<p>The formula of the OP is perfectly fine.</p>
<p>Treating the derivative as a ratio of differential (long live Leibniz), with <span class="math-container">$x$</span> representing &quot;age&quot;,
<span class="math-container">\begin{align}
\frac{d \ln w}{dx} &amp;= b_2 + 2b_3x \implies \frac 1 w \cdot \frac {d w}{dx} = b_2 + 2b_3x\\ 
\\
&amp;\implies \frac x w \cdot \frac {d w}{dx} = \frac{dw/dx}{w/x} = (b_2 + 2b_3x)\cdot x,
\end{align}</span></p>
<p>where now the left-hand-side is the definition of point-elasticity (&quot;marginal over average&quot;).</p>
<p>The confusingly so-called &quot;semi-elasticity&quot; (confusingly because it is not an elasticity) comes from writing:</p>
<p><span class="math-container">$$\frac{d \ln w}{dx} = b_2 + 2b_3x \implies \frac {d w / w}{dx} = b_2 + 2b_3x,$$</span></p>
<p>and considering <em>a one-unit change</em> in <span class="math-container">$x$</span>, which means that we will set <span class="math-container">$dx=1$</span> and obtain</p>
<p><span class="math-container">$$\frac {d w }{w} \approx \frac {\Delta w}{w} = \% \Delta(w) \approx b_2 + 2b_3x.$$</span></p>
<p>So <span class="math-container">$(b_2 + 2b_3x)\cdot x$</span> is the proper elasticity, while the &quot;semi-elasticity&quot; <span class="math-container">$b_2 + 2b_3x$</span> in essence is <strong>the percentage change in the wage if age increases by one year</strong> (and since we have included a squared term, this percentage change depends also on the current age).</p>

CC BY-SA 4.0
2024-01-26T00:48:42.010Z
2024-01-26T00:48:42.010Z
Alecos Papadopoulos
<p>No that would not be correct definition of elasticity. First, mathematically in multivariate function elasticity is defined as follows:</p>

<p><span class="math-container">$$ EL_x =\frac{ f_x '(x,y)}{f(x,y)}x$$</span> or in your case it would be: <span class="math-container">$$ \frac{ \partial \ln [w(age,Y,T,Mar)]}{\partial age} \frac{age}{\\ln [w(age,Y,T,Mar)]}$$</span></p>

<p>However, even if you would plug in the expressions in this formula you would get an elasticity of a <strong>log of wage</strong> with respect to age - I am not sure if thats what you actually want. You could exponentiate the original function but that would be messy as hell. </p>

<p>One way how to estimate the elasticity directly from the regression would be using log-log form for variables of interest. In your case (assuming there are no people with 0 age in your sample): </p>

<p><span class="math-container">$$ln(w_i)=\beta_1+\beta_2 \ln (age) +\beta_4Y_i+\beta_5T_i+\beta_6Mar_i+\epsilon_i$$</span></p>

<p>In this specification <span class="math-container">$\beta_2$</span> will already give you an elasticity. This specification also already controls for possible non-linear effect of an age so no quadratic term is necessary. This being said logs are only rarely used for age, but you could estimate this specification as an additional robustness check and still keep the main specification with quadratic term. </p>

<p>Alternatively, already in your original equation which is in log-level form:</p>

<p><span class="math-container">$$ln(w_i)=\beta_1+\beta_2age+\beta_3age_i^2+\beta_4Y_i+\beta_5T_i+\beta_6Mar_i+\epsilon_i$$</span></p>

<p>all the beta coefficients are already semi-elasticities (as also pointed out by @chan1142). So if you would only care about semi-elasticity you will already get your result from your original model as in your case semi elasticity of wages with respect to age is <span class="math-container">$\beta_2+2\beta_3age_i$</span>.</p>

