<p>The derivative of $log(x)$ is $1/x$, which implies that $dlog(x)=(dx)/x$. Therefore, $e=dlog(y)/dlog(x)$ is the same as $e=(dy)/y/(dx)/x$. This ratio $e$ is known as the elasticity of $y$ with respect to $x$. We usually express derivatives $f′(x)$ as the change in $f$ when $x$ moves one unit, but $f(x)+1∗f′(x)$ is just an approximation (a Taylor approximation) of $f(x+1)$. Therefore, saying that a 1% change in $x$ will lead to an $e$ change in $y$ is just an approximation. It's a rate of change that is exact only at the $x$ point, but not exact as you move away from it to $x+0.01∗x$.</p>

CC BY-SA 3.0
2016-05-17T14:22:40.620Z
2016-05-17T14:22:40.620Z
Fix.B.
<p>Let's define elasticity, in a regression setting, as $\frac{\partial \log E(y|u)}{\partial \log x_j }$. How do we relate this definition to the percentage change interpretation, i.e., seeing this elasticity as the approximate change in $E(y|x)$, when $x_j$ increases by 1 percent? I'm reading Wooldridge. </p>

<p>Any help would be appreciated.</p>

CC BY-SA 3.0
2016-05-11T14:13:35.680Z
So you want the percent change of elasticity (which is a percent change)?
DornerA
CC BY-SA 3.0
2016-05-11T16:21:22.640Z
@DornerA I've edited the question. I think it now answers your question.
An old man in the sea.
CC BY-SA 3.0
2016-05-11T17:44:47.157Z
Are they not the same huber in principle? The calculus version is an exact statement, the 1% version is an (Taylor) approximation of what a 1% change would bring given the derivatives evaluated at 0%. Does this make sense?
Fix.B.
CC BY-SA 3.0
2016-05-11T18:31:46.260Z
@Fix.B. could you please elaborate a bit? thanks
An old man in the sea.
CC BY-SA 3.0
2016-05-11T18:56:47.263Z
The derivative of $log x$ is $1/x$, which implies that $d log x = (dx)/x$. Therefore $e=d log y / d log x$ is the same as $e=(dy)/y / (dx)/x$. This ratio $e$ is known as the elasticity of $y$ with respect to $x$. We usually express derivatives $f'(x)$ as the change in $f$ when $x$ moves one unit, but $f(x)+ 1 * f'(x)$ is just an approximation (a Taylor approximation) of $f(x+1)$.Therefore saying that a 1% change in $x$ will lead to an $e$ change in $y$ is just an approximation. Its a rate of change that is exact only at the $x$ point, but not exact as you move away from it to $x+0.01*x$.
Fix.B.
<econometrics>
Relationship between elasticity and interpretation of approximate percentage change
An old man in the sea.
# Relationship between elasticity and interpretation of approximate percentage change

(Asked by: username_0 on 2016-05-11)

Let's define elasticity, in a regression setting, as $\frac{\partial \log E(y|u)}{\partial \log x\_j }$. How do we relate this definition to the percentage change interpretation, i.e., seeing this elasticity as the approximate change in $E(y|x)$, when $x\_j$ increases by 1 percent? I'm reading Wooldridge.

Any help would be appreciated.

**Tags:** econometrics

> **Comments:**
> **username_1**: So you want the percent change of elasticity (which is a percent change)?  
> **username_0**: @username_1 I've edited the question. I think it now answers your question.  
> **username_2**: Are they not the same huber in principle? The calculus version is an exact statement, the 1% version is an (Taylor) approximation of what a 1% change would bring given the derivatives evaluated at 0%. Does this make sense?  
> **username_0**: @username_2 could you please elaborate a bit? thanks  
> **username_2**: The derivative of $log x$ is $1/x$, which implies that $d log x = (dx)/x$. Therefore $e=d log y / d log x$ is the same as $e=(dy)/y / (dx)/x$. This ratio $e$ is known as the elasticity of $y$ with respect to $x$. We usually express derivatives $f'(x)$ as the change in $f$ when $x$ moves one unit, but $f(x)+ 1 * f'(x)$ is just an approximation (a Taylor approximation) of $f(x+1)$.Therefore saying that a 1% change in $x$ will lead to an $e$ change in $y$ is just an approximation. Its a rate of change that is exact only at the $x$ point, but not exact as you move away from it to $x+0.01*x$.  

---

## 1 answers

### Answer by username_2 on 2016-05-17. Score: 0 | Accepted answer

The derivative of $log(x)$ is $1/x$, which implies that $dlog(x)=(dx)/x$. Therefore, $e=dlog(y)/dlog(x)$ is the same as $e=(dy)/y/(dx)/x$. This ratio $e$ is known as the elasticity of $y$ with respect to $x$. We usually express derivatives $f′(x)$ as the change in $f$ when $x$ moves one unit, but $f(x)+1∗f′(x)$ is just an approximation (a Taylor approximation) of $f(x+1)$. Therefore, saying that a 1% change in $x$ will lead to an $e$ change in $y$ is just an approximation. It's a rate of change that is exact only at the $x$ point, but not exact as you move away from it to $x+0.01∗x$.


---

