This is a statement from a finance textbook - I find it pretty clear everywhere else, but this particular part I am clueless. Hopefully you guys can figure it out.

The problem is solving: $$\sup_{(C_t,M_t)} E[\int_0^\infty e^{-\delta t} \, u(C_t,M_t/I_t) \,dt]$$ $$s.t. E[ \int_0^\infty \zeta_t \cdot (C_t+\frac{M_t}{I_t}r_t) \, dt]\leq W_0$$

where $W_0, \delta$ are constants. u is some function (as nice as it has to be) and in general $(C_t) , (I_t), (M_t) , (r_t)$ are all stochastic processes. It is just stated, without further ado, that the first order conditions are $$ e^{-\delta t} u_C(C_t,M_t/I_t) = \psi \zeta_t $$ $$ e^{-\delta t} u_M(C_t,M_t/I_t) = \psi \zeta_t r_t$$ where $\psi$ is a Lagrange multiplier, which is set such that the constraint holds. $U_C$ and $U_M$ are derivatives with respect to C and M respectively. Then he continuous using this result as a truth and never speaks of it again.

To be honest I've got no idea - I've seen lagrange used before but I don't see it giving me anything here. (p.s. sorry - I couldn't think of a better title, feel very free to change it)

Edit: Fixed some typos - it should be clear from the answer which.


(I guess you're missing $dt$ in the constraint integral? -Also, it has to be that $\zeta_t$ is not the symbol of a function, but a separate entity multiplying the parnethesis that follows it).

There are a million things that must be assumed as premises to do straightforward Lagrange optimization here, but let's say that they are indeed assumed. In that case what the author does is the standard approach, namely: 1) Formulate the Langrangean: $$L=E[\int_0^\infty e^{-\delta t} \, u(C_t,M_t/I_t) \,dt] + \psi\left(W_0- E[ \int_0^\infty \zeta_t (C_t+\frac{M_t}{I_t}r_t)dt]\right) $$

2) Take the partial derivatives with respect to $C_t,M_t$ and set them equal to zero:

$$ \frac {\partial}{\partial C_t} L=0=E[\int_0^\infty e^{-\delta t} \, \frac {\partial}{\partial C_t}u(C_t,M_t/I_t) \,dt] - \psi E[ \int_0^\infty \zeta_t \frac {\partial}{\partial C_t}(C_t+\frac{M_t}{I_t}r_t)dt]=0 $$ (Note how we inserted the partial derivative operator inside the expected value and inside the intergal, in such a carefree manner - the million assumptions I mentioned before)

$$ \Rightarrow E[\int_0^\infty e^{-\delta t} \, u_C \,dt] - E[ \int_0^\infty \psi\zeta_t 1 dt]=0 $$

Apparently, the langrange multiplier here must be atemporal, that's why it can be inserted into the integral. Also apparently, the expectation is taken with respect to the same variable(s). Then, since the limits of integration are the same, and the variable of integration also the same, we can house the two integrands under one integral:

$$ \Rightarrow E[\int_0^\infty \left(e^{-\delta t} \, u_C \,-\psi \zeta_t\right)dt] =0 $$

A sufficient condition for this expression to equal zero, is that the integrand equals zero, so we arrive at $$e^{-\delta t} \, u_C \,-\psi \zeta_t = 0 \Rightarrow e^{-\delta t} \, u_C \,=\psi \zeta_t$$

...and I cannot tell where the t-subscript on $\zeta$ went. Same approach with the partial derivative with respect to $M_t$.

  • $\begingroup$ If I could give you 10 (+1) I would - you really saved my day! Does atemporal just mean it does not depend on t? Do you know anything about what it means for the problem that the lagranger multiplier has that property? (I fixed the typos you indicated in your answer) $\endgroup$ – Henrik Aug 10 '13 at 7:12
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    $\begingroup$ Yes "atemporal" means that it is not a multiplier associated with one time period, but with all time periods of the planning horizon viewed as "one". What does that mean? I suspect $\zeta_t$ must be some discount factor. Then the atemporal multiplier measures the effect on the value function of a change in the present value of the constraint. $\endgroup$ – Alecos Papadopoulos Aug 10 '13 at 13:48
  • $\begingroup$ Thanks a bunch! You are right - $\zeta_t$ is in fact a discount factor. $\endgroup$ – Henrik Aug 10 '13 at 18:06

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