1
$\begingroup$

It is possible for the interest rate to affect the expenditure with consumption. An increase in the interest rate could, in principle, lead to increases in savings and, that way, to a decrease in consumption, given the income level. Suppose that consumption is, indeed, reduced by an increase in the interest rate. How will the IS curve be affected?

Considering the following equation for the IS curve: $$ Y=C(\overbrace{Y−T(Y)}^{+})+I(\overbrace{r}^{-})+G+NX(\overbrace{Y}^{-}) $$ I assume it would be changed to $$ Y=C(\overbrace{Y−T(Y)}^{+}, \overbrace{r}^{-})+I(\overbrace{r}^{-})+G+NX(\overbrace{Y}^{-}) $$ And thus, what was $$ \frac{dY}{dr} = \frac{dI(r)}{dr} < 0 $$ now is $$ \frac{dY}{dr} = \frac{dC(Y-T(Y), r)}{dr} + \frac{dI(r)}{dr} < 0 $$ and, therefore, the curve now is more downward-sloping than before.

Is this right? Kinda feels like I am missing some indirect effect.

Thanks!

EDIT: I was on the right track, but not correct. The trick is to apply the total derivative.

Going from $Y = C(Y-T(Y)) + I(r) + G + NX(Y)$, we have: $$ \operatorname{dY} = \underbrace{\frac{\partial C(\cdot)}{\partial Y}}_{\equiv C_Y > 0} \cdot \operatorname{dY} + \underbrace{\frac{\partial I(\cdot)}{\partial r}}_{\equiv I_r < 0} \cdot \operatorname{dr} \therefore \operatorname{dY} - C_Y \cdot \operatorname{dY} = I_r \cdot \operatorname{dr} \Leftrightarrow \boxed{\boxed{\frac{\operatorname{dr}}{\operatorname{dY}}\bigg\vert_{IS} = \frac{1-C_Y}{I_r} < 0}} $$
Now, with the new equation, we have: $$ \operatorname{dY} = \underbrace{\frac{\partial C(\cdot)}{\partial Y}}_{\equiv C_Y > 0} \cdot \operatorname{dY} + \underbrace{\frac{\partial C(\cdot)}{\partial r}}_{\equiv C_r < 0} \cdot \operatorname{dr} + \underbrace{\frac{\partial I(\cdot)}{\partial r}}_{\equiv I_r < 0} \cdot \operatorname{dr} \therefore \operatorname{dY} - C_Y \cdot \operatorname{dY} = (C_r + I_r) \cdot \operatorname{dr} \Leftrightarrow \boxed{\boxed{\frac{\operatorname{dr}}{\operatorname{dY}}\bigg\vert_{IS'} = \frac{1-C_Y}{(C_r + I_r)} < 0}} $$ Since $|C_r+I_r| > |I_r|$, the IS curve is now flatter and thus the income responds more to changes in the interest rate than before.

$\endgroup$
1
  • 1
    $\begingroup$ Note, you can answer your own question if you later found correct answer $\endgroup$
    – 1muflon1
    Mar 17 at 21:52
1
$\begingroup$

The trick is to apply the total derivative.

Going from $Y = C(Y-T(Y)) + I(r) + G + NX(Y)$, we have: $$ \operatorname{dY} = \underbrace{\frac{\partial C(\cdot)}{\partial Y}}_{\equiv C_Y > 0} \cdot \operatorname{dY} + \underbrace{\frac{\partial I(\cdot)}{\partial r}}_{\equiv I_r < 0} \cdot \operatorname{dr} \therefore \operatorname{dY} - C_Y \cdot \operatorname{dY} = I_r \cdot \operatorname{dr} \Leftrightarrow \boxed{\boxed{\frac{\operatorname{dr}}{\operatorname{dY}}\bigg\vert_{IS} = \frac{1-C_Y}{I_r} < 0}} $$
Now, with the new equation, we have: $$ \operatorname{dY} = \underbrace{\frac{\partial C(\cdot)}{\partial Y}}_{\equiv C_Y > 0} \cdot \operatorname{dY} + \underbrace{\frac{\partial C(\cdot)}{\partial r}}_{\equiv C_r < 0} \cdot \operatorname{dr} + \underbrace{\frac{\partial I(\cdot)}{\partial r}}_{\equiv I_r < 0} \cdot \operatorname{dr} \therefore \operatorname{dY} - C_Y \cdot \operatorname{dY} = (C_r + I_r) \cdot \operatorname{dr} \Leftrightarrow \boxed{\boxed{\frac{\operatorname{dr}}{\operatorname{dY}}\bigg\vert_{IS'} = \frac{1-C_Y}{(C_r + I_r)} < 0}} $$ Since $|C_r+I_r| > |I_r|$, the IS curve is now flatter and thus the income responds more to changes in the interest rate than before.

$\endgroup$

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.