Hi All: I have the Lucas, 1972b paper "Econometric Testing Of the Natural Rate Hypothesis" but unfortunately not as a pdf. ( it's from a book that contains a lot of Lucas' papers called "studies in business cycle theory"). Assuming that someone reading this question has read that paper and understood it, could you explain how equations 11) and 12) come about. Thank you very much for your help.


I am going to write the equations that lead up to the two equations that I don't follow the derivation of. My hope is that not many people have the pdf of the file so the equations will obviously help.

1) $y_t = a(P_{t} - P^{*}_{t})$ (aggregate supply function ).

7) $y_{t} + P_{t} = x_{t}$ ( aggregate demand schedule )

8) $x_{t} =\rho_{1} x_{t-1} + \rho_{2} x_{t-2} + \epsilon_{t}$ ( AR(2) for $x_{t}$ ).

9) $ P_{t}^{*} = E(P_{t+1}| x_{t}, x_{t-1}, \eta_{t}) + \eta_{t} $ ( Definition of agent's expectation of the price at time $t+1$. )

10) $ (1+a) P_{t} - a P^{*} _{t} = x_{t} $ (obtained by straightforward algebraic manipulation of (1) and (7). )

Meanings of variables:

1) $P_{t}$ is the price level at time t.

2) $P^{*}_{t}$ is the expected price level at time t+1 where the expectation is taken at time t. ( the agent's expectation of price level ). Essentially, $P_{t+1}$ is the next period's price level, the variable of which, $P^{*}_{t}$, is a forecast.

3) $y_{t}$ is the output at time t.

4) $x_{t}$ is the log of the money supply and is a shift parameter. It is part of a policy rule which gives the current value of $x_{t}$ as a function of the state of the system.

After these equations are explained, the author then says:

"We shall seek linear solutions $P_{t}$ and $P^{*}_{t}$ to equations 9 and 10: "

11) $P_{t} = \pi_{1} x_{t} + \pi_{2} x_{t-1} + \pi_{3} \eta_{t} $

12) $ P^{*}_{t} = \pi_{4} x_{t} + \pi_{5} x_{t-1} + \pi_{6} \eta_{t} $

Equations 11) and 12) are the equations whose derivation escapes me totally. Thanks a lot for your help. Mark


P.S: For any newbies wrestling with the theory behind RE, this paper is the clearest and simplest that I have found, aside from the derivation of 11) and 12). I highly recommend it. I only found the book recently and have decided that out of all the top RE people, Lucas really stands out as far as the clarity of his writings.

  • $\begingroup$ I suspect that anyone who has read the paper probably won't remember what equations (11) and (12) are. So it's a good idea to at least type those equations out in your post. $\endgroup$ – Herr K. May 4 '18 at 19:21
  • $\begingroup$ thanks for advice. I put all of the equations in the edited question. $\endgroup$ – mark leeds May 4 '18 at 20:22

Equations $11$) and $12$) are not derived, they are assumed.

Specifically, with eq. $9$) Lucas defines/assumes that the value-expected for the $t+1$ price level is the conditional expectation based on $x_t, x_{t+1},\eta_t$ (plus $\eta_t$ on its own). This is imposing the Rational Expectations Hypothesis.

He then says "we will seek linear solutions..." which is translated "assume that the functional form of $E(P_{t+1}| x_{t}, x_{t-1}, \eta_{t})$ is linear in its arguments, and then determine (or estimate) the coefficients". Certainly, this is in all likelihood an approximation to the true functional form of this conditional expectation.

So assuming that

$$E(P_{t+1}| x_{t}, x_{t-1}, \eta_{t}) = \pi_{4} x_{t} +\pi_{5} x_{t-1} + \beta\eta_{t}$$

and inserting this into $9$) we get $12$) with the mapping $\pi_{6} = 1+\beta$.

Inserting $9$) in $10$) we get

$$(1+a) P_{t} - a [\pi_{4} x_{t} + \pi_{5} x_{t-1} + \pi_{6} \eta_{t}] = x_{t}$$

$$\implies P_{t} = \frac{1+a\pi_{4}}{1+a}x_t +\frac{a\pi_{5}}{1+a}x_{t-1}+\frac{a\pi_{6}}{1+a}\eta_{t}$$

which is $11$), after compacting and mapping the above coefficients to the $\pi_1, \pi_2, \pi_3$ ones.

| improve this answer | |
  • $\begingroup$ Thanks Alecos. I will go over this carefully and let you know how it goes. $\endgroup$ – mark leeds May 14 '18 at 13:31
  • $\begingroup$ Alecos: That was a beautiful answer. Thanks so much. $\endgroup$ – mark leeds May 15 '18 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.