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Spring, 2017
Ecmt 463
Midterm Exam
Each question is worth 2 points unless specified otherwise.
[1] Let Q be the quantity demanded and P be the price in dollars. You specified two models and estimated
them: (1) ln(Q)=6.7-0.01P
(2) ln(Q)=6.5-0.2ln(P)
In model (1), an increase in price by $1 decreases Q by about (i). In model (2), the estimated coefficient
indicates that the quantity demand declines by (ii) when price rises by (iii).
(i) 1%
(ii) 0.2%
(iii) 1%
[2] You have a data on individual's average hourly earnings in dollars and their characteristics about
education in years, gender (female or male), and where they live (Northeast, Midwest, South, West). You
denote the education variable by X1, and construct dummy variables:
X2=1 if female and 0 if male
X3=1 if live in Northeast and 0 otherwise,
X4=1 if live in Midwest and 0 otherwise
X5=1 if live in South and 0 otherwise
X6=1 if live in West and 0 otherwise
You run a regression equation with the average hourly earnings as the dependent variable and found
Earnings=3.75 + 5.44X1 - 2.62X2 + 0.69X3 + 0.60X4 - 0.27X5
Note that dummy variable X6 for West is not included in the regression equation.
(a) The difference in hourly earnings between female and male is $(i) and female workers earn (ii) than
male workers.
(b) The estimated coefficient of X3 is 0.69 which means that a worker living in (iii) earns $0.69 more than
a worker living in (iv).
(c) A worker who lives in Northeast earns (v) than a worker living in Midwest by $(vi) per hour.
(i) 2.62
(ii) (more, less)
(iii) (Northeast)
(iv) West
(v) (more, less)
(vi) 0.09
[3] You specify a simple linear regression model 𝑌𝑖 = 𝛽0 + 𝛽1 𝑋𝑖 + 𝑢𝑖 , and you wish to find the estimates
of 𝛽0 and 𝛽1 which explain the variation of the dependent variable the best. The Ordinary Least Squares
estimation method defines the "best" as the estimators that lead to the smallest SSR, where the SSR stands
for the (i) and it is specified as SSR= (ii) (write the equation for (ii)).
A commonly used measure of how well the model explains the dependent variable is R2. The idea of this
measure is to compare the SSRu with intercept and regressors and the SSRr with only an intercept term
(i.e., without any regressor). The former is called the (iii) model, and the latter is called the (iv) model
where the coefficients of regessors are all restricted to be zero. The SSRu is always (v) than the SSRr. As
2
the OLS estimator minimizes the SSR, the change in SSR from SSRr to SSRu is expected to be (vi) if
regressors do a good job in explaining the dependent variable. Therefore, we measure the explanatory
power of regressors by the fraction of (vii) to SSRr. This can be written as 𝑅 2 =(viii). (write the
equation)
(i) sum of squared residuals
(ii) ∑𝑛𝑖=1(𝑌𝑖 − 𝛽0 − 𝛽1 𝑋𝑖 )2
(iii) (restricted, unrestricted)
(iv) (restricted, unrestricted)
(v) (smaller, larger)
(vi) (small, large)
(vii) SSRr-SSRu
(viii)
𝑆𝑆𝑅𝑟 −𝑆𝑆𝑅𝑢
𝑆𝑆𝑅𝑟
𝑆𝑆𝑅
= 1 − 𝑆𝑆𝑅𝑢
𝑟
[4] Let X and Y be random variables. Their expected values are denoted by 𝜇𝑥 and 𝜇𝑦 , and their
variances are denoted by 𝜎𝑥2 and 𝜎𝑦2 , and their covariance is denoted by 𝜎𝑥𝑦 . The expected value of a
random is a measure of the location of the “center” of the probability distribution and it is computed by
the (probability) weighted average of the values of the random variable. The variance is a measure of the
degree of dispersion or spread of the random outcomes around its mean. The covariance is a measure of
the degree of covariation of the two random variables.
(a) Write the definition of 𝜎𝑦2 and 𝜎𝑥𝑦 .
