Econ. 513, Time Series Econometrics  Fall 2000  Chris Sims

Mid-Term Review Answers^{∗}

(1) (a) In this 3-variable system, show that there are two unit roots, with all other roots less than one in absolute value, and 2 distinct stationary linear combinations of the variables. Standard results state that in an n-variable system, when there are m unit roots and the rest are less than one in absolute value there should be exactly n − m cointegrating vectors. What regularity condition of the usual arguments is violated here to produce the extra stationary linear combination?

x_{t} = 2x_{t−1}+1.7y_{t−1}+.3z_{t−1} − x_{t−2} − y_{t−2}+ε_{xt}
y_{t} =  .3y_{t−1}−.3z_{t−1}  +ε_{yt}
z_{t} =  .2y_{t−1}+.7z_{t−1}  +ε_{zt} .

The system matrix, once we have stacked it to recognize that lagged x and y need to be part of the state vector, is

[\begin{matrix}2 & 1.7 & .3 & −1 & −1 \\ 0 & .3 & −.3 & 0 & 0 \\ 0 & .2 & .7 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{matrix}] .

The brute force approach is to call this matrix B and ask matlab to do [v, d] = eig(B). This gives us on the diagonal of d 1, 1, .5±.1414i, and one zero. Thus the model has two unit roots (the 1’s) and three stable roots (the imaginary pair and the zero). The stationary linear combinations of the 5 variables in the stacked system are the three left eigenvectors corresponding to the stable roots. Note that because two of the roots are imaginary, the corresponding eigenvectors are imaginary, and the stable real linear combinations corresponding to them are linear combinations of their real and imaginary parts. When v is non-singular, the left eigenvectors of B are just the columns of v^{−1}, but this model has a repeated root of 1, meaning that it has two roots of 1, while there is only one distinct eigenvector corresponding to that root (so v is singular). So to find left eigenvectors we just look at the columns of the matrix vi=eig(B'). The left eigenvectors corresponding to stationary linear combinations, found this way, consist of a complex pair with non-zero entries only in positions 2 and 3, and a third, real vector with non-zero entries in position1 2, 3, and 5. The complex pair implies that y_{t} and z_{t} are both stationary, as is of course any linear combination of them. The real eigenvector, since it involves lagged x and y, carries no implications about stationary linear combinations of unlagged variables. A solution method more appropriate to an in-class exam would first note that the system given is block-recursive, so that y, z is Granger-causally prior to x. This may not be apparent from the system matrix written as above. To see this we write the system instead as

∗Copyright 2000 by Christopher A. Sims. This document may be reproduced for educational and research purposes, so long as the copies contain this notice and are retained for personal use or distributed free.

[\begin{matrix}1 − 2L + L^{2} & −1.7L + L^{2} & −.3L^{2} \\ 0 & −.3L & .3L \\ 0 & −.2L & −.7L \end{matrix}] [\begin{matrix}x_{t} \\ y_{t} \\ z_{t} \end{matrix}] = [\begin{matrix}ε_{xt} \\ ε_{yt} \\ ε_{zt} \end{matrix}] .

Thus the y, z pair is determined by the last two equations alone and we can express y, z entirely in terms of weighted averages of their own past innovations. The stability properties of the y, z pair are therefore determined by the 2 × 2 system matrix

[\begin{matrix}.3 & −.3 \\ .2 & .7 \end{matrix}] ,

which has only the complex pair of eigenvalues from the original system. This shows that there are two stationary linear combinations—any linear combination of y and z. (Of course there are infinitely many such linear combinations. There are only two linearly independent linear combinations, however.) Two show that there are two unit roots, one approach in working by hand rather than with a computer would be just to form the characteristic polynomial of the system (|B − λI|) and substitute 1 for λ. The polynomial will turn out to be zero for the value of λ, showing that there is one unit root. Then one could divide the polynomial by 1 − λ and substitute λ = 1 into the result. This would again yield zero, showing that there is a second unit root. The idea here is that, if you are told the value of the root you are checking, it is easy to verify that it is a root, even for a high-order polynomial. The regularity condition for standard cointegration results that fails here is the requirement that there be no repeated roots in the system. The conditions for the usual analysis are that all the unstable roots are the same and that they not repeat. (Usually in the literature it is assumed that they all be exactly one, but that is not essential.) Again, the requirement that they not repeat means that each occurrence of an unstable root must correspond to a distinct eigenvector.

