# How to define the natural rate output

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?

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).

• 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? Feb 18 at 9:32
• @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.
– 1muflon1
Feb 18 at 10:51
• @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)
– 1muflon1
Feb 18 at 10:53