Suggested Answers to Odd-Numbered Questions

Suggested Answers to Odd-Numbered Questions [A Guide to Econometrics, 5 th edition, 2003, by Peter Kennedy] A1(a) This Monte Carlo study examines the sampling distribution of the average of 25 numbers drawn from a distribution with mean 2 and variance 9.The average of N numbers drawn randomly from a distribution with mean  and variance  2 has a sampling distribution with mean  and variance  2 /N.(b)Abar estimates the mean of this sampling distribution, so should be close to 2.(c)Avar estimates its variance, so should be close to 9/25 = 0.36.A3(a) This Monte Carlo study examines the sampling distribution of the fraction of successes h in a sample of size 50 where the probability of success is 20%. When the true probability of success is p, for a sample of size N then h has a sampling distribution with mean p and variance p(1-p)/N.(b)hav should be an estimate of 0.2.(c)hvar should be an estimate of p(1-p)/N which here is ( 0.2)(0.8)/50 = 0.0032 (d)wav estimates the mean of the sampling distribution of the estimated variance of h, and so should be approximately 0.0032.A5. Create 44 observations from a normal distribution with mean 6 and variance 4.Calculate their average A and their median B. Repeat to obtain , say, 1,000 As and 1,000 Bs. Find the mean of the 1,000 A values and the mean of the 1,000 B values and see which is closer to 6. Calculate the variance of the 1,000 A values and the variance of the 1,000 B values and see which is smaller.A7. i) Choose values for a and b, say 2 and 6 (be sure not to have zero fall between a and b because then an infinite value of 1/x 2 would be possible and its distribution would not have a mean). ii) Get the computer to generate 25 drawings (x values) from U(2, 6). (If the computer can only draw from U(0,1), then multiply each of these values by 4 and add 2.) iii) Use the data to calculate A's estimate A* and B's estimate B*. Save them. iv) Repeat from ii) 499 times, say, to get 500 A*s and 500 B*s. v) Obtain the mean m of the A Guide to Econometrics, 5e Peter Kennedy (Answer Key All Chapters, 100% Original Verified, A+ Grade) Answers to Odd-Numbered Questions 1 / 4

distribution of 1/x 2 either algebraically (the integral from a to b of 1/(b-a)x 2

• or by

averaging a very large number (10,000, say) of 1/x 2 values. vi) Estimate the bias of A* as the difference between the average of the 500 A*s and m. Estimate the variance of A* as the variance of the 500 A*s. Estimate the MSE of A* as the average of the 500 values of (A* - m) 2 . Compute the estimates for B* in like fashion and compare.A9 The bias of * is estimated as the average of the 400 *s minus , the variance of which is the variance of * divided by 400. Our estimate of this is 0.01/400. The relevant t statistic is thus 0.04/(0.1/20) = 8 which exceeds the 5% critical value, so the null is rejected.A11(a) The expected value of x − is , but the expected value of a nonlinear function of x − is not the nonlinear function of , except in large samples.(b) Use an estimate of the formula for the variance of a nonlinear function.V*(*) = [3x −

2 ] 2 V(x −

• = 9x

−

4 (6/27) = 2x −

4 .(c) i) Choose a value for , say 2, so that  = 8. ii) Generate 27 values of x drawn from N(2,6).iii) Use the data to calculate * and V*(*). Save them. iv) Repeat from ii) to get 5000 *s and 5000 V*(*)s. v) Estimate the true variance of * by the variance of the 5000 *s and see how close the average of the 5000 V*(*)s is to this value.A13(a) This process is estimating the variance of a random variable w by three different formulas differing only by their divisors. Since dividing by N-1 produces an unbiased estimate of this variance the average of the 4000 a values should be closest to 4, the true variance of w.(b) Dividing by a larger number shrinks the variance estimate towards zero, making its variance smaller, so the variance of the c values should be smallest.(c) Step viii estimates the MSEs of the three estimators. Since dividing by N+1 produces the estimate with the smallest MSE, C should be the smallest.A15. Set variables “stick” and “switch” equal to zero. *Set variable “switchguess” equal to four. Draw a number x from a distribution uniform between zero and three. Set variable “door” equal to one if 0  x < 1, equal to two if 1  x < 2, and equal to three if 2 2 / 4

 x  3. Draw a number y from a distribution uniform between zero and three. Set “doorguess” equal to one if 0  y < 1, equal to two if 1  y < 2, and equal to three if 2  y  3. If door = 1 and doorguess = 2 or 3, set switchguess = 1; if door = 2 and doorguess = 1 or 3, set switchguess = 2; if door = 3 and doorguess = 1 or 2, set switchguess = 3. If doorguess = door, add one to “stick” and if switchguess = door, add one to “switch.” Repeat from * until, say, a thousand x values have been drawn. For any original guess, sticking should work one-third of the time, but as is evident from the Monte Carlo structure, switching should work two-thirds of the time.B1. k = 3/4, found by setting the integral of kx(2-x) from 0 to 2 equal to one. E(x) = integral from 0 to 2 of 3x 2 (2-x)/4 = 1. V(x) = Ex 2

• (Ex)

2

= 1/5.

