Before leaving on his pilgrimage, the King appoints two ministers. He gives the first minister control of instrument m, and tells him to set m so that the target variable M is equal to the target M*. He gives the second minister control of instrument f, and tells him to set f so that the target variable F is equal to the target F*.
When he returns from his pilgrimage, the King looks at the data on m and f, and M and F. He sees that M was always equal to M*, and that F was always equal to F*. He rewards both his ministers for following his orders well.
The King then learns from his spies that the two ministers had disobeyed his orders. They had swapped targets. The first minister had set m to hit the target F*, and the second minister had set f to hit the target M*. The King shrugs his shoulders, because none of his subjects could tell the difference. The court econometrician couldn't tell the difference either.
Principal-agent problems are not always interesting problems.
Let V be a vector of shocks that are visible to both ministers and U be a vector of shocks that are not visible to one or both ministers. The structure of the economy is given by:
M = M(m,f,V,U) and F = F(m,f,V,U)
If the two ministers follow the King's orders, m is set to minimise E(M-M*)2, and f is set to minimise E(F-F*)2. If they disobey orders and swap targets, f is set to minimise E(M-M*)2, and m is set to minimise E(F-F*)2.
Assume that there always exists a feasible pair {m*,f*} such that M=M* and F=F*. Both targets can be hit at the same time, if both ministers get it exactly right. (Satiation is feasible.)
We need to consider games where both ministers move simultaneously (Cournot), and games where one minister moves first (Stackelberg).
Under what circumstances would the equilibrium be the same or different if the two ministers disobeyed orders and swapped targets?
1. If f is not an argument in M(.), or if m is not an argument in F(.), then there is no well-defined equilibrium if the two ministers swap targets. At least one minister won't be able to move his target variable, if they swap targets, and so won't have a well-defined reaction function. Let us set this case aside.
2. Under certainty (where U is not an argument in either M(.) or F(.) ) it makes no difference if the two ministers swap targets. This is true whether we have the Cournot or the Stackelberg game. The proof is straightforward. If the equilibrium is anything other than {M=M*; F=F*} then both ministers would know this in advance, and at least one would have made a different choice, and so there cannot be an equilibrium except at {M=M*; F=F*}.
3. I am not sure whether we get the same result as in 2 if there is uncertainty but the two ministers have the same information sets. My hunch is we would only get the same result as in 2 if U has a symmetric distribution and both M(.) and F(.) are linear functions, so we get certainty-equivalence.
4. If the two ministers have different information sets, the equilibrium will generally be different if the two ministers swap targets.
4a. For example, if only the minister in charge of m observes Um, and only the minister in charge of f observes Uf, and if M=M(m,f,V,Um) and F=F(m,f,V,Uf), then both will always hit their target if they follow the king's orders, and both will nearly always miss their target if they swap targets.
4b. For example, if the minister in charge of f has more information than the minister in charge of m, then it would be better for the minister in charge of f to be the one who moves first, because then the minister in charge of m will learn from observing the other minister's move. And swapping targets may affect the value of the information learned.
My hunch is that 4b is what is important in assigning targets to monetary and fiscal policy. The central bank moves last. And only the fiscal authority knows what it is trying to do.
The game would be very different if one of the ministers was told to maximise something where satiation is impossible. For example, if the minister in charge of f was told to minimise E(F-F*)2 where F* was outside the feasible set. ("Try to maximise the Social Welfare Function to get my subjects as near to heavenly bliss as you can."). Because then I could not use my proof in 2 above. Maybe that's why the assignment of targets to instruments matters.
I don't understand this very well.
(I wrote this because I was thinking about an old post by JW Mason.)
Worthwhile Canadian Initiative 
JW, maybe we’re talking about different problems. What is the Jacobian in the case you’re talking about? Is it this:
Jacobian = A = [dM/dm dM/df; dF/dm dF/dm] where “;” separates the two rows of the 2×2 matrix? If that’s the case, then the Jury test just ensures that the largest eigenvalue of the 2×2 Jacobian matrix A is within the unit circle, which is the normal test for stability for a discrete time feedback system that looks like this:
x[n+1] = Ax[n]
If both eigenvalues are within the unit circle this feedback system will converge to zero (i.e. the zero vector: [0 0]’) with infinite iterations, no matter what the initial x is (x[0]). If both eigenvalues are on the unit circle, it’s a unitary operator (length preserving), no matter the initial x. If one’s inside and one’s outside, what it does depends on the initial x. If both are outside, then any initial x which is not the zero vector will eventually make it blow up.