(b) Suppose the coefficients on x(t − 2) and y(t − 2) in the first equation were both changed to -.99, instead of -1. How many stationary linear combinations and unstable roots are there in this case? Does this modified system satisfy the regularity conditions that underly the standard cointegration analysis? The lower two equations are unchanged, and they still determine y and z by themselves, so the same two stable complex roots are still present. The only question is what has happened to the two unit roots. If you form |B − λI|, which is feasible here even by hand because there are so many zeros in the matrix, you are likely to notice that it can be written as

(−λ) · (λ^{2} − 2λ + .99) |[\begin{matrix}.3 − λ & −.3 \\ .2 & .7 − λ \end{matrix}]| ,

where the 2×2 determinant is that of the system matrix for the 2×2 subsystem. Failing that, you could have multiplied out the whole 5th order polynomial, observing immediately that λ = 0 is a root, and then dividing by λ^{2} − λ + .27, which is the characteristic polynomial of the 2 × 2 subsystem. The roots of λ^{2} − 2λ + .99 are 1.1 and .9. Thus this system has a single non-stationary root, and it fits the regularity criteria for cointegration theory: There is one unstable root, and two stable linear combinations in this three-variable system. As before, any linear combination of y and z alone is stationary. Any linear combination of the three variables with a nonzero coefficient on x is explosive.

(c) Suppose that in addition to the changes in (1b), the coefficient on y(t−1) in the first equation is decreased to 1.69 instead of 1.70. How many stationary linear combinations and unstable roots are there in this case? Does this modified system satisfy the regularity conditions that underly the standard cointegration analysis? Though this is not what I thought I was doing when I set up the problem, this change has no effect on the roots of the system or on the previous section’s argument that there are now two linearly independent stable linear combinations (involving y and z only) and that x is unstable.

(2) What’s the Jeffreys prior for θ in the first-order MA model

y_{t} = ε_{t} − θε_{t−1} ?

If you use the exact full sample likelihood, you’ll get a nasty expression. So instead assume all the pre-sample y’s are known and conditioned on, or else aim at a limiting result that applies in large samples.

The log of the joint pdf of ε_{1} …, ε_{T} is

− ∑_{t=1}^{T} \frac{ε_{t}(θ)^{2}}{2σ^{2}} − T log σ .

Since

ε_{t}(θ) = y_{t} + θy_{t−1} + θ^{2}y_{t−2} + …,

the Jacobian of the transformation from y_{1}, …, y_{T} to ε_{1}, …, ε_{T} is 1 and the log pdf for these y’s, conditional on all the y_{t}’s for t < 1, is obtained just by substituting expressions in terms of y for ε in the ε pdf. The expected second derivative of the log likelihood with respect to θ then can be seen to be

−(1/σ^{2})E [(∑_{t=1}^{T} ∑_{s=1}^{∞} sθ^{s−1}y_{t−s})^{2}] = \frac{−1}{1 − θ^{2}} .

The Jeffreys prior element corresponding to θ is then proportional to 1/\sqrt{1 − θ^{2}}. If one also considers separately a Jeffreys prior on σ, or else if one forms a joint Jeffreys prior for θ and σ, one ends up with a 1/σ factor as well.

(3) Consider an island on which the population’s income comes from coconut trees, which are privately owned, and haircuts. The (unrealistic) trees last forever need no fertilizer or care, and can’t be planted. However, there are good and bad years for coconuts, so the quantity of coconuts harvested varies randomly, though in a serially correlated way, from year to year. People who own few or no coconut trees provide haircuts to those who own more, and are paid in coconuts.

