<p>Any time series can be decomposed in four terms:</p>

<ul>
<li>trend: the underlying change in the variable if no other factor is present. In the simplest case, think of an upward slopping line. The growth rate of this trend is it's slope. </li>
<li>cycle: the cyclical component of the series, defined by ups and downs like a wave, with a certain "frequency" and "amplitude". Think of this like a wave.</li>
<li>seasonal component: usually a within year cycle. The standard textbook example is that of ice cream sales, which pick up in summer. This is different from the longer cycle described above, and perhaps less smooth (e.g. as ice cream sales are low 3/4 of the year)</li>
<li>noise: natural randomness due to measurement errors or other unobserved yet minor factors, including shocks (e.g. a very hot weather in a week, increasing ice cream sales)</li>
</ul>

<p>For a series $y_t$, this is something like</p>

<p>$$ y_t = d_t + c_t + s_t + e_t $$</p>

<p>Importantly, all the variables are defined in the same units <strong>by construction</strong>. E.g. if this is GDP, they are all in value added units (real, nominal, whatever). </p>

<p>A general case of a decomposition is (assume just trend and cycle):</p>

<p>$$ f(y_t) = d^{'}_{t} + c^{'}_{t} $$</p>

<p>In this case, the trend growth of <strong>$f(y_t)$</strong> is the growth rate of $d^{'}_t$. But this is <strong>not</strong> the same as the trend growth rate of $y_t$. </p>

<p>Consider an example. Let us <strong>construct</strong> a series $y_t$, where $d_t=d_0(1+g)^t$ and $c_t=sin(t)$, and $y_t=d_t+c_t$. This is a trend-cycle decomposition where the growth rate of the trend is $g$:</p>

<p>$$ \frac{d_{t+1}-d_t}{d_t} = \frac{d_0(1+g)^{t+1}-d_0(1+g)^t}{d_0(1+g)^t} = \cdots = g $$</p>

<p>Now, transform the process into log, and let us postulate a new decomposition:</p>

<p>$$ \log y_t = D_t + C_t $$</p>

<p>where $D_t = D_0(1+G)^t$, and $C_t$ is an unknown function. This is by definition correct, but $C_t$ might be an irregular, perhaps not algebraically defined function (like $sin(t)$). In this new case, it is correct, by definition, that the trend growth rate <strong>of the log of $y_t$</strong> is $G$, computed with the standard formula. </p>

<p>The central question is: can we find a general method to calculate the trend growth rate <strong>of $y_t$</strong> ($g$) using <strong>only</strong> some transformation of $D_t$? (as in your attempt using the exponential). In other words, can we use <strong>only</strong> $D_0(1+G)^t$ to obtain $g$?</p>

<p>(Here I highlight the point of why would you ever go down this route, if you can decompose $y_t$ in levels and solve your problem immediately; but anyway, let us continue).</p>

<p>The answer seems to be negative. It is trivial to see that</p>

<p>$$ e^{D_0(1+G)t} e^{C_t} = d_0(1+g)^t + sin(t) $$</p>

<p>where you cannot find a direct function between $G$ and $g$ which does not involve the use of $C_t$ and $sin(t)$. In other words, you need to use the information of <strong>both</strong> decompositions to solve the problem. But this is circular. To make the link between logs and levels, you need to make the decomposition in levels. But if you do the latter, the answer is there already. No need to go further.  </p>

<p>Finally, consider the very special case where the series $y_t$ only has a constant growth rate (trend):</p>

<p>$$ y_t = d_t = d_0(1+g)^t $$</p>

<p>Then,</p>

<p>$$ \log y_t = D_t = D_0(1+G)^t = \log \left(d_0(1+g)^t\right) $$</p>

<p>The following formula recovers the original growth rate of $y_t$ in terms of the trend from the series in log:</p>

<p>$$ \dfrac{e^{D_{t+1}} - e^{D_t}}{e^{D_t}} = \frac{d_{t+1} - d_t}{d_t} = \cdots = g $$</p>

<p>The result can be generalised to any $f(y_t)$, where the transformation of the growth rate calculation has to be $f^{-1}(\cdot)$.</p>

