<p>That is not correct  conclusion and there are several problems with the formulation you use.</p>
<ol>
<li><span class="math-container">$A$</span> is not a shock - it is a technology parameter of production function (see Woodford Interest and Prices pp 148). Heck, <span class="math-container">$A$</span> cannot even be zero. Following Woodford:</li>
</ol>
<blockquote>
<p><span class="math-container">$A_t&gt; 0$</span> is a time-varying exogenous technology factor</p>
</blockquote>
<p>there can be shocks to <span class="math-container">$A_t$</span> that increase it or decrease it but <span class="math-container">$A_t$</span> itself is strictly positive at all times.</p>
<ol start="2">
<li><p>It is actually not <span class="math-container">$s(Y^n,Y^n,ξ)=μ^{−1}$</span> but <span class="math-container">$s(Y^n_t,Y^n_t;ξ)$</span>. This might seem trivial but note <span class="math-container">$s(.)$</span> has only 2 arguments (<span class="math-container">$Y^n_t$</span> and <span class="math-container">$Y^n_t;\xi$</span>  not 3 arguments <span class="math-container">$Y^n_t$</span>, <span class="math-container">$Y^n_t$</span> and <span class="math-container">$\xi$</span>). In addition, the solution to <span class="math-container">$s(Y^n,Y^n,ξ)=μ^{−1}$</span> 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 <em>conditional</em> on vector of shocks.</p>
</li>
<li><p>Your definition of utility function is extremely non-standard, you define it as:</p>
</li>
</ol>
<p><span class="math-container">$$ u= \frac{\xi_t C_t^{1-\sigma}}{1-\sigma} $$</span></p>
<p>but why are the shocks <span class="math-container">$\xi_t$</span> even multiplied by consumption in the first place? In Woodford the equation (1.1) is given as:</p>
<p><span class="math-container">$$E_0 \left[ \sum^{\infty}_{t=0} \beta^t \left( u_t(C_t, M_t/P_t; \xi_t) - \int^{1}_{0} v(h_t(i);\xi_t)di \right) \right]$$</span></p>
<ul>
<li>first as you can see <span class="math-container">$C_t$</span> is not even conditional on shocks <span class="math-container">$\xi_t$</span>, the real money holding is <span class="math-container">$M_t/P_t; \xi_t$</span> is conditional on vector of shocks.</li>
<li>you can't just assume <span class="math-container">$u_t(C_t, M_t/P_t; \xi_t) = \xi_t C_t^{1-\sigma}$</span>, I mean that makes no economic sense whatsoever, why should utility be product of vector of shocks and consumption?</li>
</ul>
<p>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. <span class="math-container">$ M_t/P_t; \xi_t$</span> does not mean  <span class="math-container">$M_t/P_t$</span> has mean zero even if <span class="math-container">$\xi_t$</span> is given by Gaussian distribution with mean zero).</p>

CC BY-SA 4.0
2021-02-18T09:32:31.467Z
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?
qwerty-qwertz
CC BY-SA 4.0
2021-02-18T10:51:34.400Z
@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
CC BY-SA 4.0
2021-02-18T10:53:33.373Z
@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
CC BY-SA 4.0
2021-02-17T12:37:18.857Z
2021-02-17T12:37:18.857Z
1muflon1
<p>I'm following Woodford's Interest &amp; Prices to derive the microfundations for a New Keynesian model with staggereed prices.</p>
<p>I defined the utility function and disutility function (1.1 at page 144) as</p>
<p><span class="math-container">$$ u= \frac{\xi_t C_t^{1-\sigma}}{1-\sigma}, v= \frac{h_t(j)^{1+\eta}}{1+\eta}, $$</span></p>
<p>where <span class="math-container">$\xi$</span> is a consumption disturbance.</p>
<p>Then I have defined the production function ( 1.7 at page 148) as</p>
<p><span class="math-container">$$ y_t(j)= A_t * h_t(j). $$</span></p>
<p>With this functional forms we can write the real marginal cost (1.10 at page 149) as</p>
<p><span class="math-container">$$ s_t(y,Y,\xi) = \frac{v_h(f^{-1}(y_t(j) /A); \xi)}{u_c(Y_t; \xi)A} \frac{1}{f'(f^{-1}(y_t(j)/A))} = \frac{y_t(j)^\eta \cdot C_t^{1/\sigma}}{A_t  \cdot\xi_t } .$$</span></p>
<p>Woodford says that with flexible prices we can write the natural rate of output as (1.14 at page 151)</p>
<p><span class="math-container">$$s(Y^n, Y^n, \xi)= \mu^{-1}.$$</span></p>
<p>Therefore I get that the natural rate of output is equal to
<span class="math-container">$$ Y_t^n =( \mu^{-1} * A_t * \xi_t )^{\eta/ \sigma}. $$</span></p>
<p>However the problem with this definition is that in the steady state where the shocks(<span class="math-container">$\xi$</span> and <span class="math-container">$A$</span>) are zero also the natural rate of output will be zero.</p>
<p>Is this formulation correct? Is there another way to express the natural rate of output?</p>

<macroeconomics><microeconomics><dsge>
How to define the natural rate output
qwerty-qwertz
# How to define the natural rate output

(Asked by: username_0 on 2021-02-16)

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?

**Tags:** macroeconomics, microeconomics, dsge

---

## 1 answers

### Answer by username_1 on 2021-02-17. Score: 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.

2. 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.
3. 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).

> **Comments:**
> **username_0**: 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?  
> **username_1**: @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.  
> **username_1**: @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)  

---

