Why log(A) can be seen as %ΔA in economics?

Question

We usually see deduce in economitrics changes $lnY$ to $\%\Delta Y$, say we have Cobb-Douglas function like this $Y=AK^αL^{1−α}$, in one of the book I read, it is changed to :

But why can we do like this ?

Answer 1

This may help: $$ \Delta Y\simeq \Delta A (K^\alpha L^{1-\alpha})+\Delta K (A\alpha K^{\alpha-1}L) + \Delta L (A (1-\alpha) L^{-\alpha}) $$ is just the total differential of $Y=Y(A,K,L)$. Then divide everything by $Y=A K^\alpha L^{1-\alpha}$ and you get the equation.

(Another option how to look at this equation is to concider the quantity $\Delta f/f$ to be $\Delta (\ln f)$ and apply $\Delta$ to $\ln Y=\ln A + \alpha\ln K + (1-\alpha)\ln L$.)

Answer 2

If $g(x) = \ln f(x)$, then by the first order Taylor approximation, $$\Delta g \approx {dg \over dx} \Delta x \ \text{ where } \ {dg \over dx} = { f'(x) \over f(x)}$$ We have also that $f'(x) \ \Delta x \approx \Delta f$. In other words

$$\Delta g = \Delta(\ln f(x))= \frac{f'(x) \ \Delta x}{f(x)} = \frac{\Delta f}{f}$$

Drop now the crutch of $g$ and we have simply

$$\boxed{ \Delta(\ln f) = \frac{\Delta f}{f}}$$

We could also argue that

$${\Delta(\ln f) \over \Delta f} \approx { d(\ln f) \over df} = {1 \over f}$$

Hence by extension,

$$Y = AK^\alpha L^{1-\alpha} \Rightarrow \ln Y = \ln A + \alpha \ln K + (1-\alpha)\ln L \Rightarrow {\Delta Y\over Y} = {\Delta A \over A} + \alpha {\Delta K \over K} + (1-\alpha){\Delta L \over L}$$

Answer 3

Thank you @Simon S, I got the idea, let $g(x)=lnf(x)$ $$dg(x)=g'(x)dx=\frac{f'(x)}{f(x)}dx=\frac{df(x)}{f(x)}$$

Answer 4

That $\Delta\log(X_t)\approx\frac{\Delta X_t}{X_t}$ is a general approximation and has nothing to do with the Cobb-Douglas functional form: $$ \Delta\log(X_t)=\log(X_{t})-\log(X_{t-1})=\log\left(\frac{X_{t}}{X_{t-1}}\right)=\log\left(1+\frac{X_{t}-X_{t-1}}{X_{t-1}}\right)\\ \approx \frac{X_{t}-X_{t-1}}{X_{t-1}}=\frac{\Delta X_t}{X_t}\cdot $$ Going from the first line to the second line, we have used $\log(1+x)\approx x$ for $x$ small.

Now let's used this approximation: $$ \log(Y_t)=\log(A_t)+\alpha\log(K_t)+(1-\alpha)\log(L_t)\\ \implies\Delta\log(Y_t)=\Delta\log(A_t)+\alpha\Delta\log(K_t)+(1-\alpha)\Delta\log(L_t)\\ \implies\frac{\Delta Y_t}{Y_t}=\frac{\Delta A_t}{A_t}+\alpha\frac{\Delta K_t}{K_t}+(1-\alpha)\frac{\Delta L_t}{L_t} $$ where, in going from the 2nd line to the 3rd line, we have used the approximation 4 times: replace $X$ with $Y$, $A$, $K$, and $L$.