(a) Will the quantity of coconuts be Granger causally prior to the price of coconuts in terms of haircuts in this economy? The economy is specified to have the quantity of coconuts determined by a mechanism that is unaffected by human activity, including the price of coconuts. It thus is likely that lagged values of coconut price do not help predict the quantity of coconuts. This presumption could be wrong, though, if there are important variables that we cannot observe in our data set. For example, coconut harvests could be affected by rainfall and temperature, while these same variables might also affect demand for coconuts. In that case, if we did not have data on rainfall and temperature, we might find that lagged coconut prices helped predict current quantities so that GCP failed. However, the problem statement was meant to suggest that there are no such omitted variables.

(b) Can we estimate the demand for coconuts in this economy by GLS applied to a regression of the price of coconuts (in haircut units) on income (in haircut units) and the quantity of coconuts consumed? Why or why not? Of course if the quantity of coconuts is not GCP, then we will not have exogeneity of the right-hand-side variables in this regression and cannot use GLS. Even if the quantity of coconuts is GCP, though, validity of GLS in this regression requires that both GCP and income be exogenous, which would imply that income and quantity are GCP as a block. This is unlikely here, though, as random shifts in demand, by changing the price of coconuts, will also affect income measured in haircut units. It will therefore be unreasonable to maintain the assumption that the disturbance in the demand equation is unrelated to the right-hand-side income variable.

(4) Suppose our model is

y_{t} = θ_{1}y_{t−1} + ε_{t} + θ_{2}ε_{t−1} + θ_{3}ε_{t−3} ,

where as usual it is assumed that ε is i.i.d. N(0, 1).

(a) Give a formula for a surface or surfaces in ℝ^{3} (the (θ_{1}, θ_{2}, θ_{3}) parameter space) along which the likelihood function for this model will be constant. The model is the same for all θ’s for which

\frac{1 + θ_{2}L + θ_{3}L^{2}}{1 − θ_{1}L}

is the same. This will happen for all θ’s in which θ_{1} is one of the roots of the numerator polynomial, while the other root is held fixed. Equations defining this condition are, e.g.

θ_{1} = \frac{−θ_{2} + \sqrt{θ_{2}^{2} − 4θ_{3}}}{2}

\frac{−θ_{2} − \sqrt{θ_{2}^{2} − 4θ_{3}}}{2} = r .

For each value of r, this pair of equations defines a 1-dimensional manifold (i.e. a curve) in 3-space along which likelihood is constant.

(b) Suppose a MLE for this model is θ̂ = (.9, −2, .96). Find another value of the θ vector that you know will have exactly the same likelihood. Which of these two estimates is not a fundamental representation? (I.e., for which is the variance of ε not the one-step-ahead forecast error variance?) The roots of the numerator polynomial 1 − 2L + .96L^{2} are 1.2 and .8. We know, therefore, that this polynomial does not have a convergent inverse in positive powers of L. Since the ε’s are i.i.d., this cannot be the fundamental representation, because it implies that we cannot represent ε in terms of current and past y’s. A representation that implies the same ACF but does make the numerator polynomial invertible would replace the 1.2 root with 1/1.2 = .833333. However, this representation would imply a different error variance. The problem statement should have made explicit the fact that to achieve the same likelihood one would have to modify the unit variance on ε as well as the values of the θ’s. The idea is that if A(L) = 1 − 2L + .96L^{2} = (1 − .8L)(1 − 1.2L), then

A(L)A(L^{−1})′ = (1 − .8L)(1 − 1.2L)(1 − .8L^{−1})(1 − 1.2L^{−1})
= (1 − .8L)(L^{−1} − 1.2)(1 − .8L^{−1})(L − 1.2)
= (1 − .8L)(1 − .8333L)(1 − .8L^{−1})(1 − .8333L^{−1}) · 1.44 ,

and therefore with θ_{2} = 1.6333, θ_{3} = .6667, Var(ε) = 1.44, we get exactly the same ACF and hence likelihood as with the original estimates. Because the roots of the numerator and denominator polynomials are now all outside the unit circle, this model is fundamental and the innovation error variance is 1.44. It is worth noting that when ARMA models imply the same ACF, the one that implies the smaller innovation variance is never the fundamental representation. This is because the variance of the disturbance in an ARMA, whether fundamental or not, is always the forecast error from prediction based on past disturbances, and past y’s are represented (by the ARMA model) as functions of past disturbances. Thus the prediction based on past disturbances uses an information set at least as big as does prediction based on past y’s alone. So every ARMA residual variance is no bigger than the true innovation error variance.

