<p>Taking the ratio of the rearranged second and third equations of</p>
<blockquote>
<p>The comparative static identities:
<span class="math-container">$$
\begin{array}{r}B-x^{*} P_{x}-y^{*} P_{y} \equiv 0 \\ U_{x}\left(x^{*}, y^{*}\right)-\lambda^{*} P_{x} \equiv 0 \\ U_{y}\left(x^{*}, y^{*}\right)-\lambda^{*} P_{y} \equiv 0\end{array}
$$</span></p>
</blockquote>
<p>we get
<span class="math-container">$$
\frac{U_{x}\left(x^{*}, y^{*}\right)}{U_{y}\left(x^{*}, y^{*}\right)} =
\frac{P_x}{P_y}. \tag{MRS}
$$</span>
(This is a very famous condition in microeconomics, the LHS being the marginal rate of substitution.)</p>
<p>Multiplying
<span class="math-container">$$0 = U_{x} d x^{*}+U_{y} d y^{*}$$</span>
by
<span class="math-container">$$-\frac{P_x}{U_x}$$</span>
we get
<span class="math-container">$$0 = -P_x d x^{*} - \frac{P_x}{\frac{U_x}{U_y}} d y^{*}.$$</span>
It follows from the (MRS) equation above that
<span class="math-container">$$
P_y = \frac{P_x}{\frac{U_x}{U_y}},
$$</span>
thus
<span class="math-container">$$ 0 = -P_{x} d x^{*}-P_{y} d y^{*}.$$</span></p>

