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.
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 $f_i$.