2
$\begingroup$

I am studying "The Simple Mathematics of Income Determination", by Paul Samuelson (from book is called "Macroeconomia (artigos selecionados)", by APEC-CAEN. It is the 1st chapter of the book, the 13th page), and I am having some troubles in understanding a calculus passage.

First, Samuelson states the income equation:

$(1)$ $ Y = C (Y - \bar{W}) + \bar{I} + \bar{G} $

Where $Y$ corresponds to income, $C$ corresponds to consumption, $\bar{W}$ corresponds to taxes (actually, I don't exactly know the odd reason why the letter W was picked as taxes instead of a more common letter as T), $\bar{I}$ corresponds to investment and $\bar{G}$ corresponds to government spending). For purpose of simplification, these three last variables are treated as constants.

Then, $Y$ is differentiatied in $\bar{G}$ (which I suppose it is an implicit differentiation, in the form of $\frac{dy}{dx} = - \frac{F_x}{F_y}$):

$(2)$ $\frac{dY}{d\bar{G}} = \frac{1}{1 - C'(Y - \bar{W})}$

But my doubt emerges in the following passage:

$(3)$ $\frac{dY}{d\bar{(-W)}} = \frac{C'(Y - \bar{W})}{1 - C'(Y - \bar{W})} = \frac{dY}{d\bar{G}} - 1 $

I realize that Samuelson did the same implicit differentiation as in step 2 and I can visualize that

$\frac{C'(Y - \bar{W})}{1 - C'(Y - \bar{W})} = \frac{1}{1 - C'(Y - \bar{W})} * C'(Y - \bar{W}) = \frac{dY}{d\bar{G}} * C'(Y - \bar{W})$

but, how does $\frac{dY}{d\bar{G}} * C'(Y - \bar{W})$ turns to $\frac{dY}{d\bar{G}} - 1 $? I mean, how did this $-1$ replace the multiplying term $C'(Y - \bar{W})$

I tried to get it, but unsuccessfully. I hope you could shed some light on it.

$\endgroup$
1
  • $\begingroup$ By the way, there is another pertinent information about it. In the following paragraph after the step 3, it is said that "in words, a reduction of taxes and a increase in government spending increase income, but the taxes reduction effect on income is smaller than the government spending effect on income by 1". It certainly clarifies the meaning of "dY/dG - 1", but I still cannot see it mathematically. $\endgroup$
    – PGabriel96
    May 22 at 20:05
4
$\begingroup$

We start from the equation: $$ Y = C(Y - \overline{W}) + \overline{I} + \overline{G} $$ First, we totally differentiate with respect to $(-\overline{W})$ (taking into account that $Y$ is a function of $\overline{W}$, $\overline{I}$ and $\overline{G}$) $$ dY = C'(Y - \overline{W})dY + C'(Y - \overline{W}) d(-\overline{W}). $$ Bringing the $dY$ terms on one side gives: $$ (1 - C'(Y - \overline{W}))dY = C'(Y - \overline{W}) d(-\overline{W}),\\ \to \frac{dY}{d(-\overline{W})} = \frac{C'(Y - \overline{W})}{1 - C'(Y - \overline{W})} $$ Next, we totally differentiate with respect to $\overline{G}$: $$ dY = C'(Y - \overline{W})dY + d \overline{G} $$ Again gathering terms gives: $$ (1 - C'(Y - \overline{W})) dY = d \overline{G}\\ \to \frac{dY}{d \overline{G}} = \frac{1}{1 - C'(Y - \overline{W})} $$ Therefore: $$ \begin{align*} \frac{dY}{d \overline{G}} - 1 &= \frac{1}{1 - C'(Y - \overline{W})} - 1,\\ &=\frac{1 - 1 +C'(Y - \overline{W})}{1 - C'(Y - \overline{W})},\\ & = \frac{C'(Y - \overline{W})}{1 - C'(Y - \overline{W})},\\ &= \frac{dY}{d(- \overline{W})} \end{align*} $$

$\endgroup$
1
  • $\begingroup$ Thank you, tdm, that was exactly what I needed! $\endgroup$
    – PGabriel96
    May 23 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.