(5) You have issued an open invitation to 10 friends to come to a picnic Saturday. It’s Friday night, and you have to buy beer. The weather report is that the probability of rain is 50%, which you take as an objectively correct conditional probability, and if it rains you expect fewer will show up. If not, more are likely to come. However exactly how many is uncertain in either case. Neither you nor any of your friends ever drinks more than two cans of beer at a picnic, and any leftover beer will all have to be thrown away. Your loss function is convex and depends positively on how much you spend on beer and on the difference between the number of cans it turns out your guests would like to consume and the number you have actually bought.

(a) Show that, assuming beer is bought by the can (with no quantity discounts), the purchase of any number of cans of beer up to a certain maximum is an admissible decision, and describe how any such decision can be characterized as a Bayesian decision rule. The space of possible outcomes is characterized by the pair (n, r, δ), where n is the number of cans of beer those who come (including you, though the problem statement is ambiguous about this) would like to drink, r is 0 if it doesn’t rain and 1 if it does, and δ, your decision, is the number of cans of beer you buy. Your loss function has the form f(n − δ) + pδ, where f is convex and has its minimum at 0. If everyone who you invited came, and all would like to drink the maximum two cans, losses would be f(22 − δ) + pδ. The minimum of this function w.r.t. δ occurs at f′(22 − δ) = p, which, because of convexity of f, occurs at some point δ̄ ≤ 22, if it occurs at all. (Here, since f is actually defined only at integers, we interpret f′(x) as the slope of a tangent to f at x.) It is possible that f′(x) < p for all x < 22. In this case the optimum is δ = 0 no matter what happens. Setting δ equal to δ̄ is admissible, because it is the best decision in the event that people at the party actually want 22 beers. Similarly, if no one comes to the party except you, and you turn out not to feel like drinking any beer, n = 0 and the minimum of the loss function is clearly at δ = 0. Since this event is also possible, δ = 0 is admissible. Finally, any δ between 0 and δ̄ is admissible because it will satisfy f′(n−δ) = p for n = 22+δ −δ̄. Obviously any of these admissible choices of δ can be supported as Bayesian rules based on a prior that puts probability one on a particular value of n.

(b) Suppose that beer can only be bought in 12-packs, and that if you purchase two cases (48 cans) you get a discount of 20% on the full purchase price. Show that under these conditions there is a quantity to purchase that is admissible that was not admissible in (5a). Is it still true that every admissible decision rule is Bayesian? I meant to make this 24 cans for the 20% discount, not 48. Since there is no way the party needs more than 22 cans of beer, and since 48 cans clearly cost more than 24, despite the discount, it can’t pay to take advantage of the discount. The choices for δ are now limited to 0, 12 and 24 instead of to the integers, but because of my mistake in specifying the quantity discount, the structure of this part of the problem is really no different from the first part. It is true, though, that it could be optimal to purchase 24 cans of beer, which would have been inadmissible without the limit to purchasing by the case. It could be that f(22 − 24) − f(22 − 12) < 0, so that at a low enough cost per 12-pack, under certainty that n = 22, purchasing two 12-packs will be optimal, and hence admissible. It is also in this case Bayesian for a prior that puts probability one on n = 22.

