Infinit and Finite Sum Equations

General DSGE Modeling Replication Attempts
1

Dear Prof @jpfeifer

Apologies in advance for deriving the equation here, but the question is a bit involved.

I am trying again to replicate Mendoza (2002) Sudden Stop model using his original paper, where one of the FOC equations read the following:

U_{C_{t}}*(1-\lambda/\mu) = \exp(-v(t))*\mathbb{E}_{t}[(R*P_{t}^{c}/P_{t+1})*U_{C_{t+1}} (1)
where

  • Discount factor \tilde{\beta}= \exp(- \sum_{i=0}^{t-1} v_{t}(C_{i})

  • v_{t}(C_{i})= \beta* \ln(1+ C_{i}). (\tilde{\beta} is a function, \beta is a constant)

  • U_{C_{t}} = \tilde{\beta}_{t}* u(C_{t}) + \sum_{i=0}^{\infty} \mathbb{E}_{t}[(\tilde{\beta}_{t+1+i}* v_{t}^{'}(C_{t})*u(C_{t+1+i})]

In order to write eq.1 recursively and since v_{t}^{'}(C_{t}) does not depend on i then:

U_{C_{t}} = \tilde{\beta}_{t}* u^{'}(C_{t}) + v^{'}(C_{t})* \sum_{i=0}^{\infty} \mathbb{E}_{t}[(\tilde{\beta}_{t+1+i})*u(C_{t+1+i})]

Therefore:

U_{C_{t}} = \tilde{\beta}_{t}* u^{'}(C_{t}) + v_{t}^{'}(C_{t})* S_{t}

S_{t}= \mathbb{E}_{t} [\tilde{\beta}_{t+1}*u^(C_{t+1})] + \sum_{i=0}^{\infty} \mathbb{E}_{t}[(\tilde{\beta}_{t+2+i})*u(C_{t+2+i})]

where

S_{t}= \mathbb{E}_{t} [\tilde{\beta}_{t+1}*u^{'}(C_{t+1})] + S_{t+1}

U_{C_{t}} is derivative of lifetime utility.

u_{C_{t}} is utility function form.

  1. Do you agree that equation (1) can be rewritten as U_{C_{t}} = \tilde{\beta}_{t}* u^{'}(C_{t}) + v_{t}^{'}(C_{t})* S_{t} ? Am I right here ?

  2. I am left with the Discount factor \tilde{\beta}= \exp(- \sum_{i=0}^{t-1} v(C_{i})), I can do the same as above, however, how can I instruct dynare to stop the sum at t-1 where t=0 denotes current date.

\tilde{\beta}= \exp(- \sum_{i=0}^{t-1} v(C_{i}))

H_{t}= \sum_{i=0}^{t-1} v(C_{i})) = v(C_{0}) + v(C_{1}) + v(C_{t-1}).... = v(C_{t}) + H_{t+1}

Assuming I want to write \tilde{\beta}_{t+1} u^{'}(C_{t+1}) :

How to implement this in Dynare ?:
@# define k = 1
@# define k = t-1

model,
@#for k in 1: k
… FOC eq. here …

\tilde{\beta}_{t+1} * u^{'}(C_{t+1}) how to write this line ?

@#endfor

  • please note that t starts at 0, k starts at 1 since \tilde{\beta}_{t+1} is \tilde{\beta}_{1} at t=0.
2

Can you please go over the equations again and fix the bracketing.

3

Dear Prof @jpfeifer,

Herein a second go where I fixed some parenthesis, sorry if I have been sloppy a bit. The main equation is the following FOC and I am referring to as eq (1) throughout the text.

U_{C_{t}}*(1-\lambda/\mu) = \exp(-v(t))*\mathbb{E}_{t}[(R*P_{t}^{c}/P_{t+1})*U_{C_{t+1}}]

where

  • Discount factor \tilde{\beta_{\tau}}= \exp(- \sum_{\tau=0}^{t-1} v(C_\tau))

  • \tilde{\beta_{1}} = exp(-v(t)) = exp(v(C_{0})) ; \tilde{\beta_{0}}=1

  • v(C_{\tau}) = \beta* \ln(1+ C_{\tau}). (\tilde{\beta} is a function, \beta is a constant)

  • U_{C_{t}} = \tilde{\beta}_{t}* u^{'}(C_{t}) + \sum_{i=0}^{\infty} \mathbb{E}_{t}[(\tilde{\beta}_{t+1+i}* v_{t}^{'}(C_{t})*u(C_{t+1+i})]

U_{C_{t}} is derivative of lifetime utility.

u_{C_{t}} is utility function form.

