Problem Set 5 Answers. ( ) 2. Yes, like temperature. See the plot of utility in the notes. Marginal utility should be positive.

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Business John H. Cochrane Problem Set Answers Part I A simple very short readings questions. + = + + + = + + + + = ( ). Yes, like temperature. See the plot of utility in the notes. Marginal utility should be positive. h i. = ( +) ( ) + = [ + ] If people are risk neutral, prices are just a constant times expected payoffs.. + = + + + + + +. k ( )k ( ) ( ) ( ) = = ( ) That s a problem, since the market Sharpe ratio is about 8/6=.. premium puzzle. This is called the equity Part I B. The plot is below. (a) ln( )+ ln( + ) = ln( + ) = ln( ) + = ln( ) Isetupagridfor, a couple of different chosen artistically, and set matlab to work on it. (b) The slope is in general [ ( + ) ] = + + + = + = + + = = + + The third to last equation is important indifference curve slope = ratio of marginal utilities = marginal rate of substitution. At + = the slope is /.9 -.. So the slope is a bit steeper than -.

(c) My solution is a little more general; I wait until the end to plug ( ) =ln( ) and the numbers in to the answer. You probably got there quicker by specializing earlier. You also might have avoided the Lagrangian by substituting, e.g. + =. = + + max { +} ( )+ ( + ) = ( )+ ( + ) + + From the first two, In the log case : ( ) = + : ( + ) = : + + = ( )= ( + ) = + = + Substituting into the budget constraint, + ( ) = = + = 9 = $ 8 + = = 9 + 9 = $ 9 In order to plot the associated indifference curve, I found the for this optimal choice ln( )+ ln( +) = ln( 8) + 9 ln( 9) = The budget constraint is of course + + = NowwehavealltheingredientsforThePlot: + = ( )= ( )

maximum budget const indiff curves 8 c t + 6 6 7 8 9 c t (d) Notice that + =, the investor consumes = 9 % more in the second period. The attraction of the % interest rate outweighs the impatience of = 9 and the investor is induced to save.. The point here is that the relative risk aversion coefficient is a measure of how willing people are to take a gamble. (a) i. ii. iii. ( ) = ( + )+ ( ) ( ) ( )+ ( ) + ( ) + ( ) ( ) + ( ) ( ) ( ) = ( ) ( ) = [ ( + )] ( )+ ( ) ( ) ( ) = ( ) ( ) = ( ) ( ) ( ) = = bribe variance 6

(b) Note ( ) = + ;so ( ) = (( ) ) =. Meanwhile, [( ) ] =. So we can restate the answer, how much extra mean do you need to compensate you for taking on variance, with both mean and variance expressed as percentages of expected consumption ( ) (( ) ) = ( ) (( ) ) ( ) = ( ) = ( ) = ( ) ( ) = That s why we call the risk aversion coefficient! The bribe is, but we usually leave out the / Part II A. Here are my regressions. Data sample 96.. regression of mean xs return on mean log beme and mean log size const size beme R b.66 -..66.76 As expected, more size means less returns, more beme means higher returns. These are about the same size coefficients as reported by FF Table. A % increase in size means a -bp decrease in monthly return; a percent increase in beme (i.e. beme from. to.) means a 7 bp increase in monthly return.. Here s the table. actual mean return.8.79.8..6.8.7.9.9..9.78.79.88.8.6.7.7.86.87.6..9.7.6 regression on average log size and average log beme.6.8.89.97...7.8.88..8.66.7.8.97..6.7.79.9...6.7.8 As usual, it s clearer in a plot: 7

mean returns from cs regression mean returns from mean beme,size mean mo return %.. mean mo return %.. size value size value I also made another plot of fitted vs. actual means.. sv Actual mean.8.6. sv sv sv sv sv sv svsv sv sv sv sv sv sv sv svsv sv sv sv sv sv sv sv....6.8. Fitted mean a + b*e(size) + c*e(bm) As you can tell in the numbers and graph, the cross sectional regression is doing a pretty good job. This plot looks very much like the mean return plot we started with. If the regression were perfect, we d see a degree line. The major failing is, no surprise, sv, the small growth portfolio. This portfolio does much worse than a linear extrapolation of being small and being growth suggests it should.. The regression with a cross term. ( )= + ( )+ ( )+ ( ) ( )+ regression of mean xs return on mean log beme and mean log size and cross term const size beme s x bm R b.8 -.8 -.9 -.68.87 8