CC BY-SA 4.0
2021-11-24T10:09:22.753Z
2021-11-24T10:09:22.753Z
Giskard
<p>Much of this is setting up the problem. So if you're familiar it's likely best to start from the very bottom and work up if needed.  The question is asking about the income and substitution effect.</p>
<p>Constrained max setup where <span class="math-container">$U = U(x,y) = (x+2)(y+1)$</span> with <span class="math-container">$U_x, U_y &gt; 0$</span> and subject to budget constraint: <span class="math-container">$x P_x + y P_y = B$</span>.</p>
<p>The comparative static identities:
<span class="math-container">$$
\begin{array}{r}B-x^{*} P_{x}-y^{*} P_{y} \equiv 0 \\ U_{x}\left(x^{*}, y^{*}\right)-\lambda^{*} P_{x} \equiv 0 \\ U_{y}\left(x^{*}, y^{*}\right)-\lambda^{*} P_{y} \equiv 0\end{array}
$$</span>
I take the total differentials,
<span class="math-container">$$
\begin{aligned}-P_{x} d x^{*}-P_{y} d y^{*} &amp;=x^{*} d P_{x}+y^{*} d P_{y}-d B \\-P_{x} d \lambda^{*}+U_{x x} d x^{*}+U_{x y} d y^{*} &amp;=\lambda^{*} d P_{x} \\-P_{y} d \lambda^{*}+U_{y x} d x^{*}+U_{y y} d y^{*} &amp;=\quad \lambda^{*} d P_{y} \end{aligned}
$$</span>
then for the effect of <span class="math-container">$P_x$</span>, we set <span class="math-container">$dP_y = dB = 0$</span> and <span class="math-container">$dP_x \neq 0$</span> and divide all three equations by <span class="math-container">$dP_x$</span>. With Cramer's rule on
<span class="math-container">$$
\left[\begin{array}{ccc}0 &amp; -P_{x} &amp; -P_{y} \\ -P_{x} &amp; U_{x x} &amp; U_{x y} \\ -P_{y} &amp; U_{y x} &amp; U_{y y}\end{array}\right]\left[\begin{array}{c}\left(\partial \lambda^{*} / \partial P_{x}\right) \\ \left(\partial x^{*} / \partial P_{x}\right) \\ \left(\partial y^{*} / \partial P_{x}\right)\end{array}\right]=\left[\begin{array}{c}x^{*} \\ \lambda^{*} \\ 0\end{array}\right]
$$</span>
we get
<span class="math-container">$$
\begin{aligned}\left(\frac{\partial x^{*}}{\partial P_{x}}\right) &amp;=\frac{1}{|J|}\left|\begin{array}{ccc}0 &amp; x^{*} &amp; -P_{y} \\ -P_{x} &amp; \lambda^{*} &amp; U_{x y} \\ -P_{y} &amp; 0 &amp; U_{y y}\end{array}\right| \\ &amp;=\frac{-x^{*}}{|J|}\left|\begin{array}{ll}-P_{x} &amp; U_{x y} \\ -P_{y} &amp; U_{y y}\end{array}\right|+\frac{\lambda^{*}}{|J|}\left|\begin{array}{cc}0 &amp; -P_{y} \\ -P_{y} &amp; U_{y y}\end{array}\right| \\ &amp; \equiv T_{1}+T_{2} \quad\left[T_{i} \text { means the } i \text { th term }\right] \end{aligned}
$$</span>
First it was shown that by setting effective income loss from a change in <span class="math-container">$P_x$</span> to zero: i.e. the first equation of the total differential below we set <span class="math-container">$x^* dP_x = 0$</span>, we compensate the income effect and get:
<span class="math-container">$$
\left(\frac{\partial x^{*}}{\partial P_{x}}\right)_{\text {compensated }}=\frac{1}{|J|}\left|\begin{array}{ccc}0 &amp; 0 &amp; -P_{y} \\ -P_{x} &amp; \lambda^{*} &amp; U_{x y} \\ -P_{y} &amp; 0 &amp; U_{y y}\end{array}\right|=\frac{\lambda^{*}}{|J|}\left|\begin{array}{cc}0 &amp; -P_{y} \\ -P_{y} &amp; U_{y y}\end{array}\right|=T_{2}
$$</span>
And can express the derivative: <span class="math-container">$\left(\frac{\partial x^{*}}{\partial P_{x}}\right)=T_{1}+T_{2}=\underbrace{-\left(\frac{\partial x^{*}}{\partial B}\right) x^{*}}_{\text {income effect }}+\underbrace{\left(\frac{\partial x^{*}}{\partial P_{x}}\right)_{\text {compensated }}}_{\text {substitution effect }}$</span></p>
<p><strong>Text Question:</strong> When studying the effect of <span class="math-container">$dP_x$</span> alone, the first equation (of the total differentials) reduces to <span class="math-container">$-P_{x} d x^{*}-P_{y} d y^{*}=x^{*} d P_{x}$</span>, and when we compensate for the consumer's effective income loss by dropping <span class="math-container">$x^{*} d P_{x}$</span>, equation becomes <span class="math-container">$-P_{x} d x^{*}-P_{y} d y^{*}=0$</span>. Show that last result can be obtained from a compensation procedure where we keep consumer's optimal utility level <span class="math-container">$U^*$</span> (rather than effective income) unchanged, so that the term <span class="math-container">$T_2$</span> can be interpreted as <span class="math-container">$(\partial x^{*} / \partial P_x)_{U^* = \text{constant}}$</span> [Hint: Make use of <span class="math-container">$\frac{U_{x}}{U_{y}}=\frac{P_{x}}{P_{y}}$</span>]</p>
<p><strong>Text Answer:</strong> Optimal utility is <span class="math-container">$U^{*}=U^{*}\left(x^{*}, y^{*}\right)$</span>.  Thus <span class="math-container">$d U^{*}=U_{x} d x^{*}+U_{y} d y^{*}$</span> where <span class="math-container">$U_x, U_y$</span> evaluated at optimum.  When <span class="math-container">$U^*$</span> constant, we have <span class="math-container">$dU^* = 0$</span> or <span class="math-container">$U_{x} d x^{*}+U_{y} d y^{*}=0$</span>.  From the above expression for <span class="math-container">$\left(\frac{\partial x^{*}}{\partial P_{x}}\right)$</span> we have <span class="math-container">$\frac{U_{x}}{U_{y}}=\frac{P_{x}}{P_{y}}$</span>. Thus we can express <span class="math-container">$d U^{*}=0$</span> by <span class="math-container">$P_{x} d x^{*}+P_{y} d y^{*}=0$</span>, or <span class="math-container">$-P_{x} d x^{*}-P_{y} d y^{*}=0$</span></p>
<p><strong>Q:</strong> How exactly does <span class="math-container">$ 0 =\boldsymbol{U}_{x} d x^{*}+\boldsymbol{U}_{y} d y^{*}$</span> turn into <span class="math-container">$-P_{x} d x^{*}-P_{y} d y^{*}=0$</span>?</p>

