I'm following Woodford's Interest & Prices to derive the microfundations for a New Keynesian model with staggereed prices.

I defined the utility function and disutility function (1.1 at page 144) as

$$ u= \frac{\xi_t C_t^{1-\sigma}}{1-\sigma}, v= \frac{h_t(j)^{1+\eta}}{1+\eta}, $$

where $\xi$ is a consumption disturbance.

Then I have defined the production function ( 1.7 at page 148) as

$$ y_t(j)= A_t * h_t(j). $$

With this functional forms we can write the real marginal cost (1.10 at page 149) as

$$ s_t(y,Y,\xi) = \frac{v_h(f^{-1}(y_t(j) /A); \xi)}{u_c(Y_t; \xi)A} \frac{1}{f'(f^{-1}(y_t(j)/A))} = \frac{y_t(j)^\eta \cdot C_t^{1/\sigma}}{A_t \cdot\xi_t } .$$

Woodford says that with flexible prices we can write the natural rate of output as (1.14 at page 151)

$$s(Y^n, Y^n, \xi)= \mu^{-1}.$$

Therefore I get that the natural rate of output is equal to $$ Y_t^n =( \mu^{-1} * A_t * \xi_t )^{\eta/ \sigma}. $$

However the problem with this definition is that in the steady state where the shocks($\xi$ and $A$) are zero also the natural rate of output will be zero.

Is this formulation correct? Is there another way to express the natural rate of output?


1 Answer 1


That is not correct conclusion and there are several problems with the formulation you use.

  1. $A$ is not a shock - it is a technology parameter of production function (see Woodford Interest and Prices pp 148). Heck, $A$ cannot even be zero. Following Woodford:

$A_t> 0$ is a time-varying exogenous technology factor

there can be shocks to $A_t$ that increase it or decrease it but $A_t$ itself is strictly positive at all times.

  1. It is actually not $s(Y^n,Y^n,ξ)=μ^{−1}$ but $s(Y^n_t,Y^n_t;ξ)$. This might seem trivial but note $s(.)$ has only 2 arguments ($Y^n_t$ and $Y^n_t;\xi$ not 3 arguments $Y^n_t$, $Y^n_t$ and $\xi$). In addition, the solution to $s(Y^n,Y^n,ξ)=μ^{−1}$ gives you the natural rate of output not the equation itself. In this context how I read the Woodford semicolon (;) is used to separate variables and shock parameters. Hence, it is just to indicate that the ξ is not a variable but given by the shock parameters of the economy (i.e. indicate variable conditional on vector of shocks.

  2. Your definition of utility function is extremely non-standard, you define it as:

$$ u= \frac{\xi_t C_t^{1-\sigma}}{1-\sigma} $$

but why are the shocks $\xi_t$ even multiplied by consumption in the first place? In Woodford the equation (1.1) is given as:

$$E_0 \left[ \sum^{\infty}_{t=0} \beta^t \left( u_t(C_t, M_t/P_t; \xi_t) - \int^{1}_{0} v(h_t(i);\xi_t)di \right) \right]$$

  • first as you can see $C_t$ is not even conditional on shocks $\xi_t$, the real money holding is $M_t/P_t; \xi_t$ is conditional on vector of shocks.
  • you can't just assume $u_t(C_t, M_t/P_t; \xi_t) = \xi_t C_t^{1-\sigma}$, I mean that makes no economic sense whatsoever, why should utility be product of vector of shocks and consumption?

I don't think you will get the same result with some more reasonable utility function and take into account that the variables are conditional on the vector of shocks (e.g. $ M_t/P_t; \xi_t$ does not mean $M_t/P_t$ has mean zero even if $\xi_t$ is given by Gaussian distribution with mean zero).

  • $\begingroup$ thanks for your answer. 1)What would a more reasonable utility function be? would \xi_t + C_t^{1-\sigma} be ok? 2) can't A_t be interpreted as an AR1 process? $\endgroup$ Feb 18, 2021 at 9:32
  • $\begingroup$ @qwerty-qwertz 1. No, first you have to integrate here over all individuals so for consumption part you would want something like $C_t = [\int_{i=0}^1 C_{ti}^{(\sigma-1)/\sigma)}]^{\sigma/(\sigma-1)}$. The shocks have to be introduced through real money holdings. I guess easiest, but not necessary the most rigorous way is to just simply assume that $Y=M/P$ and without gov and investment $Y\equiv C$. Then you can assume that $P= P+\xi$. Of course, there are multiple other ways how you could do it so take that as a pointer rather than saying it has to be done like that. $\endgroup$
    – 1muflon1
    Feb 18, 2021 at 10:51
  • $\begingroup$ @qwerty-qwertz 2. No $A_t$ should not be AR process because it is a stock variable... it will likely have unit root along trend instead of reverting to some mean. $\Delta A_t$ could be modeled as AR process I suppose, but it depends on whether you want to assume growth is exogenous or endogenously given (in the latter case it would be more complex) $\endgroup$
    – 1muflon1
    Feb 18, 2021 at 10:53

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