Isaac's Errata for 2nd edition of Farmer's Macroeconomics ========================================================= p.37, first paragraph This paragraph is written in a very unclear manner. To make it clearer, try this: Delete: The term (S-I) represents private saving Insert: The term (S-I) represents net flows from the private sector into the financial markets Delete: Equation 2.9 demonstrates that there are two ways the government can generate a surplus. Insert: Equation 2.9 suggests two ways the government can finance a deficit. p.39, last paragraph Farmer states: "Since then [1992], the budget deficit and the trade deficit have both fallen." To make sense of this you need to compare the *averages* for the 1980s and 1990s. p.50, problem 3 "Gross investment" is gross *private* domestic investment. p.51, problem 7 The question asks for the capital stock in 2002 but only provides data up to 1999. Compute the capital stock for 2000 based on this data. p.65, Table 3.1 The entries in this table are out of alignment. See the corrected table in my online notes. p.69, Table 3.2 Definition of the unemployed should say "full-time or part-time", not just "full-time". See http://www.bls.gov/cps/cps_htgm.htm and http://www.bls.gov/opub/mlr/1995/10/art3exc.htm p.97, Box 4.5, employment rate graph: There is a typo in the axis labeling. If you look at the right axis, you will see that the labels have all been shifted up by 20%. If you fix this, the graph matches the one on p.67. You might ask, "match?", since you will do not see the trend. But ask why, given our discussion of representing data in class. The answer of course it that by starting at 0% and going to 100% on the right axis (after fixing the labels), all the variation is squeezed into a very small visual range. It would be be better to show only 50%-70%; then you see the trend easily. E.g., you can look at: http://research.stlouisfed.org/fred2/series/EMRATIO?cid=12 p.322, Table 14.2 The table refers to data for 1950-1979, but the text describes the same data as applying to 1946-1979, as does figure 14.6. p. 324, Figure 14.7 The debt dynamics equation in this figure wrongly signs the intercept. Figure 14.7 The steady-state value in this figure should be should be -3.34 (not -0.33). Footnote 7, p.338 Typo: Elasticity definition is missing a 'Y' after the first Delta. Footnote 12, p.350 Typo: in the penultimate line, Farmer writes 'gE gQ' when he means 'gN gQ'. Equation 15.10, p.348 Minor typo: The solution for k-bar uses the glyph 'a' in stead of the glyph 'alpha'. Figure 15.4 The numbers in the pie chart are correct, but the proprtions shaded are wrong. Appendix to ch.15, p.353 Typo: The second equation show an incorrect '+' instead of the correct '=' right before kt. Problem 10, p.354 'n' is undefined. (It is the population growth rate.) Box 16.1, p.364 The error here is missing discussion. The data source is not specified. Panel A shows I/Y = 15% in contrast to the nearly 20% shown in figure 15.5b. Figure 16.7, p.375 The dynamic adjustment for Economy A is drawn incorrectly. (It "bounces" off the Economy B line instead of the 45 degree line.) A Few Problem Answers and Hints =============================== 10.13.a. Recall that the multiplier is M/H = (1+cu)/(rd+cu). We assume all reserves are required reserves. So the multiplier is m = (1+0.2)/(0.1 + 0.2) = 3. So M = 3 * H = 3 * 100M = 300M. 10.13.b. The multiplier is m = (1+0.15)/(0.05 + 0.15) = 1.15/0.2 = 5.75. Given: D = 10,000 So CU = cu * D = 0.15 * 10,000 = 1,500. So R = rd * D = 0.05 * 10,000 = 500. So MB = CU + R = 1500 + 500 = 2,000. So M = CU + D = 1,500 + 10,000 = 11,500. Or, M = m * MB = 5.75 * 2,000 = 11,500. 10.13.c. Now the multiplier is (1+0.25)/(0.05 + 0.25) = 1.25/0.3 = 25/6. It has shrunk because people hold more cash. We still have MB = 2,000. But now, M = mult * MB = 25/6 * 2,000 = 50,000/6. If you still want M=11,500, you need H = 11,500/(25/6)=2760, a rise of 760. 11.7 Start with the loanable funds diagram. Note that the **nominal** interest rate is on the vertical axis. But S and I depend on the **real** interest rate. Therefore both S and I down up by the change in expected inflation. There is no real changes implied by this diagram. To match the outcome in the loanable fund market (lower i with unchanged Y) the IS curve must shift down by the same amount. 12.1 Start with the discussion of 11.7. This time expected inflation rises. This case is illustrated in your textbook. Start with the loanable funds diagram. Note that the **nominal** interest rate is on the vertical axis. Therefore both S and I shift up by the change in expected inflation. There is no real changes implied by this diagram. However, when we move to the IS curve, this means that it also shifts up by the change in expected inflation. Again, the **nominal** interest rate is on the vertical axis. This time, however, there is an interaction with the LM curve. Unless the LM curve is vertical, the upward shift in the IS curve produces real changes. (E.g., an increase in Y.) 12.5 On the exam, I will not be asking about how such changes affect the IS and LM slopes. That said, if investment is interest sensitive, a large change in Y (and thus saving) can be offset by a small change in i, and the IS curve is therefore shallow (in Y,i-space). 12.10 The monetarists (Mark I) think that money can have powerful effects on the economy in the short run, and that a volatile money supply can correspondingly contribute to business cycles. Is that enough of a clue? 12.15.a Assume no foreign sector; zero expected inflation. IS: Y = C + I + G => Y = [100 + 0.8(Y-50)] + (50-25i) + 50 => 0.2 Y = 160 - 25i => i = 160/25 - Y/125 LM: M/P = L(i,Y) => 400/2 = Y - 100 i => i = Y/100 - 2 15.9 You can start with the most general version of the dynamic adjustment equation, as given in the appendix to chapter 15. k' = k(1-d)/(1+gE) + k^a sA/(1+gE) Solving for the steady state (when k'=k) we have: k(gE + d) = k^a sA => k = [sA/(gE+d)]^[1/(1-a)] For this problem you have special values: s=0.16, d=0.10, gE=0, A=1, a=0.5 So in the steady state, k = [s/d]^[1/(1-0.5)] = 1.6^2 = 2.56 15.10 Use the same formula as for 15.9. That is, in the steady state k = [sA/(gE+d)]^[1/(1-a)] For this problem you have special values: s=0.45, d=0.10, gE=0.05, a=0.5. We will set A=1, but you can keep it in your algebra. So in the steady state, k = [0.45/(0.15)]^[1/(1-0.5)] = 3^2 = 9 Copyright (c) 2007, Alan G. Isaac