Econ 371 Problem Set #1 Answer Sheet 2.1 In this question, you are asked to consider the random variable Y, which denotes the number of heads that occur when two coins are tossed. a. The first part of the question asks you to derive the probability distribution for Y. For a discrete random variable, this is just a listing of all the possible outcomes and the probability that they can occur. Assuming that the two coins are fair, there are only four possible results when we flip the two coins: Heads,Heads (Y = 2) Heads,Tails (Y = 1) Tails,Heads (Y = 1) Tails,Tails (Y = 0) which happen with equal probability. The corresponding probability distribution is given by the following table. outcome (number of heads) Y = 0 Y = 1 Y = 2 probability 0.25 0.50 0.25 b. The cumulative probability distribution function for Y corresponds to just computing the probabilities of Y being less than or equal to a given value outcome (number of heads) Y < 0 0 Y < 1 1 Y < 2 Y > 2 cumulative probability 0 0.25 0.75 1 c. Finally, you are asked to derive the mean and variance of Y. These are given by: and E(Y ) = V ar(y ) = E[(Y µ Y ) 2 ] = k P r(y = y i )y i = k p i y i = (0.25) 0 + (0.50) 1 + (0.25) 2 = 1 k p i (y i µ Y ) 2 = (0.25) (0 1) 2 + (0.50) (1 1) 2 + (0.25) (2 1) 2 = (0.25) 1 + (0.50) 0 + (0.25) 1 = 0.5 2.3 In this question, you are asked to compute various characteristics of two random variables, W and V that are functions of the random variables in Table 2.2 (X and Y ). Specifically, we have: W = 3 + 6X V = 20 7Y a. First, you are asked to compute E(W ) and E(V ). From equation (2.12) in the text, we know that: E(W ) = 3 + 6E(X) (1) 1
but, using Table 2.2 and our formula for computing means, we have that E(X) = k X p i x i = (0.30) 0 + (0.70) 1 = 0.7 Substituting this into (1) yields. E(W ) = 3 + 6E(X) = 3 + 6(0.7) = 7.2 (2) Similarly, E(Y ) = k Y p i x i = (0.22) 0 + (0.78) 1 = 0.78, so that E(V ) = 20 7E(Y ) = 20 7(0.78) = 14.54. (3) b. In part b you are asked to compute the corresponding variances for W and V. Equation (2.13) in the text tells us that: σw 2 = 6 2 σx 2 = 36σX 2 σv 2 = ( 7) 2 σy 2 = 49σY 2 Turning to our formulas for variances, however, we know that: V ar(x) = k X p i (x i µ X ) 2 = (0.30) (0 0.7) 2 + (0.70) (1 0.7) 2 = (0.30) (0.49) + (0.70) (0.09) = 0.21 So that Similarly, σ 2 W = 6 2 σ 2 X = 36(0.21) = 7.56 (4) V ar(y ) = k Y p i (y i µ Y ) 2 = (0.22) (0 0.78) 2 + (0.78) (1 0.78) 2 = (0.22) (0.6084) + (0.78) (0.0484) = 0.1716 So that σ 2 V = 49 2 σ 2 Y = 49(0.1716) = 8.4084 (5) c. Finally, you are asked to compute σ W V and corr(w, V ) There are two ways to do this (at least). First, you can derive the formula for the covariance term using: Cov(W, V ) = Cov(3 + 6X, 20 7Y ) = 6( 7)Cov(X, Y ) (6) 2
But, using the formula in equation (2.24) in the text, we have that: So that Cov(X, Y ) = σ XY = E [(X µ X )(Y µ Y )] = k X k Y j=1 x i µ X )(y j µ Y )P r(x = x i, Y = y j ) = (0 0.7)(0 0.78)(0.15) + (1 0.7)(0 0.78)(0.07) + (0 0.7)(1 0.78)(0.15) + (1 0.7)(1 0.78)(0.63) = 0.0819 0.01638 0.0231 + 0.04158 = 0.084 Cov(W, V ) = Cov(3 + 6X, 20 7Y ) = 6( 7)(0.084) = 3.528. (7) Alternatively, since knowing the values of X and Y tells the corresponding values for W and V, we can use this information to construct the probability distribution table for W and V. Specifically, we have that: