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1 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 ππ = π½0 + π½1 ππ + π’π , and you wish to find the estimates of π½0 and π½1 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 π 2 =(viii). (write the equation) (i) sum of squared residuals (ii) βππ=1(ππ β π½0 β π½1 ππ )2 (iii) (restricted, unrestricted) (iv) (restricted, unrestricted) (v) (smaller, larger) (vi) (small, large) (vii) SSRr-SSRu (viii) πππ π βπππ π’ πππ π πππ = 1 β πππ π’ π [4] Let X and Y be random variables. Their expected values are denoted by ππ₯ and ππ¦ , and their variances are denoted by ππ₯2 and ππ¦2 , and their covariance is denoted by ππ₯π¦ . The expected value of a random is a measure of the location of the βcenterβ 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 ππ¦2 and ππ₯π¦ . ππ¦2 = πΈ[π β πΈ(π)]2 , ππ₯π¦ = πΈ([π β ππ₯ ][π β ππ¦ ]) (b) Show that ππ¦2 = πΈ[π 2 ] β ππ¦2 ππ¦2 = πΈ[(π β ππ¦ )2 ]=E[Y2-2ππ¦ Y+ππ¦2 ]= E[Y2]-2ππ¦ E(Y)+ ππ¦2 =E(Y2)- 2ππ¦2 +ππ¦2 = E(Y2)- ππ¦2 (c) Show that ππ₯π¦ = πΈ(ππ) β ππ₯ ππ¦ ππ₯π¦ =E[(X-ππ₯ )(Y-ππ¦ )]=πΈ(ππ β ππ¦ π β ππ₯ π + ππ₯ ππ¦ ) = πΈ(ππ) β ππ¦ πΈ(π) β ππ₯ πΈ(π) + ππ₯ ππ¦ = πΈ(ππ) β ππ₯ ππ¦ [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 ππ¦ = πΈ(π) = 1 × 0.3 + 2 × 0.15 = 0.6 ππ¦2 = πΈ(π 2 ) β ππ¦2 = 1 × 0.3 + 4 × 0.15 β 0.62 = 0.54 πππ£(π, π) = πΈ(ππ) β ππ₯ ππ¦ = 1 × 0.2 + 2 × 0.1 β 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) π(π¦|π₯) = π(π₯,π¦) π(π₯) (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)=0.4×0.3β 0.1. [6] You wish to estimate the probability ο² 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 πΜ be an estimator of ο². (a) An estimator πΜ is an unbiased estimator of ο²ο if (E(πΜ)=ο²). (b) You decided to estimate ο² by the fraction of the number of heads in n tosses, that is, compute the π β ππ estimate of ο² by the sample average, πΜ = πΜ = π=1 . Show that this estimator is unbiased. π 1 1 1 1 πΈ(πΜ) = πΈ (π βππ=1 ππ ) = π βππ=1 πΈ(ππ ) = π βππ=1 π = π (ππ) = π (4 points)