One measure of success, the has gone up to.87 from.7. You can see how with the cross term the dark blue fitted value can go down for growth stocks and up for value stocks, and give therefore a much better fit. One way to see what this does, is that it allows a greater size effect for value than for growth stocks. I.e. ( )= +[ + ( )] ( )+ ( )+ mean returns from cs regression mean returns from mean beme,size, and cross term mean mo return %.. mean mo return %.. size value size value. sv Actual mean.8.6. sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv sv....6.8. Fitted mean a + b*e(size) + c*e(bm) + d*e(s)*e(bm) Though not perfect, you can see that the pattern of Table A is not really just a size effect and a beme effect, but there is an important interaction as well. The betas will have to have the same pattern if they are to explain these returns. 9

. Yes, you are allowed to do anything you want in these regressions Remember, we re just describing returns, we re not explaining returns here.. The content of these regressions is simply whether the pattern in Table panel A can be matched by a linear function of suitably transformed B/M and size. If so, that is a convenience, but no theory of anything hangs on the answer. Table A itself is a parameter nonlinear fit. Is a linear function of transformed BM and size a good description of average returns? Pretty much yes, except for the small growth anomaly. But unlike betas, there is no economic reason why average returns should be linear functions of BM and size. We can use nonlinear functions, or cross terms if we want. By contrast you are not allowed to use cross terms or nonlinear functions in ( )= + + + +( ) + ; = The factor model says average returns depend linearly on betas. Period. That s because there is a deep portfolio interpretation. We form the portfolio You can t do that with cross terms.. A reminder: the cross sectional regression regression of mean xs return on mean log beme and mean log size const size beme R b.66 -..66.76 Now, the forecasting regressions pooled forecasting regression of return on log beme and log size const size beme R b.6 -.6.. You see roughly the same coefficient on size. The beme coefficient is the same sign but about twice as big. Why? pooled forecasting regression of sample mean x for each portfolio b.66 -..66 pooled forecasting regression of x - sample mean x for each portfolio b.7 -.8.6 Aha! The forecast regression using the sample mean gives exactly the same result as the cross sectional regression! But we see in both coefficients that if portfolio A has an unusually high b/m, relative to portfolio A s long-term average, that is a signal that A will have a strong return the next year, and this signal is even stronger than the signal that portfolio A has, on average, a higher b/m than portfolio B. This is what FF left on the table. We can do better by looking at change in value over time, not just across stocks. This is a pooled regression with portfolio dummes. Taking out the portfolio averages and thus looking only at variation over time in the right hand variables + = + [ ( )] + [ ( )] + +

is mechanically the same as putting in portfolio dummies Additional discussion (not in the problem set) + = + + + + You can also do the opposite. Let s take out the average size and BM across portfolios at each moment in time. This is the same thing as adding time dummies, + = + + + + Now the regression ignores variation over time, but only asks if portfolio i has more than portfolio j, how does that affect returns pooled forecasting regression, time means b -.8 -.. pooled forecasting regression, removing time means b.7 -..89 In this problem you have run a cross-sectional regression and a pooled time-series cross-sectional regression. Part B. const size beme Cross sectional regression, regress_jc t statistics b.66 -..66 se.68..7 t 9. -.8 7.6 Conventional s,t statistics conv se.6..6 t 9. -.97 7. Corrected se, t statistics using Sigma/T corr se.9.7.8 t.8 -..86 Fama MacBeth coefficients, standard errors, t statistics FMB b.87 -.. se..7.8 t.889 -.6. We saw the regression above. The T statsitics are huge, boy these effects are strong. The "conventional" statistics ( ) ( ) use the variance of the residual, where the regression package uses /(N-k) in place of /N. As you see this makes a small difference in t statistics. I hope you remembered to take a square root The corrected t statistics are half or less of the uncorrected statistics! In cross-sectional regressions with thousands of individual stocks, a factor of is more common. Do not run cross sectional regressions

with t statistics that are not corrected for cross-sectional correlation. It is one of the most common mistakes even in published articles. The FMB coefficients are just about the same as the cross-sectional coefficients! There is close to a theoremhere. Ifthexvariablesarethesameovertime(ifwehadused ( ) not )thenfmband cross sectional coefficients would have been exactly the same. Variation of over time gives rise to the small difference. The FMB standard errors and ts are just about the same as the corrected ones! You can see a great similarity in the way its constructed. FMB was invented before this formula was invented, but survives because everyone knows it so well.