CC BY-SA 3.0
2017-05-24T19:49:56.457Z
Thanks for the answer! But why can't we say  (why doesn't it make any sense) to say that there exist a trend $d_t$ 
and cyclical component $c_t$ such that $\log(y_t)=d_t+c_t$ (assuming that $s_t=e_t=0$ for the moment)? And is your answer to my two questions "yes"? Note: Changed the wording in my original post.
Elias
CC BY-SA 3.0
2017-05-24T20:06:58.377Z
@Anna It "makes sense", but since this is a mere additive decomposition, all elements have the same units. Like when decomposing GDP as C + I + G + NX. Think of decomposing a distance (km) into several sub-sections. They are also in km. If you want one of the elements to be in miles, you need to add a parameters that makes the conversion. In this case the formula for the trend does not change, as the parameter cancels out (trend in miles and km is the same), but a non-linear transformation (log) is different. Perhaps my point is, why would you ever use log and level when doing a decomposition?
luchonacho
CC BY-SA 3.0
2017-05-24T20:14:20.980Z
@luchanacho I don't get your point! Suppose we know for a fact that $\log(y_t)=d_t+c_t$. Then is it correct to say that $[\exp(d_{t+1})-\exp(d_t)]\Big/\exp(d_t)$ is the trend growth rate of $y_t$? That is my question. Does the term "trend growth rate" refer to the growth rate in the trend term $d_t$, that is $(d_{t+1}-d_t)/d_t$? Or does it refer to the growth rate in $y_t$ given that $c_t=0$? Why I would use a log and level when doing a decomposition is another question as I see it.
Elias
CC BY-SA 3.0
2017-05-24T21:16:23.347Z
@Anna What do you mean by "we know"? This is a choice you make. You have a series, and want to decompose it into a trend and a cycle. You do not "know" anything before-hand. This is about separating a series into two factors. You are free to decide.
luchonacho
CC BY-SA 3.0
2017-05-25T13:19:54.623Z
This is a formal, mathematical question about a definition of a term and I wrote "suppose we know". What is the meaning of the concept "trend growth rate" in mathematical terms? That is my question. Example: Suppose we know that $\log(y_t)=d_t+c_t$. Then what is the definition of "the trend growth rate"? You can also suppose that we have no data, so that this is a pure technical question.
Elias
CC BY-SA 3.0
2017-05-24T19:42:56.633Z
2017-05-25T19:33:48.650Z
2017-05-25T19:33:48.650Z
luchonacho
<p>I am a bit unsure about how the "trend growth rate" is <strong><em>defined</em></strong> in macroeconomics. Is it to correct to say that if a time series $\{y_t\}$ may be decomposed as $$y_t=d_t+c_t,$$ where $d_t$ is a deterministic trend and $c_t$ a cyclical component, then $$(d_{t+1}-d_t)/d_t$$ is the "trend growth rate", i.e. the growth rate in $y_t$ assuming $c_t=0$? And if it can be decomposed as $$\log(y_t)=d'_t+c'_t,$$ with another trend and cyclical component, then $$[\exp(d'_{t+1})-\exp(d'_t)]/\exp(d'_t)$$ is the "trend growth rate", i.e. the growth rate in $y_t$ assuming $c'_t=0$?</p>

<p>Sometimes I feel that the definition is $(d_{t+1}-d_t)/d_t$ and $(d'_{t+1}-d'_t)/d'_t$ in the two cases, referring to the growth rate in the trend component and not to the growth rate in the dependent variable $y_t$ ignoring the cyclical component (i.e., setting $c_t=0$ for all $t$). </p>

<p>That is, I interpret "trend growth rate" as the growth rate in $y_t$ ignoring the cyclical component. Is that a correct interpretation?</p>

<macroeconomics><definition>
Definition: What Is The "Trend Growth Rate"?
Elias
# Definition: What Is The "Trend Growth Rate"?

(Asked by: username_0 on 2017-05-24)

I am a bit unsure about how the "trend growth rate" is ***defined*** in macroeconomics. Is it to correct to say that if a time series $\{y\_t\}$ may be decomposed as $$y\_t=d\_t+c\_t,$$ where $d\_t$ is a deterministic trend and $c\_t$ a cyclical component, then $$(d\_{t+1}-d\_t)/d\_t$$ is the "trend growth rate", i.e. the growth rate in $y\_t$ assuming $c\_t=0$? And if it can be decomposed as $$\log(y\_t)=d'\_t+c'\_t,$$ with another trend and cyclical component, then $$[\exp(d'\_{t+1})-\exp(d'\_t)]/\exp(d'\_t)$$ is the "trend growth rate", i.e. the growth rate in $y\_t$ assuming $c'\_t=0$?

Sometimes I feel that the definition is $(d\_{t+1}-d\_t)/d\_t$ and $(d'\_{t+1}-d'\_t)/d'\_t$ in the two cases, referring to the growth rate in the trend component and not to the growth rate in the dependent variable $y\_t$ ignoring the cyclical component (i.e., setting $c\_t=0$ for all $t$).

That is, I interpret "trend growth rate" as the growth rate in $y\_t$ ignoring the cyclical component. Is that a correct interpretation?

**Tags:** macroeconomics, definition

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## 1 answers

### Answer by username_1 on 2017-05-24. Score: 2 | Accepted answer

Any time series can be decomposed in four terms:

* trend: the underlying change in the variable if no other factor is present. In the simplest case, think of an upward slopping line. The growth rate of this trend is it's slope.
* cycle: the cyclical component of the series, defined by ups and downs like a wave, with a certain "frequency" and "amplitude". Think of this like a wave.
* seasonal component: usually a within year cycle. The standard textbook example is that of ice cream sales, which pick up in summer. This is different from the longer cycle described above, and perhaps less smooth (e.g. as ice cream sales are low 3/4 of the year)
* noise: natural randomness due to measurement errors or other unobserved yet minor factors, including shocks (e.g. a very hot weather in a week, increasing ice cream sales)

For a series $y\_t$, this is something like

$$ y\_t = d\_t + c\_t + s\_t + e\_t $$

Importantly, all the variables are defined in the same units **by construction**. E.g. if this is GDP, they are all in value added units (real, nominal, whatever).

