<p>Write <span class="math-container">$i=I(y,a)$</span> for the IS function (expressed as some function of <span class="math-container">$y$</span>). Likewise, let <span class="math-container">$i=L(y)$</span> be the LM function. Assume the following properties:</p>

<ul>
<li><p><span class="math-container">$\frac{\partial I(y,a)}{\partial a}&gt;0$</span> : <span class="math-container">$a$</span> is a parameter that causes the IS curve to shift up.</p></li>
<li><p><span class="math-container">$\frac{\partial I(y,a)}{\partial y}&lt;0$</span>: the IS function slopes downwards</p></li>
<li><p><span class="math-container">$\frac{\partial L(y)}{\partial y}&gt;0$</span> : the LM function slopes upwards</p></li>
</ul>

<p>For any <span class="math-container">$a$</span>, we find the equilibrium <span class="math-container">$y^*(a)$</span> at the intersection of the two functions:
<span class="math-container">$$I(y^*(a),a)- L(y^*(a))=0.$$</span>
Applying the implicit function theorem:
<span class="math-container">$$\frac{\partial I}{\partial a}+\frac{\partial y^*}{\partial a}\left(\frac{\partial I}{\partial y^*}-\frac{\partial L}{\partial y^*}\right)=0,$$</span>
which implies
<span class="math-container">$$\frac{\partial y^*}{\partial a}=-\frac{\frac{\partial I}{\partial a}}{\frac{\partial I}{\partial y^*}-\frac{\partial L}{\partial y^*}}.$$</span></p>

<p>The numerator is positive by hypothesis. Thus, for <span class="math-container">$y^*$</span> to increase with <span class="math-container">$a$</span> we require that the denominator be negative: <span class="math-container">$\frac{\partial I}{\partial y^*}&lt;\frac{\partial L}{\partial y^*}$</span>. This is clearly true because <span class="math-container">$I$</span> slopes downward and <span class="math-container">$L$</span> slopes upward. But <span class="math-container">$y^*$</span> would also increase even if <span class="math-container">$I$</span> sloped upward—provided itdoes so more gradually than <span class="math-container">$L$</span>.</p>

CC BY-SA 4.0
2018-11-18T19:06:53.247Z
2018-11-24T15:18:11.350Z
2018-11-24T15:18:11.350Z
Ubiquitous
<p>I have a super basic question on the shift in supply and demand curves ( I haven't seen this stuff since high school). More generally, this is about solving systems of simultaneous equations.</p>

<p>We have two curves, an IS curve and an LM curve. We know from theory
that the IS curve is downward sloping in <span class="math-container">$i-y$</span> space and the LM
curve is upward sloping in <span class="math-container">$i-y$</span> space. Let's take a basic example.
The IS curve is represented by
<span class="math-container">$$
i=-y
$$</span>
and the <span class="math-container">$LM$</span> curve is represented by 
<span class="math-container">$$
i=y
$$</span>
 Equilibrium would imply values of <span class="math-container">$y$</span> and <span class="math-container">$i$</span> that solve these
equations simultaneously, or <span class="math-container">$i=y=0.$</span> Now suppose government activity
increased, shifting the <span class="math-container">$IS$</span> curve to the right. In this framework,
this would result in a horizontal translation of <span class="math-container">$2$</span> units of the
<span class="math-container">$IS$</span> curve. 
<span class="math-container">$$
i=-y+2
$$</span>
 while the <span class="math-container">$LM$</span> curve remains unchanged. The new equilibrium is given
by <span class="math-container">$y=1$</span> and <span class="math-container">$1.$</span> Notice that the initial increase of <span class="math-container">$2$</span> in the
IS curve resutlts in equilibrium increasing by only <span class="math-container">$1.$</span> Intuitively,
this is because as <span class="math-container">$y$</span> increases, so does <span class="math-container">$i$</span> according to <span class="math-container">$LM$</span>
relation. And, as <span class="math-container">$i$</span> increases, <span class="math-container">$y$</span> decreases according to <span class="math-container">$IS$</span>
relation. As both these relations have to be satisfied, some of the
initial increase is moderated. My question is: what ensures that <span class="math-container">$y$</span>
in fact doesn't decrease? If the <span class="math-container">$LM$</span> curve determines interest rates
that are super sensitive to increases in <span class="math-container">$y,$</span> can it not be that
<span class="math-container">$i$</span> increases sufficiently that <span class="math-container">$y$</span> in equilibrium decreases? Graphically,
this is not possible- in the extreme case, the <span class="math-container">$LM$</span> curve is vertical,
<span class="math-container">$y$</span> does not change. But, what condition ensures that <span class="math-container">$y$</span> does not decrease?</p>

