Question on calibration of discount factor

Dear Professor Pfeifer,

I was trying to calibrate the discount factor \beta in a model with specified trend. The first order condition on household’s deposit is:

{{\tilde{\lambda }}_{t}}=\frac{\beta {{{\tilde{\lambda }}}_{t+1}}{{R}_{t}}}{{{\pi }_{t+1}}\varphi _{t+1}^{y}}

in which \varphi _{t}^{y} is growth rate of output.

Then the steady state equation displays:

R=\frac{\varphi _{{}}^{y}}{\beta }

According to the real data, the steady state R is equal to 1.01, which means 4% yearly interest rate. And the steady state of quarterly growth rate of output \varphi _{}^{y} is equal to 1.0214. However, this imply that the discount factor \beta is larger than 1. Did I misunderstand anything? How should I deal with this situation?

Is the problem lies in the household’s utility function \max {{E}_{0}}\sum\limits_{t=0}^{\infty }{{{\beta }^{t}}\left[ \log {{c}_{t}}-\frac{1}{1+\gamma }{{\left( {{n}_{t}} \right)}^{1+\gamma }} \right]}? Should I change it to \max {{E}_{0}}\sum\limits_{t=0}^{\infty }{{{\left( \beta \varphi _{{t}}^{y} \right)}^{t}}\left[ \log {{c}_{t}}-\frac{1}{1+\gamma }{{\left( {{n}_{t}} \right)}^{1+\gamma }} \right]} or something?

Thank you for your time on this.

There seems to be a difference between the long-run steady state value and the sample average. You are right that theoretically the growth rate must not be bigger than the interest rate. The typical solution would be to pick realistic values for both parameters that are not necessarily equal to the data moments. In the long-run, the growth rate is likely to come down, I guess.

Dear Professor Pfeifer,

Thank you for your reply. I then have three more confusing problems on this. Would you please take a look?

  1. How should I “pick realistic values that are not necessarily equal to the data moments”? i.e. what is basis to calibrate the value of growth rate in the long-run?

  2. How should I deal with the observation equation on output data? The observation equation of output is written as:

{\tilde{y_{t}}}^{obs}={{\tilde{y}}_{t}}-{{\tilde{y}}_{t-1}}+{{\tilde{\varphi }}_{t}^{y}}

The residual of {\tilde{y_{t}}}^{obs} is log(1.0214) according to the data. If the steady state of {\varphi }_{t}^{y} is not set as 1.0214, the residual of the equation will not equal to zero.

  1. I read in a mathematical appendix NERI_APPENDIX_E.pdf (164.2 KB) that the authors add a growth rate parameter into the discount factor(which becomes as {\beta}G_c in household’s utility function), and this will change the first order condition on deposit and make the steady state equation back to

R=\frac{1}{\beta }.

Would this be the solution of my probelm? And why do they add this growth rate parameter into the discount factor?

Thanks again for your time on this. I really appreciated your kindness.

Dear Professor Pfeifer,

Sincerely looking forward to your advice. Thank you!

  1. Usually, the literature will guide you on realistic values. But before doing that, you should think hard about whether a model approximated in the neighborhood of a balanced growth path is the right model for your economy. It seems you are rather far away from the steady state.
  2. Again, you are confusing the data mean and the steady state. Take the Solow or Ramsey model starting with a capital intensity below steady state. In this case, the average growth rate will be higher than the steady state growth rate due to capital accumulation. See also
  1. That growth adjustment is necessary when you not have log utility. See

Dear Professor Pfeifer,

Thank you for your reply!

  1. My model is based on an emerging economy which has a sustained and rapid growth in these years, so you are definitely right that the economy might not be approximated to a balanced growth path. But if I want to consider the balanced growth path in my model, is there any way to set a proper growth rate in the long-run?

  2. In case that I set the long-run growth rate \varphi^{y} as 1.005, which represents 2% yearly growth rate, meanwhile the data I can observe and use to do the estimation displays a 1.0214 growth rate, could I add a constant term into the observation equation to make it even? just like
    {\tilde{y_{t}}}^{obs}={{\tilde{y}}_{t}}-{{\tilde{y}}_{t-1}}+{{\tilde{\varphi }}_{t}^{y}}+\mu?