A general case of a decomposition is (assume just trend and cycle):

$$ f(y\_t) = d^{'}\_{t} + c^{'}\_{t} $$

In this case, the trend growth of **$f(y\_t)$** is the growth rate of $d^{'}\_t$. But this is **not** the same as the trend growth rate of $y\_t$.

Consider an example. Let us **construct** a series $y\_t$, where $d\_t=d\_0(1+g)^t$ and $c\_t=sin(t)$, and $y\_t=d\_t+c\_t$. This is a trend-cycle decomposition where the growth rate of the trend is $g$:

$$ \frac{d\_{t+1}-d\_t}{d\_t} = \frac{d\_0(1+g)^{t+1}-d\_0(1+g)^t}{d\_0(1+g)^t} = \cdots = g $$

Now, transform the process into log, and let us postulate a new decomposition:

$$ \log y\_t = D\_t + C\_t $$

where $D\_t = D\_0(1+G)^t$, and $C\_t$ is an unknown function. This is by definition correct, but $C\_t$ might be an irregular, perhaps not algebraically defined function (like $sin(t)$). In this new case, it is correct, by definition, that the trend growth rate **of the log of $y\_t$** is $G$, computed with the standard formula.

The central question is: can we find a general method to calculate the trend growth rate **of $y\_t$** ($g$) using **only** some transformation of $D\_t$? (as in your attempt using the exponential). In other words, can we use **only** $D\_0(1+G)^t$ to obtain $g$?

(Here I highlight the point of why would you ever go down this route, if you can decompose $y\_t$ in levels and solve your problem immediately; but anyway, let us continue).

The answer seems to be negative. It is trivial to see that

$$ e^{D\_0(1+G)t} e^{C\_t} = d\_0(1+g)^t + sin(t) $$

where you cannot find a direct function between $G$ and $g$ which does not involve the use of $C\_t$ and $sin(t)$. In other words, you need to use the information of **both** decompositions to solve the problem. But this is circular. To make the link between logs and levels, you need to make the decomposition in levels. But if you do the latter, the answer is there already. No need to go further.

Finally, consider the very special case where the series $y\_t$ only has a constant growth rate (trend):

$$ y\_t = d\_t = d\_0(1+g)^t $$

Then,

$$ \log y\_t = D\_t = D\_0(1+G)^t = \log \left(d\_0(1+g)^t\right) $$

The following formula recovers the original growth rate of $y\_t$ in terms of the trend from the series in log:

$$ \dfrac{e^{D\_{t+1}} - e^{D\_t}}{e^{D\_t}} = \frac{d\_{t+1} - d\_t}{d\_t} = \cdots = g $$

The result can be generalised to any $f(y\_t)$, where the transformation of the growth rate calculation has to be $f^{-1}(\cdot)$.

> **Comments:**
> **username_0**: Thanks for the answer! But why can't we say  (why doesn't it make any sense) to say that there exist a trend $d_t$  and cyclical component $c_t$ such that $\log(y_t)=d_t+c_t$ (assuming that $s_t=e_t=0$ for the moment)? And is your answer to my two questions "yes"? Note: Changed the wording in my original post.  
> **username_1**: @Anna It "makes sense", but since this is a mere additive decomposition, all elements have the same units. Like when decomposing GDP as C + I + G + NX. Think of decomposing a distance (km) into several sub-sections. They are also in km. If you want one of the elements to be in miles, you need to add a parameters that makes the conversion. In this case the formula for the trend does not change, as the parameter cancels out (trend in miles and km is the same), but a non-linear transformation (log) is different. Perhaps my point is, why would you ever use log and level when doing a decomposition?  
> **username_0**: @luchanacho I don't get your point! Suppose we know for a fact that $\log(y_t)=d_t+c_t$. Then is it correct to say that $[\exp(d_{t+1})-\exp(d_t)]\Big/\exp(d_t)$ is the trend growth rate of $y_t$? That is my question. Does the term "trend growth rate" refer to the growth rate in the trend term $d_t$, that is $(d_{t+1}-d_t)/d_t$? Or does it refer to the growth rate in $y_t$ given that $c_t=0$? Why I would use a log and level when doing a decomposition is another question as I see it.  
> **username_1**: @Anna What do you mean by "we know"? This is a choice you make. You have a series, and want to decompose it into a trend and a cycle. You do not "know" anything before-hand. This is about separating a series into two factors. You are free to decide.  
> **username_0**: This is a formal, mathematical question about a definition of a term and I wrote "suppose we know". What is the meaning of the concept "trend growth rate" in mathematical terms? That is my question. Example: Suppose we know that $\log(y_t)=d_t+c_t$. Then what is the definition of "the trend growth rate"? You can also suppose that we have no data, so that this is a pure technical question.  

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