Question on Taylor Rule in the model with trend

General DSGE Modeling
1

Dear Professor Pfeifer,

Thank you for reading this post, and I’ve got two problems.

  1. I read that in your paper Fiscal News and Macroeconomic Volatility.pdf (527.0 KB) the Taylor Rule equation is written as
    \frac{{{R}_{t}}}{R}={{\left( \frac{{{R}_{t-1}}}{R} \right)}^{{{\rho }_{R}}}}{{\left( {{\left( \frac{{{\Pi }_{t}}}{{\bar{\Pi }}} \right)}^{{{\phi }_{{{R}_{\Pi }}}}}}{{\left( \frac{{{Y}_{t}}}{{{Y}_{t-1}}}\frac{1}{{{\mu }^{y}}} \right)}^{{{\phi }_{{{R}_{Y}}}}}} \right)}^{1-{{\rho }_{R}}}}\exp \left( \xi _{t}^{R} \right) \quad \quad(23).
    I’m confused why it’s \mu^{y} instead of \mu^{y}_{t} in the equation and why \mu^{y} is the denominator.
    As I thought that Taylor Rule equation is written as \frac{{{R}_{t}}}{R}={{\left( \frac{{{R}_{t-1}}}{R} \right)}^{{{\rho }_{R}}}}{{\left( {{\left( \frac{{{\Pi }_{t}}}{{\bar{\Pi }}} \right)}^{{{\phi }_{{{R}_{\Pi }}}}}}{{\left( \frac{{{Y}_{t}}}{{{Y}_{t-1}}} \right)}^{{{\phi }_{{{R}_{Y}}}}}} \right)}^{1-{{\rho }_{R}}}}\exp \left( \xi _{t}^{R} \right) in the model without trend, and then transform to \frac{{{R}_{t}}}{R}={{\left( \frac{{{R}_{t-1}}}{R} \right)}^{{{\rho }_{R}}}}{{\left( {{\left( \frac{{{\Pi }_{t}}}{{\bar{\Pi }}} \right)}^{{{\phi }_{{{R}_{\Pi }}}}}}{{\left( \frac{{{y}_{t}}{{\Gamma }_{t}}}{{{y}_{t-1}}{{\Gamma }_{t-1}}} \right)}^{{{\phi }_{{{R}_{Y}}}}}} \right)}^{1-{{\rho }_{R}}}}\exp \left( \xi _{t}^{R} \right), and finally get \frac{{{R}_{t}}}{R}={{\left( \frac{{{R}_{t-1}}}{R} \right)}^{{{\rho }_{R}}}}{{\left( {{\left( \frac{{{\Pi }_{t}}}{{\bar{\Pi }}} \right)}^{{{\phi }_{{{R}_{\Pi }}}}}}{{\left( \frac{{{y}_{t}}\mu _{t}^{y}}{{{y}_{t-1}}} \right)}^{{{\phi }_{{{R}_{Y}}}}}} \right)}^{1-{{\rho }_{R}}}}\exp \left( \xi _{t}^{R} \right)( \Gamma_{t} is the growth of output) . Did I misunderstand anything?

  2. Another question is about parameter calibration.How should I calibrate \eta in household’s utility function
    \max {{E}_{0}}\sum\limits_{t=0}^{\infty }{\beta _{{}}^{t}\left[ \log {{c}_{t}}-\frac{\eta }{1+\gamma }{{\left( {{l}_{t}} \right)}^{1+\gamma }} \right]}?
    I see in some papers this parameter is just set to 1.

Thanks again for your time, and sincerely appreciate your kindness.

2
  1. The equations presented are not yet detrended. The term \frac{Y_t}{Y_{t-1}}\frac{1}{\mu_y} punishes deviations of output growth from its steady state growth rate \mu_y. In steady state \frac{Y_t}{Y_{t-1}}=\mu_y, so the term \frac{Y_t}{Y_{t-1}}\frac{1}{\mu_y}=1.
  2. Typically, people set \eta to imply a particular amount of hours worked in steady state, for example 1/3.
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Dear Professor Pfeifer,

Thank you very much for your reply.

If I understand it correctly, the detrended equation should be written as
\frac{{{R}_{t}}}{R}={{\left( \frac{{{R}_{t-1}}}{R} \right)}^{{{\rho }_{R}}}}{{\left( {{\left( \frac{{{\Pi }_{t}}}{{\bar{\Pi }}} \right)}^{{{\phi }_{{{R}_{\Pi }}}}}}{{\left( \frac{{{y}_{t}}\mu _{t}^{y}}{{{y}_{t-1}}\mu^{y}} \right)}^{{{\phi }_{{{R}_{Y}}}}}} \right)}^{1-{{\rho }_{R}}}}\exp \left( \xi _{t}^{R} \right). Is that right ?

4

Yes, that is correct.