Log linearization of Tobin Q equation (SW, 2003)

General DSGE Modeling
1

Hi everybody,
I am struggling with the following log linearization (it is the FOC for K(t) in SW, 2003); in particular, given that functional form for the capital utilization function, I was wondering whether I have to take first order taylor expansion of the entire function when log linearizing(in that case, I use the entire function evaluated at utilization rate=1 which is the stady steady state value of capital utilization as a pivotal point) or just of the exponential term (in that case, I just use utilization rate = 1 as a pivotal point). I’ve tried both ways, but I am not able to obtain equation 30. Would anyone be so helpful to give me some insights? Thank everybody for the attention
tobin q log linearization.pdf (443.3 KB)

2

You need to do a full Taylor of the whole equation.

3

What do you mean exactly by doing a full Taylor expansion of the whole equation? I have always done log linearization on the single terms (ie. x’=log(x(t)) - log(x) ==>x(t)=xexp(x’)==>x(t)=x(1+x’)) and I have obtained the rest of the equations in the SW paper this way. In this particular case, I do not kwon how to treat the exponential term in the variable capital utilization function, as if I applied the aforementioned method, I wouldn’t get rid of the exponential. Thanks for being helpful.

4

I case like this it’s easier to do a Taylor approximation than the Uhlig trick.
{e^{\psi \left( {{u_{t + 1}} - 1} \right)}} \approx \psi {e^{\psi \left( {u - 1} \right)}}\left( {{u_{t + 1}} - u} \right) = \psi {e^{\psi \left( {u - 1} \right)}}u\frac{{\left( {{u_{t + 1}} - u} \right)}}{u} = u\psi {e^{\psi \left( {u - 1} \right)}}{\hat u_{t + 1}}

5

Since the steady state value of capital utilization rate is 1 by assumption, is it correct to claim that the resulting value for that expression is u(hat)(t+1)? If so, can I procceed with the Uhlig trick in the rest of equation? I have tried so but I get something very different from the SW paper.
Thank you in advance

6

Given that the steady state is 1, percentage deviations and absolute deviations are the same. Thus, the answer should be yes.