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2020-01-30T11:15:54.123Z
But what if I want full elasticity of ln(wage) with respect to age, not just semi-elasticity?
Bob Charles
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2020-01-30T11:16:26.490Z
@BobCharles I address that in 3rd paragraph (or 4th depending on whether you count that one sentence after 1st paragraph as separate paragraph)
1muflon1
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2020-01-30T11:23:14.753Z
So my expression (2+23)() is under no circumstance correct? Just wondering because it was listed in the solutions to the worksheet.
Bob Charles
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2020-01-30T11:31:04.207Z
@BobCharles I don’t know how exactly was the question formatted and other details but the answer that I gave you was taken from the Essential Mathematics for economic analysis textbook by Sydeater et al which is quite popular textbook. Also the answers that log-log form gives you elasticity and log-level form semi elasticity is based also on the latest edition of Verbeeks guide to modern econometrics and I recall seeing it in Wooldridge textbook as also in Stock and Watson textbook. This being said there might be special cases when that’s the solution, also it might be the solution of just...
1muflon1
CC BY-SA 4.0
2020-01-30T11:32:39.997Z
Applying the formulas I gave you (I did not checked for that) maybe once you exponentiate both sides and do calculations that will be the result. For example if you assume that ln (w(age, ...)=1 that would be the correct answer for elasticity of log of wage wrt age, not sure if it’s justified in this case though
1muflon1
CC BY-SA 4.0
2020-03-03T16:58:11.183Z
@BobCharles Seems like this answer has helped you. It would be great if you could accept it as an answer, or suggest ways that @ 1muflon1 could edit it so that you would accept it.
Art
CC BY-SA 4.0
2020-01-30T08:15:35.790Z
2020-01-30T08:21:17.547Z
2020-01-30T08:21:17.547Z
1muflon1
<p>Given the following regression: <span class="math-container">$ln(w_i)=\beta_1+\beta_2age+\beta_3age_i^2+\beta_4Y_i+\beta_5T_i+\beta_6Mar_i+\epsilon_i$</span> I am asked to calculate the elasticity of wages with respect to age. Is the following attempt valid?</p>

<p><span class="math-container">$$\frac{\partial ln(w)}{{\frac{\partial age}{age}}}=\frac{\partial ln(w)}{\partial age}age=(\beta_2+2\beta_3age)(age)$$</span></p>

<p>Is this consistent with the definition of an elasticity as percent change over percentage change? Thank you.</p>

CC BY-SA 4.0
2020-01-30T04:53:47.467Z
People usually use a semi-elasticity: dln(w)/dage, which is b2+2*b3*age in your case. It’s perfectly fine to use it. If you must calculate the elasticity, I think your formula works. But it is uncommon to report it.
chan1142
<econometrics><mathematical-economics><statistics><real-analysis>
Calculating elasticity between terms in a regression equation
Bob Charles
# Calculating elasticity between terms in a regression equation

(Asked by: username_0 on 2020-01-30)

Given the following regression: $ln(w\_i)=\beta\_1+\beta\_2age+\beta\_3age\_i^2+\beta\_4Y\_i+\beta\_5T\_i+\beta\_6Mar\_i+\epsilon\_i$ I am asked to calculate the elasticity of wages with respect to age. Is the following attempt valid?

$$\frac{\partial ln(w)}{{\frac{\partial age}{age}}}=\frac{\partial ln(w)}{\partial age}age=(\beta\_2+2\beta\_3age)(age)$$

Is this consistent with the definition of an elasticity as percent change over percentage change? Thank you.

**Tags:** econometrics, mathematical-economics, statistics, real-analysis

> **Comments:**
> **username_1**: People usually use a semi-elasticity: dln(w)/dage, which is b2+2*b3*age in your case. It’s perfectly fine to use it. If you must calculate the elasticity, I think your formula works. But it is uncommon to report it.  

---

## 2 answers

### Answer by username_2 on 2024-01-26. Score: 1

The formula of the OP is perfectly fine.

Treating the derivative as a ratio of differential (long live Leibniz), with $x$ representing "age",
\begin{align}
\frac{d \ln w}{dx} &= b\_2 + 2b\_3x \implies \frac 1 w \cdot \frac {d w}{dx} = b\_2 + 2b\_3x\\
\\
&\implies \frac x w \cdot \frac {d w}{dx} = \frac{dw/dx}{w/x} = (b\_2 + 2b\_3x)\cdot x,
\end{align}

where now the left-hand-side is the definition of point-elasticity ("marginal over average").