<macroeconomics><supply-and-demand>
Question on shifts in Supply and Demand Curves
ChinG
# Question on shifts in Supply and Demand Curves

(Asked by: username_0 on 2018-11-16)

I have a super basic question on the shift in supply and demand curves ( I haven't seen this stuff since high school). More generally, this is about solving systems of simultaneous equations.

We have two curves, an IS curve and an LM curve. We know from theory
that the IS curve is downward sloping in $i-y$ space and the LM
curve is upward sloping in $i-y$ space. Let's take a basic example.
The IS curve is represented by
$$
i=-y
$$
and the $LM$ curve is represented by
$$
i=y
$$
Equilibrium would imply values of $y$ and $i$ that solve these
equations simultaneously, or $i=y=0.$ Now suppose government activity
increased, shifting the $IS$ curve to the right. In this framework,
this would result in a horizontal translation of $2$ units of the
$IS$ curve.
$$
i=-y+2
$$
while the $LM$ curve remains unchanged. The new equilibrium is given
by $y=1$ and $1.$ Notice that the initial increase of $2$ in the
IS curve resutlts in equilibrium increasing by only $1.$ Intuitively,
this is because as $y$ increases, so does $i$ according to $LM$
relation. And, as $i$ increases, $y$ decreases according to $IS$
relation. As both these relations have to be satisfied, some of the
initial increase is moderated. My question is: what ensures that $y$
in fact doesn't decrease? If the $LM$ curve determines interest rates
that are super sensitive to increases in $y,$ can it not be that
$i$ increases sufficiently that $y$ in equilibrium decreases? Graphically,
this is not possible- in the extreme case, the $LM$ curve is vertical,
$y$ does not change. But, what condition ensures that $y$ does not decrease?

**Tags:** macroeconomics, supply-and-demand

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## 1 answers

### Answer by username_1 on 2018-11-18. Score: 3

Write $i=I(y,a)$ for the IS function (expressed as some function of $y$). Likewise, let $i=L(y)$ be the LM function. Assume the following properties:

* $\frac{\partial I(y,a)}{\partial a}>0$ : $a$ is a parameter that causes the IS curve to shift up.
* $\frac{\partial I(y,a)}{\partial y}<0$: the IS function slopes downwards
* $\frac{\partial L(y)}{\partial y}>0$ : the LM function slopes upwards

For any $a$, we find the equilibrium $y^\*(a)$ at the intersection of the two functions:
$$I(y^\*(a),a)- L(y^\*(a))=0.$$
Applying the implicit function theorem:
$$\frac{\partial I}{\partial a}+\frac{\partial y^\*}{\partial a}\left(\frac{\partial I}{\partial y^\*}-\frac{\partial L}{\partial y^\*}\right)=0,$$
which implies
$$\frac{\partial y^\*}{\partial a}=-\frac{\frac{\partial I}{\partial a}}{\frac{\partial I}{\partial y^\*}-\frac{\partial L}{\partial y^\*}}.$$

The numerator is positive by hypothesis. Thus, for $y^\*$ to increase with $a$ we require that the denominator be negative: $\frac{\partial I}{\partial y^\*}<\frac{\partial L}{\partial y^\*}$. This is clearly true because $I$ slopes downward and $L$ slopes upward. But $y^\*$ would also increase even if $I$ sloped upward—provided itdoes so more gradually than $L$.


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