<microeconomics><mathematical-economics><optimization><comparative-statics>
Comparative Statics: Income Effect
aisync
# Comparative Statics: Income Effect

(Asked by: username_0 on 2021-11-24)

Much of this is setting up the problem. So if you're familiar it's likely best to start from the very bottom and work up if needed. The question is asking about the income and substitution effect.

Constrained max setup where $U = U(x,y) = (x+2)(y+1)$ with $U\_x, U\_y > 0$ and subject to budget constraint: $x P\_x + y P\_y = B$.

The comparative static identities:
$$
\begin{array}{r}B-x^{\*} P\_{x}-y^{\*} P\_{y} \equiv 0 \\ U\_{x}\left(x^{\*}, y^{\*}\right)-\lambda^{\*} P\_{x} \equiv 0 \\ U\_{y}\left(x^{\*}, y^{\*}\right)-\lambda^{\*} P\_{y} \equiv 0\end{array}
$$
I take the total differentials,
$$
\begin{aligned}-P\_{x} d x^{\*}-P\_{y} d y^{\*} &=x^{\*} d P\_{x}+y^{\*} d P\_{y}-d B \\-P\_{x} d \lambda^{\*}+U\_{x x} d x^{\*}+U\_{x y} d y^{\*} &=\lambda^{\*} d P\_{x} \\-P\_{y} d \lambda^{\*}+U\_{y x} d x^{\*}+U\_{y y} d y^{\*} &=\quad \lambda^{\*} d P\_{y} \end{aligned}
$$
then for the effect of $P\_x$, we set $dP\_y = dB = 0$ and $dP\_x \neq 0$ and divide all three equations by $dP\_x$. With Cramer's rule on
$$
\left[\begin{array}{ccc}0 & -P\_{x} & -P\_{y} \\ -P\_{x} & U\_{x x} & U\_{x y} \\ -P\_{y} & U\_{y x} & U\_{y y}\end{array}\right]\left[\begin{array}{c}\left(\partial \lambda^{\*} / \partial P\_{x}\right) \\ \left(\partial x^{\*} / \partial P\_{x}\right) \\ \left(\partial y^{\*} / \partial P\_{x}\right)\end{array}\right]=\left[\begin{array}{c}x^{\*} \\ \lambda^{\*} \\ 0\end{array}\right]
$$
we get
$$
\begin{aligned}\left(\frac{\partial x^{\*}}{\partial P\_{x}}\right) &=\frac{1}{|J|}\left|\begin{array}{ccc}0 & x^{\*} & -P\_{y} \\ -P\_{x} & \lambda^{\*} & U\_{x y} \\ -P\_{y} & 0 & U\_{y y}\end{array}\right| \\ &=\frac{-x^{\*}}{|J|}\left|\begin{array}{ll}-P\_{x} & U\_{x y} \\ -P\_{y} & U\_{y y}\end{array}\right|+\frac{\lambda^{\*}}{|J|}\left|\begin{array}{cc}0 & -P\_{y} \\ -P\_{y} & U\_{y y}\end{array}\right| \\ & \equiv T\_{1}+T\_{2} \quad\left[T\_{i} \text { means the } i \text { th term }\right] \end{aligned}
$$
First it was shown that by setting effective income loss from a change in $P\_x$ to zero: i.e. the first equation of the total differential below we set $x^\* dP\_x = 0$, we compensate the income effect and get:
$$
\left(\frac{\partial x^{\*}}{\partial P\_{x}}\right)\_{\text {compensated }}=\frac{1}{|J|}\left|\begin{array}{ccc}0 & 0 & -P\_{y} \\ -P\_{x} & \lambda^{\*} & U\_{x y} \\ -P\_{y} & 0 & U\_{y y}\end{array}\right|=\frac{\lambda^{\*}}{|J|}\left|\begin{array}{cc}0 & -P\_{y} \\ -P\_{y} & U\_{y y}\end{array}\right|=T\_{2}
$$
And can express the derivative: $\left(\frac{\partial x^{\*}}{\partial P\_{x}}\right)=T\_{1}+T\_{2}=\underbrace{-\left(\frac{\partial x^{\*}}{\partial B}\right) x^{\*}}\_{\text {income effect }}+\underbrace{\left(\frac{\partial x^{\*}}{\partial P\_{x}}\right)\_{\text {compensated }}}\_{\text {substitution effect }}$