The probability distribution for W and V can be used directly to compute X = 0 W = 3 X = 1 W = 9 Y = 0 V = 20 0.15 0.07 Y = 1 V = 13 0.15 0.63 Cov(W, V ) using equation 2.24 in the book, which will give us Cov(W, V ) = 3.528. compute Cov(W, V ), the correlation results from using the equation: corr(w, V ) = However you Cov(W, V ) 3.528 = = 0.4425 (8) V ar(w )V ar(v ) (7.56)(8.4084) 2.6 This question considers the joint probability distribution between employment status and college graduation. a. The first question asks you to compute the mean of the employed random variable Y. This would be given by: b. The unemployment rate is given by E(Y ) = k Y p i y i = (0.05) (0) + (0.95) (1) = 0.95 Unemployment rate = Number employed Number in labor force = P r(y = 0) = 1 P r(y = 1) = 1 E(Y ) c. In order to compute the conditional means in this question, we need to know the conditional probabilities. These can be calculated using equation (2.17). For the current problem: P r(y = 0 X = 0) = P r(y = 1 X = 0) = P r(y = 0 X = 1) = P r(y = 1 X = 1) = P r(x = 0, Y = 0) P r(x = 0) P r(x = 0, Y = 1) P r(x = 0) P r(x = 1, Y = 0) P r(x = 1) P r(x = 1, Y = 1) P r(x = 1) = 0.045 0.754 = 0.0597 = 0.709 0.754 = 0.9403 = 0.005 0.246 = 0.0203 = 0.241 0.246 = 0.9797 3
The conditional means then follow using equation (2.18) in the text: E(Y X = 0) = (0) P r(y = 0 X = 0) + (1) P r(y = 1 X = 0) = 0.9403 E(Y X = 1) = (0) P r(y = 0 X = 1) + (1) P r(y = 1 X = 1) = 0.9797 d. The unemployment rate, conditional on the college graduation status, is computed in much the same way as the unemployment rate was computed in part (b), only now we are conditioning on graduation status. Thus, Similarly Unemployment rate X = 1 = Number employed among college graduates Number of college graduates in labor force = P r(y = 0 X = 1) = 1 P r(y = 1 X = 1) = 1 E(Y X = 1) = 0.0203 Unemployment rate X = 0 = Number employed among non-college graduates Number of non-college graduates in labor force = P r(y = 0 X = 0) = 1 P r(y = 1 X = 0) = 1 E(Y X = 0) = 0.0597 e. This question asks first what the probability of a randomly selected individual is unemployed. This is given by the P r[y = 0] = 0.05. You are then asked to compute conditional probabilities. That is, you are asked to compute the probability that an individual is a college graduate, given that you know they are unemployed. Mathematically, this corresponds to: P r(x = 1 Y = 0) = P r(x = 1, Y = 0) P r(y = 0) = 0.005 = 0.10 (9) 0.050 Thus, there is only a ten percent chance that a randomly selected unemployed person is a college graduate. The probability that a randomly selected unemployed person is a non-college graduate is then P r(x = 0 Y = 0) = 1 P r(x = 1 Y = 0) = 0.90 (10) f. In order to check independence, we need to determine if equation (2.22) in the text holds; i.e., However, we know that this does not hold, since P r(y = y X = x) = P r(y = y). (11) P r(y = 0 X = 0) = 0.0597 P r(y = 0) = 0.05. (12) 2.12 This question asks you to look up a number of the probabilities associated with specific distributions. a. The first question asks for P r(y > 1.75) if Y is distributed t 15. This corresponds to the 1-sided significance level in Table 2. Looking under the 15 degrees of freedom case, we find that the 1-sided significance level of 5% has a critical level of 1.75. That is, P r(y > 1.75) = 0.05, which is what we were looking for. b. The second question asks for P r( 1.99 Y 1.99) if Y is distributed t 90. However, this can be rewritten as: P r( 1.99 Y 1.99) = 1 [P r(y < 1.99) + P r(y > 1.99)] (13) 4