  3. I’m not sure if I get the meaning of “log utility”. It seems that they have “log utility” in that mathematical appendix but still add growth adjustment for discount factor.

Thanks again for reading all these, and sincerely appreciate your kindness.

  1. You could simply set the growth rate to the steady state in advanced economies, e.g. 2 percent. The worry still is that the transition dynamics will matter.
  2. You could of course simply demean the data to eliminate any difference in average growth rates. But you need to decide whether that makes sense with respect to the question you are trying to investigate.
  3. You can see in Section 3.1 of the document that with log utility the G_C actually drops out in the end. That’s the point of log utility. Otherwise, it would still appear.

Dear Professor Pfeifer,

Thank you for your reply!

For the third question, I’m still a little confused. As you said “that growth adjustment is necessary when you not have log utility”, so I guess the growth adjustment is not necessary when I set log utility just like the form in that appendix document. But why they add growth adjustment G_C in that document?

And also, adding this growth adjustment will remake the steady state R=\frac{\varphi _{{}}^{y}}{\beta } back to R=\frac{1}{\beta }. This will avoid a big value of \beta. So I wonder if it (not adding the growth adjustment) is the reason I got a big value of \beta. Did I misunderstand anything?

As always, thank you for your time on this.

  1. It very much depends on your setup. That’s the reason the Neri et al. paper starts with the proper growth adjustment and then goes on to show that with log utility the steady state relations are not affected.
  2. Yes, with log utility your problem should vanish. See also

Dear Professor Pfeifer,

Thank you for your reply!

In Neri et al. paper, the growth rate G_C is a parameter as they do not contain stochastic shock in it. But the growth rate in my model is a variable. How should I add it as a growth adjustment?
\max {{E}_{0}}\sum\limits_{t=0}^{\infty }{{{\left( \beta \varphi _{{t}}^{y} \right)}^{t}}\left[ \log {{c}_{t}}-\frac{1}{1+\gamma }{{\left( {{n}_{t}} \right)}^{1+\gamma }} \right]}
\max {{E}_{0}}\sum\limits_{t=0}^{\infty }{{{\left( \beta \varphi _{}^{y} \right)}^{t}}\left[ \log {{c}_{t}}-\frac{1}{1+\gamma }{{\left( {{n}_{t}} \right)}^{1+\gamma }} \right]} (\varphi _{}^{y} is the steady state of \varphi _{t}^{y}) ?

Thanks again for your time on this.

You would need to start from scratch with the proper detrending. My hunch is that the result will involve \varphi_t^n.

Dear Professor Pfeifer,

Again, thank you for your reply!

  1. Do you mean that I should use
    \max {{E}_{0}}\sum\limits_{t=0}^{\infty }{{{\left( \beta \varphi _{{t}}^{y} \right)}^{t}}\left[ \log {{c}_{t}}-\frac{1}{1+\gamma }{{\left( {{n}_{t}} \right)}^{1+\gamma }} \right]}?

  2. Another problem comes out is that: Neri et al. use observation equation as

data_CC = log(exp(c) + exp(c1)) - CC_SS + TRENDY ;

in their code Iacoviello_2010.mod (17.8 KB)US_data_65Q106Q4.m (36.3 KB). It seems that they did not choose log-difference form. What is the dealing process of getting “data_CC” from raw data in this case (they did not seem to tell this in the paper)?

  1. And also, would you please recommend any authoritative paper about the DSGE model with specified trend? Because I find that I have a lot to learn on this type of model and I really want to get the right thing.

Thank you so much for your time on this.

Dear Professor Pfeifer,

Sincerely looking forward to your reply. Thank you very much!

  1. You need to do the full detrending. The growth factor would not be raised to the power of t.
  2. I don’t know what is going on in their model. They assumed linear trends in their model.
  3. Have a look at Fernández-Villaverde/Rubio-Ramírez (2006):