# Question on Taylor Rule in the model with trend

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.

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.

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 ?

Yes, that is correct.