𝜎𝑦2 = 𝐸[𝑌 − 𝐸(𝑌)]2 ,
𝜎𝑥𝑦 = 𝐸([𝑋 − 𝜇𝑥 ][𝑌 − 𝜇𝑦 ])
(b) Show that 𝜎𝑦2 = 𝐸[𝑌 2 ] − 𝜇𝑦2
𝜎𝑦2 = 𝐸[(𝑌 − 𝜇𝑦 )2 ]=E[Y2-2𝜇𝑦 Y+𝜇𝑦2 ]= E[Y2]-2𝜇𝑦 E(Y)+ 𝜇𝑦2 =E(Y2)- 2𝜇𝑦2 +𝜇𝑦2 = E(Y2)- 𝜇𝑦2
(c) Show that 𝜎𝑥𝑦 = 𝐸(𝑋𝑌) − 𝜇𝑥 𝜇𝑦
𝜎𝑥𝑦 =E[(X-𝜇𝑥 )(Y-𝜇𝑦 )]=𝐸(𝑋𝑌 − 𝜇𝑦 𝑋 − 𝜇𝑥 𝑌 + 𝜇𝑥 𝜇𝑦 ) = 𝐸(𝑋𝑌) − 𝜇𝑦 𝐸(𝑋) − 𝜇𝑥 𝐸(𝑌) + 𝜇𝑥 𝜇𝑦
= 𝐸(𝑋𝑌) − 𝜇𝑥 𝜇𝑦
[5] The following table shows the joint probability f(x,y)=P(X=x, Y=y) of daily beer consumption and
gender of TAMU students. For example, joint probability f(0,1) indicates that, when a person is selected
randomly, the probability that the person is a female who drinks one can of beer per day is 0.1.
Female (X=0)
Male (X=1)
f(y)
no beer
(Y=0)
0.25
1 can of beer
(Y=1)
0.1
2 cans of beer
(Y=2)
0.05
0.3
0.55
0.2
0.3
0.1
0.15
(a) Fill in the blank cells of marginal probabilities.
(b) Compute the expected value and variance of Y, and compute cov(X,Y).
f(x)
0.4
0.6
1.0
3
𝜇𝑦 = 𝐸(𝑌) = 1 × 0.3 + 2 × 0.15 = 0.6
𝜎𝑦2 = 𝐸(𝑌 2 ) − 𝜇𝑦2 = 1 × 0.3 + 4 × 0.15 − 0.62 = 0.54
𝑐𝑜𝑣(𝑋, 𝑌) = 𝐸(𝑋𝑌) − 𝜇𝑥 𝜇𝑦 = 1 × 0.2 + 2 × 0.1 − 0.6 × 0.6 = 0.04
(c) Suppose we know that the randomly selected person is a female. What is the probability that she
drinks one can of beer per day? This is the conditional probability of Y=1 given X=0, which is written as
f(y|x)=P(Y=y|X=x). How do we compute the conditional probability? Conditional probabilities must
satisfy two conditions:
conditional probabilities must sum to one, for example, f(0|0)+f(1|0)+f(2|0)=1
conditional probabilities must retain the relative likelihood of drinking habit of a female student, e.g.,
f(0|0)=5×f(2|0) because f(0,0)=5×f(0,2)
These considerations lead to the definition of conditional probability which is given by (i). Applying this
definition, we find that, given a randomly chosen student being a male, the probability f(0|1) that he does
not drink beer is (ii). It is straightforward to show f(1|1)=(iii) and f(2|1)=(iv).
(i) 𝑓(𝑦|𝑥) =
𝑓(𝑥,𝑦)
𝑓(𝑥)
(ii) f(0|1)=0.3/0.6=1/2
(iii) f(1|1)=0.2/0.6=1/3
(iv) f(2|1)=1/6
Using these conditional probabilities, we can compute the conditional expectation E(Y|X=1):
E(Y|X=1)=0×1/2+1×1/3+2×1/6=4/6=2/3
(e) Random variables X and Y are statistically independent if information on X does not affect the
probabilities of random variable Y. This implies f(y|x)=(i). This also implies f(x,y)=(ii).
(i) f(y)
(ii) f(x)f(y)
Show that the second implication follows from the first implication.
f(y|x)=f(x,y)/f(x) and f(y|x)=f(y). Therefore, f(y)=f(x,y)/f(x), which gives f(x,y)=f(x)f(y)
(f) Are X and Y in the table above statistically independent? Show your work.
They are not statistically independent because P(X=0, Y=1)=0.1 while P(X=0)P(Y=1)=0.4×0.3≠0.1.
[6] You wish to estimate the probability  of head in coin tossing. You toss the coin n times. Let Yi be the
random variable for the outcome of i-th toss, Yi=1 if H and Yi=0 if T. Let 𝜃̂ be an estimator of .
(a) An estimator 𝜃̂ is an unbiased estimator of if (E(𝜃̂)=).
(b) You decided to estimate  by the fraction of the number of heads in n tosses, that is, compute the
𝑛
∑
𝑌𝑖
estimate of  by the sample average, 𝜃̂ = 𝑌̅ = 𝑖=1
. Show that this estimator is unbiased.
𝑛
1
1
1
1
𝐸(𝜃̂) = 𝐸 (𝑛 ∑𝑛𝑖=1 𝑌𝑖 ) = 𝑛 ∑𝑛𝑖=1 𝐸(𝑌𝑖 ) = 𝑛 ∑𝑛𝑖=1 𝜃 = 𝑛 (𝑛𝜃) = 𝜃
(4 points)