B3. Try supply of 100. All units are sold and profit on each unit is $5, so expected profit is $500. Try supply of 200. All units are sold with a probability of 0.95, with profit of $1,000; with probability 0.05 only 100 units are sold, with a profit $500 on the units sold but a loss of $1,000 on the units not sold for a net loss of $500. Expected profit is thus 0.95(1000) + 0.05(-500) = 925. Try supply of 300. All units are sold with a probability of 0.85, with a profit of $1,500; with probability 0.1 only 200 units are sold, with net profit of zero; and with probability 0.05 only 100 units are sold, with a net loss of $1,500.Expected profit is thus 0.85(1500) + 0.1(0) + 0.05(-1500) = 1200. For supply of 400 expected profit is 1100. That the supply of 300 is the supply that maximizes expected profit can be verified by calculating expected profit for supplies of 299 and 301. The variance of profit if supply is 300 is calculated as 0.85(1500-1200) 2

+ 0.1(-1200)

2 +

0.05(-1500-1200)

2

= 585,000.

B5. The probability of finding a minimum price of $2 is 1/8, namely the probability of checking three stores all of which have a price of $2.Expected price is thus 2*(1/8) + 1*(7/8) = 9/8.B7(a) True. If * is unbiased in small samples it is also unbiased in large samples.(b) False. Asy Var (*) = (1/T)limT(4/T + 16 2 /T 2

) = 4/T

(c) True, because * is asymptotically unbiased and the limit as T goes to infinity of its variance is zero.(d) Uncertain. Don't know about other estimators, or if * is MLE. 3 / 4

B9(a) Both have mean 5%. 1 has variance 30 2 *6 = 5400. 2 has variance 3*10 2

*6 = 1800.

(b) Variance of 2 becomes 10 2

*6 + 10

2

(6 – 2*3 + 6) = 1200.

(c) False. In example of part a) above, positive correlation will raise the variance of 2 above 1800, but it will not make it as large as 5400. Perfect correlation would make the variance 5400.(d) False. Although it will have zero variance, it will have an expected return of 5%.B11(a) The expected value of x is (-10)/2; setting the mean of the data x − equal to the mean of the distribution we get  mm = 2x −

• 10.

(b) V( mm

• = 4V(x)/N = (-10)

2 /3N.B13. The formula for the variance of a nonlinear function of a random vector produces

V(**) = *

2

V(*) + *

2 V(*) + 2**C(*, *) which when evaluated at estimated values yields 0.49, so the estimated standard error is 0.7.B15. V(y) = [2V(x) + 2C(xt, xt-1)]/4. V(x) =  2 /0.36, calculated by squaring both sides of the expression for xt, taking expected values and solving for V(x) = E(xt 2

• = E(xt-1

2 ).C(xt, xt-1) = 0.8V(x), calculated by multiplying through the expression for xt by xt-1 and taking expected values. Using these values, V(y) = 2.5  2 .C1(a) Unbiased; it is as likely to be too steep as to be too flat, depending on the chosen observation. i = yi/xi =  + i/xi, so Ei = , where i denotes the chosen observation.(b) Variance is  2 /(xi) 2 , the variance of i/xi.(c) Choose the observation with the largest x value, to minimize variance.(d) No. As the sample size grows, the variance does not shrink to zero.C3(a) All three are unbiased. The variance of b*** is (8 + 4)/2 2 = 3.0, the smallest of the three, so it is preferred.(b) Any weighted average of the two unbiased estimates is unbiased, so one should choose that weighted average that minimizes variance. Choose the estimator ab* + (1-a)b** whose variance, 8a 2

• 4(1-a)

2 , is minimized for a = 1/3.(c) Pool the two data sets and run a single regression.5(a) b*** is unbiased, so choose a to minimize its variance: a = Vb**/(Vb*+Vb**).(b) When b* is a very precise estimator (Vb* small), more weight is given to b* in the weighted average, and similarly when b** is very precise more weight is given to b**.

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