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

CC BY-SA 4.0
2021-03-17T22:02:19.377Z
2021-03-17T22:02:19.377Z
Pedro Cunha
<p>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?</p>
<p>Considering the following equation for the IS curve:
<span class="math-container">$$
Y=C(\overbrace{Y−T(Y)}^{+})+I(\overbrace{r}^{-})+G+NX(\overbrace{Y}^{-})
$$</span>
I assume it would be changed to
<span class="math-container">$$
Y=C(\overbrace{Y−T(Y)}^{+}, \overbrace{r}^{-})+I(\overbrace{r}^{-})+G+NX(\overbrace{Y}^{-})
$$</span>
And thus, what was
<span class="math-container">$$
\frac{dY}{dr} = \frac{dI(r)}{dr} &lt; 0
$$</span>
now is
<span class="math-container">$$
\frac{dY}{dr} = \frac{dC(Y-T(Y), r)}{dr} + \frac{dI(r)}{dr} &lt; 0 
$$</span>
and, therefore, the curve now is more downward-sloping than before.</p>
<p>Is this right? Kinda feels like I am missing some indirect effect.</p>
<p>Thanks!</p>
<p>EDIT: I was on the right track, but not correct. The trick is to apply the total derivative.</p>
<p>Going from <span class="math-container">$Y = C(Y-T(Y)) + I(r) + G + NX(Y)$</span>, we have:
<span class="math-container">$$
\operatorname{dY} = \underbrace{\frac{\partial C(\cdot)}{\partial Y}}_{\equiv C_Y &gt; 0} \cdot \operatorname{dY} + \underbrace{\frac{\partial I(\cdot)}{\partial r}}_{\equiv I_r &lt; 0} \cdot \operatorname{dr} \therefore \operatorname{dY} - C_Y \cdot \operatorname{dY} = I_r \cdot \operatorname{dr} \Leftrightarrow \boxed{\boxed{\frac{\operatorname{dr}}{\operatorname{dY}}\bigg\vert_{IS} = \frac{1-C_Y}{I_r} &lt; 0}}
$$</span><br />
Now, with the new equation, we have:
<span class="math-container">$$
\operatorname{dY} = \underbrace{\frac{\partial C(\cdot)}{\partial Y}}_{\equiv C_Y &gt; 0} \cdot \operatorname{dY} + \underbrace{\frac{\partial C(\cdot)}{\partial r}}_{\equiv C_r &lt; 0} \cdot \operatorname{dr} + \underbrace{\frac{\partial I(\cdot)}{\partial r}}_{\equiv I_r &lt; 0} \cdot \operatorname{dr} \therefore \operatorname{dY} - C_Y \cdot \operatorname{dY} = (C_r + I_r) \cdot \operatorname{dr} \Leftrightarrow \boxed{\boxed{\frac{\operatorname{dr}}{\operatorname{dY}}\bigg\vert_{IS'} = \frac{1-C_Y}{(C_r + I_r)} &lt; 0}}
$$</span>
Since <span class="math-container">$|C_r+I_r| &gt; |I_r|$</span>, the IS curve is now flatter and thus the income responds more to changes in the interest rate than before.</p>

CC BY-SA 4.0
2021-03-17T21:52:49.430Z
Note, you can answer your own question if you later found correct answer
1muflon1
<macroeconomics><interest-rate><consumption>
IS curve, interest rate and consumption
Pedro Cunha
# IS curve, interest rate and consumption

(Asked by: username_0 on 2021-03-15)

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.

**Tags:** macroeconomics, interest-rate, consumption

> **Comments:**
> **username_1**: Note, you can answer your own question if you later found correct answer  

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## 1 answers

### Answer by username_0 on 2021-03-17. Score: 1 | Accepted answer

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.


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