In order to write U_{C_{t}} recursively and since v_{t}^{'}(C_{t}) does not depend on i then:

U_{C_{t}} = \tilde{\beta}_{t}* u^{'}(C_{t}) + v^{'}(C_{t})* \sum_{i=0}^{\infty} \mathbb{E}_{t}[(\tilde{\beta}_{t+1+i})*u(C_{t+1+i})]

Therefore:

U_{C_{t}} = \tilde{\beta}_{t}* u^{'}(C_{t}) + v_{t}^{'}(C_{t})* S_{t}

S_{t}= \mathbb{E}_{t} [\tilde{\beta}_{t+1}*u^(C_{t+1})] + \sum_{i=0}^{\infty} \mathbb{E}_{t}[(\tilde{\beta}_{t+2+i})*u(C_{t+2+i})]

Therefore:

S_{t}= \mathbb{E}_{t} [\tilde{\beta}_{t+1}*u(C_{t+1})] + S_{t+1}

  1. **Do you agree that it is sufficient to define ** U_{C_{t}} = \tilde{\beta}_{t}* u^{'}(C_{t}) + v_{t}^{'}(C_{t})* S_{t} then FOC equation (1) is well defined ? Am I right here ?

  2. In order to get the values of \tilde{\beta}_{t+2+i} as it is updated to \tilde{\beta}_{t+3+i} , \tilde{\beta}_{t+4+i} …in the S_{t} equation , then If I dropped i and defined:

\tilde{\beta}_{t+1}= exp^{log((1+C-N^\delta/\delta)} Which is \tilde{\beta}_{1} then S will understand to update \tilde{\beta}_{t+2} as exp^{log((1+C_{1}-N_{1}^\delta/\delta)} and so on so far ?

4

@jpfeifer Any advise if you please

5

Dear Prof,

In terms of Tilda definition, Tilda is in recursive form in order to account for an endogenous discount factor as follows:

\tilde{\beta_{\tau}}= \exp(- \sum_{\tau=0}^{t-1} v(C_\tau))

Therefore, \tilde{\beta_{t}}=\tilde{\beta_{0}}=1

\tilde{\beta}_{t+1}=\tilde{\beta_{1}}= exp^{-B*log(1+C_{0}+N^\delta_{0}/\delta)}

I defined the variable tilda which is effectively \tilde{\beta}_{t+1} since discounting factor kicks in at t+1, namely in the following equation:

S_{t}= \mathbb{E}_{t} [\tilde{\beta}_{t+1}*u(C_{t+1})] + \sum_{i=0}^{\infty} \mathbb{E}_{t}[(\tilde{\beta}_{t+2+i})*u(C_{t+2+i})]. Equation (1)

Where S comes from the following FOC form:

U_{C_{t}} = \tilde{\beta}_{t}* u^{'}(C_{t}) + v_{t}^{'}(C_{t})* S_{t}

To update tilda calculations in Equation (1):

tilda=exp^{-B*log(1+C_{0}+N^\delta_{0}/\delta)}*tilda(-1)

where:

tilda(t) represents \tilde{\beta}_{t+1}=\tilde{\beta_{1}}=exp^{-B*log(1+C_{0}-N^\delta_{0}/\delta)}

Does the Latter form of Tilda works for updating Tilda as the discount factor defined above ?

Thank you very much professor.
Best,
Mendo.mod (7.7 KB)

6

In case you get time for the above Prof. @jpfeifer , with many thanks.

8

I am working on a paper where government adjust deficit given debt sustainability where the forward variables represents infinite summation also there are examples in the manual that might help.

9

Thank you Alia. It’s about endogenous discount factor that ends at t-1. @jpfeifer Any advise if you may

10

Mendo2.mod (8.3 KB)

Hi again Prof.,

Hope all is well,

I redid the derivation of the FOC and came across two conflicting equations:

The first order condition and the same equation in Mendoza paper (2002).
Uc = exp(-B*log(1+C-(N^deltaa)/deltaa))*(P*R/P(+1))*Uc(+1)

The second is an endogenous discount factor tilda which enters another equation for computing an infinite sum (attached .mod file)
tilda = exp(-B*log(1+C(-1)-((N(-1)^deltaa)/deltaa)))*tilda(-2)

In the Steady state, the two equations leads to different values of
exp(-B*log(1+C-(N^deltaa)/deltaa)) of 1/R and 1 respectively. Did I defined tilda equation wrongly ?