However the system I was referring to didn’t use A (the Jacobian) in this way. Since I solved first one row, and then the other, using the results of one to feed into the the other, I essentially broke the problem down into two 1×1 systems:
x1[n+1] = K1 + G1x1[n]
x2[n+1] = K2 + G2x2[n]
Where Ki and Gi are constructed from the elements of the original Jacobian (and the original desired result Y in the cases of the Kis). You don’t need to do a 2×2 Jury test on that, you just have to be sure that |G1| < 1 and |G2| < 1 (essentially two 1×1 Jury tests). Now it turns out that G1 = G2 … let’s call that g, so you can just do one test: |g| < 1. It’s also true we can make a new 2×2 discrete time system out of that, but it’ll be an uncoupled system (i.e. the off diagonals will be zero):
x[n+1] = Gx[n] + K, where G = [g 0; 0 g], and K = [K1 K2]’;
Doing your 2×2 Jury test (test for eigenvalues inside the unit circle) on 2×2 diagonal matrix G will give you
1 – 2g + g^2 > 0
g^2 < 1
1 + 2g + g^2 > 0
Well the 1st and 3rd lines can be factored as (g – 1)^2 and (g + 1)^2 resp., which are squares, so it’s always true they’re > 0 (provided g isn’t +/- 1) which just leaves g^2 < 1 which is the same as requiring |g| < 1, assuming the Jacobian (A) is all real valued in the first place. That was my original test.
So anyway, that’s a long aside, but my point is that iteratively calculating
x[n+1] = Ax[n] equation 1
with A the Jacobian is a lot different than iteratively calculating
x[n+1] = Gx[n] + K equation 2
Which is what I thought you were trying to do.
Please let me know how it’s going… I’m interested to know if I have the correct Jacobian here, and if so why you want to calculate equation 1 iteratively rather than equation 2. Recall, that I originally set up the problem like this:
Y = Ax
So it’s the elements of both Jacobian A and desired result Y that go into K, but only elements of A that go into g (and thus G).
Also, you can try what I coded up here (at the very bottom… scroll past all the junk at the top):
http://brown-blog-5.blogspot.com/p/extra.html
You can try it out in Scilab for youself. I’m pretty sure it will always work for you. (Scilab is a free download). The solutions always iteratively converge to x = inverse(A)Y if it passes my convergence test, and they don’t if it doesn’t.
Tom-
I shouldn’t have said your solution was not right. Sorry. I was wrong for two reasons. First, you are right, the problem as I’ve set it up is slightly different from the way you did. I have both variables adjusting simultaneously on the basis if the previous period’s values, as opposed to one and then the other. Exactly as you say. Second, for reasonable parameter values (at least, if each instrument moves it’s own target in the right direction without overshooting) the first and third conditions of the Jury test will am
Hi JW, your last sentence got cut off there. But let me construct a new feedback system where both x1 and x2 are solved for simultaneously assuming the other will stay fixed. In that case we have
x1[n+1] = (y1 – a12x2[n])/a11
x2[n+1] = (y2 – a21x1[n])/a22
Which we can rewrite as X[n+1] = BX[n] + C
Where B = [0 -a12/a11; -a21/a22 0], C = [y1/a11 y2/a22]’
trace(B) = 0
determinant(B) = a12a21/(a11a22) = d
1 – trace + determinant > 0 implies d > -1
1 + trace + determinant > 0 implies d > -1
d < 1
So again, the condition for stability is the same since d = g = a12a21/(a11a22). If there is a steady state solution it’s X = inverse(I-B)C
If you work that out, you’ll see it’s the correct answer (same as above), i.e. inverse(A)*Y
Am I getting warmer, or still not matching what you’re after?
… and I think if you start with this instead:
x2[n+1] = (y1 – a22x1[n])/a12
x1[n+1] = (y2 – a11x2[n])/a21
Then your condition becomes |a11a22/(a12a21)| < 1
To see that’s true just match up the coefficients with the above. It doesn’t matter what Y is for stability. Now swap y1 w/ y2 to see the steady state solution is the same too.
Sorry, commenting glitch and then I had to run to class. What I was going to say was that the first and third conditions will always be satisfied, so everything comes down to the determinant being less than one. Which is just another way of describing the condition you gave originally. So yes, exactly what I was after.
JW, are you sure the 1st and 3rd don’t contribute? What I have above is that the 1st and 3rd are redundant, but they put the lower bound on the determinant, and the 2nd provides the upper bound:
-1 < d < 1
… and what this all says is it doesn’t really matter whether we solve first for x1 using row 1, assume that x1 is now fixed and solve for x2 using row 2, etc. Or if we solve for x2 using row 2, then assume x2 is fixed and solve for x1 using row 1, etc. Or if we solve for x1 using row 1 assuming x2 is fixed and simultaneously solve for x2 using row 2 assuming x1 is fixed (your case), etc. All three are almost exactly the same, have the same test for convergence, and converge at the same rate.
What matters is the assignment, like you started off stating: i.e. solving x1 using row 1 or row 2, etc. And then the test and convergence rate in one case just become reciprocals in the other assignment case.
The don’t contribute if you set up the problem the way I think you should. Writing b for the budget balance, i for the interest rate, Y for output and D for the change or level of debt, then if dY/db < 0, dD/db < 0, dY/di < 0, and dD/di > 0 (all of which we have good reason to believe), then the determinant is always positive. So we don’t need the lower bound condition.
Ah, OK. Well great… thanks for a fun conversation.