The confusingly so-called "semi-elasticity" (confusingly because it is not an elasticity) comes from writing:

$$\frac{d \ln w}{dx} = b\_2 + 2b\_3x \implies \frac {d w / w}{dx} = b\_2 + 2b\_3x,$$

and considering *a one-unit change* in $x$, which means that we will set $dx=1$ and obtain

$$\frac {d w }{w} \approx \frac {\Delta w}{w} = \% \Delta(w) \approx b\_2 + 2b\_3x.$$

So $(b\_2 + 2b\_3x)\cdot x$ is the proper elasticity, while the "semi-elasticity" $b\_2 + 2b\_3x$ in essence is **the percentage change in the wage if age increases by one year** (and since we have included a squared term, this percentage change depends also on the current age).


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### Answer by username_3 on 2020-01-30. Score: 0

No that would not be correct definition of elasticity. First, mathematically in multivariate function elasticity is defined as follows:

$$ EL\_x =\frac{ f\_x '(x,y)}{f(x,y)}x$$ or in your case it would be: $$ \frac{ \partial \ln [w(age,Y,T,Mar)]}{\partial age} \frac{age}{\\ln [w(age,Y,T,Mar)]}$$

However, even if you would plug in the expressions in this formula you would get an elasticity of a **log of wage** with respect to age - I am not sure if thats what you actually want. You could exponentiate the original function but that would be messy as hell.

One way how to estimate the elasticity directly from the regression would be using log-log form for variables of interest. In your case (assuming there are no people with 0 age in your sample):

$$ln(w\_i)=\beta\_1+\beta\_2 \ln (age) +\beta\_4Y\_i+\beta\_5T\_i+\beta\_6Mar\_i+\epsilon\_i$$

In this specification $\beta\_2$ will already give you an elasticity. This specification also already controls for possible non-linear effect of an age so no quadratic term is necessary. This being said logs are only rarely used for age, but you could estimate this specification as an additional robustness check and still keep the main specification with quadratic term.

Alternatively, already in your original equation which is in log-level form:

$$ln(w\_i)=\beta\_1+\beta\_2age+\beta\_3age\_i^2+\beta\_4Y\_i+\beta\_5T\_i+\beta\_6Mar\_i+\epsilon\_i$$

all the beta coefficients are already semi-elasticities (as also pointed out by @username_1). So if you would only care about semi-elasticity you will already get your result from your original model as in your case semi elasticity of wages with respect to age is $\beta\_2+2\beta\_3age\_i$.

> **Comments:**
> **username_0**: But what if I want full elasticity of ln(wage) with respect to age, not just semi-elasticity?  
> **username_3**: @BobCharles I address that in 3rd paragraph (or 4th depending on whether you count that one sentence after 1st paragraph as separate paragraph)  
> **username_0**: So my expression (2+23)() is under no circumstance correct? Just wondering because it was listed in the solutions to the worksheet.  
> **username_3**: @BobCharles I don’t know how exactly was the question formatted and other details but the answer that I gave you was taken from the Essential Mathematics for economic analysis textbook by Sydeater et al which is quite popular textbook. Also the answers that log-log form gives you elasticity and log-level form semi elasticity is based also on the latest edition of Verbeeks guide to modern econometrics and I recall seeing it in Wooldridge textbook as also in Stock and Watson textbook. This being said there might be special cases when that’s the solution, also it might be the solution of just...  
> **username_3**: Applying the formulas I gave you (I did not checked for that) maybe once you exponentiate both sides and do calculations that will be the result. For example if you assume that ln (w(age, ...)=1 that would be the correct answer for elasticity of log of wage wrt age, not sure if it’s justified in this case though  
> **username_4**: @BobCharles Seems like this answer has helped you. It would be great if you could accept it as an answer, or suggest ways that @ username_3 could edit it so that you would accept it.  

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