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I am reading Dynamic scoring: A back-of-the-envelope guide by Mankiv and Weinzierl (here) and on page 1420, I don't get the FOC in equation $(10)$, which says $r=...$. I do get the FOC with $v'(n)=...$ by using Lagrange.

I found the same FOC in a paper from Ferede (Dynamic Scoring in the Ramsey Growth Model, here) and he says, that it

is obtained by combining the first order conditions of the utility maximization with respect to capital and consumption (page 5).

But I only observe $\lambda=-c^{-\gamma}e^{gt(1-\gamma)}e^{(1-\gamma)v(n)}$ from the FOC w.r.t. consumption and $-\lambda[(1-\tau_k)r-g]=0$ from the FOC w.r.t. capital.

How am I supposed to get $\dot{n}$ and $\dot{c}$ there?

Well let me show you, what I have got: The Lagrange-function is given by:

$$L=\frac{1}{1-\gamma}[c^{1-\gamma}e^{gt(1-\gamma}e^{(1-\gamma)v(n)}-1]-\lambda [(1-\tau_n)wn + (1-\tau_k)rk - c - gk +T - \dot{k}]$$

So the FOC w.r.t. consumption is given by $\frac{\partial L}{\partial c}=c^{-\gamma}e^{gt(1-\gamma)} e^{(1-\gamma)v(n)}+\lambda=0$, while the FOC w.r.t. to capital is $-\lambda[(1-\tau_k)r-g]=0$.

So the equation for $\dot{\lambda}$ is based on the equation $\frac{\partial \lambda}{\partial t}=\frac{\partial \lambda}{\partial c} \frac{\partial c}{\partial t}$. And you are fully differentiating it.

I hope someone can help me.

NEW: So again: $\lambda=e^{-pt}u'(c)$. So you are indeed right with your equation after your words "Then substitute back in the FOC for consumption", but then you substitute it to the FOC wrt to k: $\dot{\lambda}=\lambda[g-(1-\tau)r]$, so we get $\gamma \dot{c}/c - (1-\gamma)(g+v'(n)\dot{n})+p=g-(1-\tau)r$, so that the signs don't fit anymore and this unfortunately leads to something else than equation $(10)$.

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  • $\begingroup$ So you should be using a Hamiltonian here, not a Lagrangian, because you're objective function is an integral over time. $\endgroup$
    – VCG
    Commented Aug 22, 2016 at 18:58
  • $\begingroup$ I wrote it all out. Your problem came from incorrect FOCs and yes you need to totally differentiate, because it's a foc so the choice variables are functions of all the parameters. $\endgroup$
    – VCG
    Commented Aug 22, 2016 at 19:09

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So Lets use a present value Hamiltonian here:

$H=e^{-pt}u(c)+\lambda(\dot k)$

FOC: wrt c: $0=e^{-pt}u'(c)-\lambda$

wrt k: $-\dot\lambda=\lambda[(1-\tau)r -g]$

Take the FOC from consumption and differentiate wrt to time:

$\dot\lambda =-\gamma c^{-\gamma-1}\dot c e^{-pt+gt(1-\gamma)}e^{(1-\gamma)v(n)}+c^{-\gamma}e^{-pt+ gt(1-\gamma)}e^{(1-\gamma)v(n)}[(1-\gamma)(g+v'(n)\dot n)-p]$

Then substitute back in the FOC for consumption:

$\dot\lambda =-\gamma c^{-1}\dot c\lambda +\lambda[(1-\gamma)(g+v'(n)\dot n)-p]$

Then substitute this guy into the Foc for capital:

$g-(1-\tau )r=-\dot c/c\gamma+(1-\gamma)[g+v'(n)\dot n]-p$

leading to:

$r(1-\tau)=p+(\dot c/c +g)\gamma+(1-\gamma)[-v'(n)\dot n] $

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  • $\begingroup$ Thank you for your answer, but unfortunately I still don't get it completely. How do I get the $\rho$ from equation $(10)$? And when you differentiate for $n$, it is $c^{-\gamma}e^{gt(1-\gamma)}e^{(1-\gamma)v(n)} (1-\gamma)v'(n)\dot{n}$, isn't it. Could you please show me how to substitute it correctly, if I am wrong. $\endgroup$
    – Frodo361
    Commented Aug 22, 2016 at 18:35
  • $\begingroup$ Well I assumed your FOC for consumption was correct. Are you using current value or present value Hamiltonian? And I don't differentiate n, I differentiate wrt t . I did forget the g so I put that back in. $\endgroup$
    – VCG
    Commented Aug 22, 2016 at 18:40
  • $\begingroup$ It looks like you did PV Hamiltonian, and that means you're missing the $e^{pt}$ term from the foc. $\endgroup$
    – VCG
    Commented Aug 22, 2016 at 18:53
  • $\begingroup$ Thank you so much. You really helped me. The equation after your line with "Then substitute back in the FOC for consumption" is in fact $-\dot{\lambda}$ so you have to switch the signs, but after that everything is correct in my opinion. It's seems like Mankiw and Weinzierl made a mistake there, because they have in their equation $(10)$ $[...+(1-\gamma)v'(n) \dot{n}]$. You have observed it with a $-$, like Ferede also did.... $\endgroup$
    – Frodo361
    Commented Aug 22, 2016 at 19:55
  • $\begingroup$ I have posted something new in my question on the top. I think the last two equations from your answer have wrong signs... $\endgroup$
    – Frodo361
    Commented Aug 22, 2016 at 21:01

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