1
$\begingroup$

We are given that $C$ is a function of $Y_D$ and $Y_D=Y-Y\tau$. What would be the total differential of $Y=C(Y_D)$?

So far I have the following: $$ dY=C_{Y_D}(1-\tau)dY+C_{Y_D}(-Y)d\tau$$

However I am not sure how to write this expression in terms of $C_Y$ and $C_{\tau}$. Where do I go from here? Is leaving the expression in terms of $C_{Y_D}$ appropriate?

$\endgroup$
3
  • $\begingroup$ Does the function $C$ only have one argument? If so, what is the difference between $C_Y$ and $C_\tau$? $\endgroup$
    – jmbejara
    Commented Feb 13, 2016 at 20:51
  • $\begingroup$ $C_Y$ would be partial derivative of consumption C with respect to income Y and $C_\tau$ would be partial derivative with respect to $\tau$ $\endgroup$
    – cpage
    Commented Feb 13, 2016 at 20:53
  • $\begingroup$ I have replied in my answer. $\endgroup$
    – jmbejara
    Commented Feb 13, 2016 at 21:12

1 Answer 1

1
$\begingroup$

$\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\dd}{\, \mathrm{d}}$ How many arguments does $C$ have? If the only argument to $C$ is $Y_D$, then $C_Y$ and $C_\tau$ might not make sense. I think it's easy to get yourself in trouble here.

Suppose that $C$'s only argument is $Y_D$. Yet, we want to impose the restriction that $Y_D = Y(1-\tau)$. Well, as you wrote \begin{align*} \dd C = C_{Y_D} \dd Y = C_{Y_d} (1-\tau) \dd Y - C_{Y_D} Y \dd \tau \end{align*} is correct. But what is $C_Y$ and $C_\tau$? Well, technically, $$ \pd C Y = \pd C \tau = 0. $$ This is because $Y$ and $\tau$ do not directly affect $C$. I guess what you have in mind is something similar to the following. Suppose you define a function $f$ such that $$ f(Y, \tau) = C(Y(1-\tau)). $$ Then \begin{align*} f_Y &= \pd f Y = \pd C {Y_D} (1-\tau) \\ f_\tau &= \pd f \tau = -\pd C {Y_D} Y. \end{align*} So, really, $C_Y \neq f_Y$ and $C_\tau \neq f_\tau$, because $C_Y$ and $C_\tau$ are both equal to zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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