge for areferesher

P(|T|>| t|)= 2P(T>| t|)

= 2[1 P(T


49/59


Conclusion0.988 > = 0.05do not rejectH0; it should be the same as the t-test

P(|T|>| t|) = 2P(T>| t|)

= 2[1 P(T


50/59


Simple rule:

Reject p-values smaller than

Statistical Inference (Contin ed)


51/59


Confidence interval:

A rule used to construct a random interval so that

a certain percentage of all data sets, determined

by the confidence level, yields an interval thatcontains the population value

In other words, if we construct confidence

intervals for a certain parameter out of a certain

number of samples, these intervals will contain its

population value 100*(1)% of the time



52/59


Back to our example:

Earlier, we found the critical value for /2 = 0.025,which was 1.984

Note: confidence intervals are always considered to be

two-sided We have all the information we need to construct the

confidence interval

t= 0.3671 ( 0.37)

0.1941~t98

= 0.014940752



53/59


That above is the 95% confidence interval for

1

CI= [ 1 c2

se( 1), 1+ c2

se( 1)]= [ 0.3671 1.984(0.1941), 0.3671+1.984(0.1941)]

= [ 0.3671 0.3850944, 0.3671+0.3850944]

= [ 0.7521944,0.0179944]

STATA!


54/59

STATA!

Disclaimer: pure R user, never used STATA

However, I am able to read the outputs and have

constructed questions around them

STATA Output


55/59

STATA Output

STATA Output


56/59

STATA Output

SSE/ESS SSR/RSS SST/TSS any two can be used to solve for R-squared

STATA Output


57/59

STATA Output

0 1 se( i ),i= 0,1

STATA Output


58/59

STATA Output

t= i

se( i ), i= 0,1Significance test (H0: i= 0)

STATA Output


59/59

STATA Output

p-value of significance test

econ 321 midterm prep

Documents