(c) Comment: This exercise was not structured well. It was meant to remind you that admissible rules can fail to be Bayesian in cases where the available decisions do not form a convex set, which they don’t in this example. But still all the admissible rules are Bayesian in the example. We could get a more interesting (and less spuriously complicated) example by supposing that there are only three options, purchase no beer, purchase 24 cans of beer, or purchase bad beer from a store that allows returns. It might then be plausible to say the losses are n if no beer is purchased, 24 − n if 24 cans are purchased, and 14, regardless of the value of n, if bad beer is purchased. Expected losses under a prior π(n) on n are then

∑_{n=0}^{22}π(n) · n  ∑_{n=0}^{22}(1 − π(n)) · (24 − n)  14 ,

respectively, under the three options. The bad-beer choice is clearly admissible, since neither other option dominates it for all values of n. But if it were to be a Bayesian decision rule it would have to satisfy

∑_{n=0}^{22} π(n) · n > 14

∑_{n=0}^{22} π(n) · (24 − n) > 14 .

Adding these two inequalities together produces the contradiction 24 > 28, so the bad-beer choice is not Bayesian under any prior. If the set of available risk functions were convex, we would be able to choose a decision that delivers the average of the no-beer and 24-cans decisions, i.e. a 12cans decision. This would dominate the bad-beer choice and be a Bayes rule for priors symmetric around 12.

(6) Suppose we postulate that the length of recessions is exponentially distributed and i.i.d. across time. That is, the i’th recession in the data lasts t_{i} months (we pretend fractional months are allowed), and t_{i} is i.i.d. across i with pdf

λe^{−λt_{i}} .

Our sample contains 6 recessions with mean recession length 38 months.

(a) Find the MLE of λ. The likelihood for the sample is

λ^{6} exp (−λ ∑_{i=1}^{6} t_{i}) ,

from which it is easy to deduce that the MLE is λ̂ = 6/ ∑ t_{i} = 1/t̄.

(b) Show that the likelihood has the form of a Γ distribution on λ. Using this fact, and either a table of the distribution or Matlab’s gammainc function (or an equivalent computer program), find a minimum-length flat-prior 95% probability interval for λ. From the expression above for the likelihood, it is easy to see that the likelihood is proportional to a Γ(7) pdf on λ ∑ t_{i}. A minimal length 95% interval for a Γ(7) variate is (2.35,12.26). (It satisfies the criterion that the probability between those points it .95 and at the same time the height of the pdf is the same at both points.) Thus for λ the interval is the interval for the standard Γ(7) multiplied by 1/(6 ∗ 38), or (.0103,.053).

(c) Find the pdf of the sampling distribution of the sample mean recession length t̄ = ∑ t_{i}/6 as a function of λ and t̄. [Use the fact that the sum of two independent random variables with Γ(p, λ) and Γ(q, λ) distributions is distributed as Γ(p + q, λ).] Since each t_{i}λ is, according to the model, distributed as Γ(1), their sum is Γ(6), and thus t̄ is a Γ(6) multiplied by 1/(6λ). Thus the pdf for t̄ is proportional to t̄^{5} exp(−6λt̄).

(d) Display an exact 95% confidence interval for λ based on the sampling distribution of t̄. Does it differ from the flat-prior posterior interval you have already derived? The minimal length 95% probability interval for 6λt̄ is then (1.76,10.86). Thus there is 95% presample probability that λ lies in (.0077,.048), which is different from the flat-prior Bayesian interval. The difference arises entirely from the 7 degrees of freedom in the likelihood and the 6 degrees of freedom in the sampling distribution. A prior on λ proportional to 1/λ would eliminate the discrepancy, and this is the Jeffreys prior for this model.

(e) Since t̄ is a mean of i.i.d., finite-variance variables, it is asymptotically normal. Find a 95% confidence interval for t̄ based on this asymptotic approximation. Is it the same as a 95% posterior probability interval based on the second-order Taylor expansion of the log likelihood about the MLE? The asymptotic approximation, since each t_{i} has variance 1/λ^{2}, takes t̄ ∼ N(1/λ, 1/(6λ^{2})). The standard approach would then assert

P [t̄∈ (\frac{1}{λ} ± 1.96\frac{1}{\sqrt{6}λ})] ≐ .95

and thus lead to, as an interval for λ, (.0053, .0474). This is exactly the same as the Bayesian interval based on the second-order approximation to the log likelihood about the MLE. [This is always true. The Bayesian asymptotics and the classical asymptotics lead to the same confidence regions so long as both lead to normality in the asymptotic distributions.]