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My professor presented two concepts of elasticity in my last microeconomics class:

Product Elasticity: $\mu_i=\frac{\partial f(x)}{\partial x_i}\frac{x_i}{f(x)}$

Scale Elasticity: $\mu(x)=\frac{d\ln(f(tx))}{d\ln t}\Bigr\rvert_{t = 1}$

Also, he asked us to prove that $\mu(x)=\sum_{i=1}^n\mu_i$

I found this demonstration on the internet, but I can't really understand it:

$$\mu(x)=\frac{d\ln(f(tx))}{d\ln t}\Bigr\rvert_{t = 1}=\frac{\sum_{i=1}^n f_i(x)x_i}{f(x)}=\sum_{i=1}^n\mu_i$$

If someone could help me with the steps of this proof, I would really appreciate it.

Thank you very much.

Jackaba
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1 Answers1

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I assume that $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is differentiable. $x$ is a vector in $\mathbb{R}^n$. Then we first use the chain rule.

\begin{align} \mu(x)&=\frac{\mathrm{d}\ln(f(tx))}{\mathrm{d}\ln t}=\frac{\mathrm{d}\ln(f(tx))}{\mathrm{d}t}\frac{\mathrm{d} t}{\mathrm{d}\ln t} \end{align}

For the second term we use the fact that $\frac{dx}{df(x)}=\frac{1}{\frac{df(x)}{dx}}$, if all the expressions are well-defined. The first term is a directional derivative that evaluates to a scalar product of the gradient of $f$ and $x$:

\begin{align} =\nabla_x\ln(f(tx))\cdot x \frac 1{\frac 1t}= \sum_{i=1}^n \frac{\partial \ln(f(tx))}{\partial x_i} x_i \cdot t = \sum_{i=1}^n \frac 1{f(tx)}\frac{\partial f(tx)}{\partial x_i} x_i \cdot t \end{align}

Then just evaluate the expression at $t=1$:

\begin{align} =\sum_{i=1}^n \frac {x_i}{f(x)}\frac{\partial f(x)}{\partial x_i}=\sum_{i=1}^n \mu_i \end{align}

Nota bene: I believe your Professor wants to indicate the partial derivative by $f_i$.

Snake707
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