**Text Question:** When studying the effect of $dP\_x$ alone, the first equation (of the total differentials) reduces to $-P\_{x} d x^{\*}-P\_{y} d y^{\*}=x^{\*} d P\_{x}$, and when we compensate for the consumer's effective income loss by dropping $x^{\*} d P\_{x}$, equation becomes $-P\_{x} d x^{\*}-P\_{y} d y^{\*}=0$. Show that last result can be obtained from a compensation procedure where we keep consumer's optimal utility level $U^\*$ (rather than effective income) unchanged, so that the term $T\_2$ can be interpreted as $(\partial x^{\*} / \partial P\_x)\_{U^\* = \text{constant}}$ [Hint: Make use of $\frac{U\_{x}}{U\_{y}}=\frac{P\_{x}}{P\_{y}}$]

**Text Answer:** Optimal utility is $U^{\*}=U^{\*}\left(x^{\*}, y^{\*}\right)$. Thus $d U^{\*}=U\_{x} d x^{\*}+U\_{y} d y^{\*}$ where $U\_x, U\_y$ evaluated at optimum. When $U^\*$ constant, we have $dU^\* = 0$ or $U\_{x} d x^{\*}+U\_{y} d y^{\*}=0$. From the above expression for $\left(\frac{\partial x^{\*}}{\partial P\_{x}}\right)$ we have $\frac{U\_{x}}{U\_{y}}=\frac{P\_{x}}{P\_{y}}$. Thus we can express $d U^{\*}=0$ by $P\_{x} d x^{\*}+P\_{y} d y^{\*}=0$, or $-P\_{x} d x^{\*}-P\_{y} d y^{\*}=0$

**Q:** How exactly does $ 0 =\boldsymbol{U}\_{x} d x^{\*}+\boldsymbol{U}\_{y} d y^{\*}$ turn into $-P\_{x} d x^{\*}-P\_{y} d y^{\*}=0$?

**Tags:** microeconomics, mathematical-economics, optimization, comparative-statics


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## 1 answers

### Answer by username_1 on 2021-11-24. Score: 3 | Accepted answer

Taking the ratio of the rearranged second and third equations of

> The comparative static identities:
> $$
> \begin{array}{r}B-x^{\*} P\_{x}-y^{\*} P\_{y} \equiv 0 \\ U\_{x}\left(x^{\*}, y^{\*}\right)-\lambda^{\*} P\_{x} \equiv 0 \\ U\_{y}\left(x^{\*}, y^{\*}\right)-\lambda^{\*} P\_{y} \equiv 0\end{array}
> $$

we get
$$
\frac{U\_{x}\left(x^{\*}, y^{\*}\right)}{U\_{y}\left(x^{\*}, y^{\*}\right)} =
\frac{P\_x}{P\_y}. \tag{MRS}
$$
(This is a very famous condition in microeconomics, the LHS being the marginal rate of substitution.)

Multiplying
$$0 = U\_{x} d x^{\*}+U\_{y} d y^{\*}$$
by
$$-\frac{P\_x}{U\_x}$$
we get
$$0 = -P\_x d x^{\*} - \frac{P\_x}{\frac{U\_x}{U\_y}} d y^{\*}.$$
It follows from the (MRS) equation above that
$$
P\_y = \frac{P\_x}{\frac{U\_x}{U\_y}},
$$
thus
$$ 0 = -P\_{x} d x^{\*}-P\_{y} d y^{\*}.$$


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