Temple University

Department of Economics

Economics 615

Econometrics I

Midterm Exam

 

Directions: This is a closed book exam. You may have 2 1/2 hours to complete the exam. You must answer all questions. The point values are shown. Your answers must be your own work.

 

  1. (25 points) Use the following joint distribution
 

X

  

0

1

2

3

  

Y

1

1/8

0

0

1/8

1/4

2

0

1/4

1/4

0

1/2

3

0

1/8

1/8

0

1/4
   

1/8

3/8

3/8

1/8

1
  1. What is E(X)?

    2. Ex=1/8(0)+3/8(1)+3/8(2)+1/8(3)=1.5
  2. What is E(Y)?

    3. Ey=1(1/4)+2(1/2)+3(1/4)=2
  3. What is E(X|Y=2)?
    4. x P(x|y=2) xP(x|y=2)
    0 0 0
    1 1/2 1/2
    2 1/2 1
    3 0 0

    E(x|y=2)=1.5
  4. What is E(X|Y=1)?
    5. x P(x|y=1) xP(x|y=1)
    0 1/2 0
    1 0 0
    2 0 0
    3 1/2 1.5

    E(x|y=1)=1.5
  5. What is E(XY)?

    6. x, y

    xy

    P(x,y)

    xyP(x,y)

    0,1

    0

    1/8

    0

    1,1

    1

    0

    0

    2,1

    2

    0

    0

    3,1

    3

    1/8

    3/8

    0,2

    0

    0

    0

    1,2

    2

    1/4

    4/8

    2,2

    4

    1/4

    1

    3,2

    6

    0

    0

    0,3

    0

    0

    0

    1,3

    3

    1/8

    3/8

    2,3

    6

    1/8

    6/8

    3,3

    9

    0

    0

    Exy=3
  6. What is the covariance of X and Y?

    7. Cov(x,y)=Exy-ExEy=3-1.5(2)=0
  7. Are X and Y independent? How do you know? How do you reconcile your answers in a., c., d., and f.?

x and y are not independent. We conclude this from the fact that the intersection is not equal to the product of the marginals. We cannot rely on the conditional expectation or the covariance to make this inference since these are necessary conditions, but not sufficient.

 

2.  (20 points) For the probability density

  1. Is this a bona fide density? How do you know?

    2. Yes, it is a bona fide density. For all intervals we can show 0<P(A)<1 and the integral over the domain is one.

  2. Find E(X).

  3. Find Variance(X).

Var(x)=Ex2-(Ex)2

Var(x)=6-22=2

3. (10 points) If x has a normal distribution with mean 3 and variance 16, what are the following:

  1. Prob[ |x| > 7 ] ?

    2. P(|x|>7)=P(x<-7)+P(x>7)

    =P(z<-2.5)+P(z>1)=.1649
  2. Prob[ x > -2 | x<4 ] ?

4. (30 points) For the sample data 12, 15, 9, 3, 23, 1, 5, 8, 11, 17 drawn from a normal distribution with mean and variance.

  1. Compute the sample mean.

  2. Compute the sample standard deviation.

    3. xi

    12

    1.6

    2.56

    15

    4.6

    21.16

    9

    -1.4

    1.96

    3

    -7.4

    54.76

    23

    12.6

    158.76

    1

    -9.4

    88.36

    5

    -5.4

    29.16

    8

    -2.4

    5.76

    11

    .6

    .36

    17

    6.6

    43.56

    s2=45.155 s=6.7197
  3. Test the hypothesis

    4. at the 95% level using the classical framework.

    The critical values are -2.262 and 2.262. Therefore do not reject the null.
  4. Test the hypothesis

    5. at the 95% level using the classical framework.

    The critical value is 16.91, so reject the null.
  5. Using a likelihood ratio test, test the following hypothesis:

at the 95% level.

The density is , so the log likelihood is

In the restricted parameter space (under the null hypothesis) make the substitution . The log likelihood is -92.015.

In the unrestricted parameter space (under the alternate hypothesis) make the substitution . The log likelihood evaluated at the sample data is -32.21.

The test statistic is -2(-92.05-(-32.21))=119. For any reasonable level of test and two degrees of freedom we reject the null.

5.  (15 points) Suppose that in a sample of 100 observations from a normal distribution with mean and variance you are told that 25% of the observations are less than 2.5 and 55% are less than 5.3. Estimate and .

P(x<5.3)=.55, P(x<2.5)=.25