The term in square brackets above is just the 2-sided significance level in Table 2 (i.e., the probability that Y lies in the tails of the t-distribution - above 1.99 or below -1.99). Looking under the 90 degrees of freedom case, we find that the 2-sided significance level of 5% has a critical level of 1.99. That is, [P r(y < 1.99) + P r(y > 1.99)] = 0.05. This in turn implies that P r( 1.99 Y 1.99) = 1 [P r(y < 1.99) + P r(y > 1.99)] = 0.95 (14) c. The third question asks for P r( 1.99 Y 1.99) if Y is distributed N(0, 1). This can be rewritten as: P r( 1.99 Y 1.99) = P r(y 1.99) P r(y < 1.99) (15) From Table 1 we know that P r(y 1.99) = 0.9767 and P r(y < 1.99) = 0.0233, so that P r( 1.99 Y 1.99) = P r(y 1.99) P r(y < 1.99) = 0.9767 0.0233 = 0.9534. (16) d. The t-distribution and the standard normal are approximately the same as the degrees of freedom become large. e. The fifth question as for P r(y > 4.12) if Y is distributed F 4,7. This corresponds to the significance level. Looking through the three tables 5A, 5B and 5C, we see that for an F 4,7 distribution: P r(y > 2.96) = 0.10 P r(y > 4.12) = 0.05 P r(y > 7.85) = 0.01 The answer we are looking for is the second one; i.e., P r(y > 4.12) = 0.05. f. The sixth question as for P r(y > 2.79) if Y is distributed F 7,120. This corresponds to the significance level. Looking through the three tables 5A, 5B and 5C, we see that for an F 7,120 distribution: P r(y > 1.77) = 0.10 P r(y > 2.09) = 0.05 P r(y > 2.79) = 0.01 The answer we are looking for is the third one; i.e., P r(y > 2.79) = 0.01. 2.14 This question makes use of the central limit theorem, which says that as the sample size gets large, the sampling distribution for Ȳ can be approximated by N(µ Y, σ 2 ), where Ȳ This also means that Ȳ µ Ȳ σ Ȳ σ 2 Ȳ = σ2 Y n. (17) N(0, 1). (18) a. In this first question, you are told that n = 100, so that σ 2 Ȳ = σ2 Y n = 43 100 = 0.43. With this information, we know that: P r ( Ȳ 101 ) = P r ( ) Ȳ µ Ȳ 101 µ Ȳ (Ȳ ) µȳ 101 µȳ = P r σ Ȳ σ ( ) Ȳ 101 µȳ = Φ σ ( Ȳ ) 101 100 = Φ 0.43 Φ(1.525) = 0.9364. 5
b. In this second question, you are told that n = 165, so that σ 2 Ȳ = σ2 Y n = 43 165 information, we know that: P r ( Ȳ 98 ) = P r ( ) Ȳ µ Ȳ 98 µ Ȳ (Ȳ ) µȳ 98 µȳ = P r σ Ȳ σ ( ) Ȳ 98 µȳ = Φ σ ( Ȳ ) 98 100 = Φ 0.2606 =.2606. With this Φ( 3.92) 0.0000. We can then use this to calculate: P r ( Ȳ > 98 ) = 1 P r ( Ȳ 98 ) 1. (19) c. In this third question, you are told that n = 64, so that σ 2 Ȳ = σ2 Y n = 43 64 =.6719. With this information, we know that: P r ( 101 Ȳ 103) = P r ( 101 µ Ȳ Ȳ ) µȳ 103 µ Ȳ ( 101 µȳ = P r Ȳ ) µȳ 103 µȳ σ Ȳ σ Ȳ σ ( ) ( ) Ȳ 103 µȳ 101 µȳ = Φ Φ σ Ȳ σ ( ) ( Ȳ ) 103 100 101 100 = Φ Φ 0.6719 0.6719 Φ(3.66) Φ(1.22) 0.9999 0.8888 = 0.1111. 2.16 There are several ways to do this. Here is one way. Generate n draws of Y, Y 1, Y 2,... Y n. Let X i = 1 if Y i < 3.6, otherwise set X i = 0.. Notice that X i is a Bernoulli random variables with µ X = P r(x = 1) = P r(y < 3.6). Compute X. Because X converges in probability to µ X = P r(x = 1) = P r(y < 3.6), X will be an accurate approximation if n is large 6