math and interpretation problem

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Xian Chi

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Jul 18, 2010, 3:10:43 PM7/18/10
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Ladies and Gentleman, Dudes and Dudettes!

I have got a question, for most of you I guess it's pretty trivial. So
I would really appreciate it if you could help me out a little bit, as
I am not the biggest math genius.

There is a consumer maxmization problem

\max_{c(t),k(t)} \int_{0}^{\infty} e^{-pt}u(c(t))dt
s.t.
c+\dot{k}=wL^(c)+(r-\delta)(1-\tau)k+T^{c}

ok, fine the problem is easy to solve by using a Hamiltonian

H=e^{-pt}u(c(t)) +\lambda[wL^(c)+(r-\delta)(1-\tau)k+T^{c}-c]

and solving for the first order conditions

\partial H / \partial c: e^{-pt}u'(c)=\lambda (1)
\partial H / \partial k: \lambda(r-\delta)(1-\tau)=-\dot{\lambda} (2)
\partial H / \partial \lambda: ... (3)
\partial \lambda / \partial t: .... (4)

Okay this way is clear. I merge (4) and (1) and (2). Toegether with two
this gives me the two equilibrium equations. I know that (2) could be
interpreted as a sort of arbitrage condition or fisher equation.
Merging (1) and (2) would therefore give the following arbitrage
condition:

-\dot{\lambda}=e^{-pt}u'(c)(r-\delta)(1-\tau) (5)

And this is where my problem starts. The author denotes the arbitrage
condition as:

u'(c) \equiv \int_{t}^{\infty}e^{-pt}u'(c)(r-\delta)(1-\tau)

Ok this looks very similar. But where does the integral come from? How
is it possible to define it as u'(c) ? Can I just take the integral
from (5) to write \lambda instead of /dot{\lambda} and interpret
\lambda as the shadow price of consumption?

The problem is from the classic Kenneth Judd paper from 1985. Here is
the free link to the working paper version

www.kellogg.northwestern.edu/research/math/papers/572.pdf

It is about equation (2) in the text, page 4. I mean I wouldn't take
this to serious, the equilibrium conditions can be found by the
Hamiltonian anyway, but I need this arbitrage to show later that the
solution is bounded.

It would be so nice and helpful from you if you could take a look at
this problem and give me some ideas! Best regards,

yours Chi

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