Problem Sets' Answer Key

Problem Set 4

Setup

Two consumers have income $I=120$. Prices are $p_Y=1$ and, initially, $p_X=2$; later $p_X$ rises to $3$.

Consumer $T$ is a traditional consumer, with utility $u_T(x,y)=\sqrt{x}+\sqrt{y}$. Consumer $R$ is a reference-dependent consumer, with reference basket $(\bar x,\bar y)=(20,80)$ and utility $$\begin{aligned}&u_R(x,y\mid 20,80)=\sqrt{x}+\sqrt{y}\\ &\qquad+\mu(\sqrt{x}-\sqrt{20})+\mu(\sqrt{y}-\sqrt{80}),\end{aligned}$$ where $\mu(\Delta u)=\min(\Delta u,0)+\tfrac12\max(\Delta u,0)$.

As in the lecture's numerical example, ${A=(20,80)}$ is the initial basket and $B$ is the basket chosen after the price rise.

The Traditional Consumer

Problem Set 4, Questions 1–6

1. Write $\MU_x(x,y)$ and $\MU_y(x,y)$.

$$\begin{aligned}\MU_x(x,y)&=\frac{1}{2\sqrt{x}},\\ \MU_y(x,y)&=\frac{1}{2\sqrt{y}}.\end{aligned}$$

2. Evaluate both at the basket $(20,80)$.

$$\begin{aligned}\MU_x(20,80)&=\frac{1}{2\sqrt{20}}\approx 0.112,\\ \MU_y(20,80)&=\frac{1}{2\sqrt{80}}\approx 0.056.\end{aligned}$$ Since $\sqrt{80}=2\sqrt{20}$, $\MU_x$ is exactly twice $\MU_y$ at this basket.

3. With $p_X=2$, is $(20,80)$ on the budget line $2x+y=120$? Show your work.

Yes: $2\cdot 20+80=40+80=120$.

4. What is the price ratio $p_X/p_Y$?

$p_X/p_Y=2/1=2$.

5. Compute $\MU_x/p_X$ and $\MU_y/p_Y$ at $(20,80)$. Would moving one dollar from $Y$ to $X$ raise utility?

$$\begin{aligned}\frac{\MU_x}{p_X}&=\frac{1}{2\sqrt{20}}\cdot\frac12=\frac{1}{4\sqrt{20}}\approx 0.056,\\ \frac{\MU_y}{p_Y}&=\frac{1}{2\sqrt{80}}=\frac{1}{4\sqrt{20}}\approx 0.056.\end{aligned}$$ No. The two are equal, so the dollar adds as much utility through $X$ as it takes away through $Y$.

6. Compute $\mathrm{MRS}=\MU_x/\MU_y$ at $(20,80)$, and show $\mathrm{MRS}=\sqrt{y/x}$. Compare with the price ratio and conclude.

$$\begin{aligned}\mathrm{MRS}&=\frac{\MU_x}{\MU_y}=\frac{1/(2\sqrt{x})}{1/(2\sqrt{y})}\\ &=\frac{\sqrt{y}}{\sqrt{x}}=\sqrt{\frac{y}{x}},\end{aligned}$$ so at $(20,80)$, $\mathrm{MRS}=\sqrt{80/20}=\sqrt{4}=2=p_X/p_Y$. The basket ${A=(20,80)}$ is on the budget line and the indifference curve is tangent to the budget line there, so $A$ is Consumer $T$'s optimal basket.

The Price of $X$ Rises

Problem Set 4, Questions 7–8

7. $p_X$ rises to $3$. Is $(20,80)$ still affordable? Show your work.

No: $3\cdot 20+80=140\gt 120$.

8. Solve $\sqrt{y/x}=3$ together with $3x+y=120$.

$\sqrt{y/x}=3$ gives $y=9x$. Then $3x+9x=12x=120$, so $x=10$ and $y=90$. Consumer $T$ moves to ${B=(10,90)}$: it buys less of $X$, now dearer, and more of $Y$.

Budget lines at $p_X=2$ (solid) and $p_X=3$ (dashed), with Consumer $T$'s indifference curves through $A$ (dark) and $B$ (light). Each curve is tangent to its budget line, and $A$ lies outside the new budget set.

The Reference-Dependent Consumer

Problem Set 4, Questions 9–12

9. Write $\MU_x$ for $x\lt 20$ and for $x\gt 20$.

For $x\lt 20$ the change is a loss, ${\mu(\sqrt{x}-\sqrt{20})}={\sqrt{x}-\sqrt{20}}$, so the $x$-terms of $u_R$ are ${2\sqrt{x}-\sqrt{20}}$. For $x\gt 20$ it is a gain, ${\mu(\sqrt{x}-\sqrt{20})}={\tfrac12(\sqrt{x}-\sqrt{20})}$, so they are ${\tfrac32\sqrt{x}-\tfrac12\sqrt{20}}$. Hence $$\MU_x=\begin{cases}\dfrac{1}{\sqrt{x}}, & x\lt 20,\\ \dfrac{3}{4\sqrt{x}}, & x\gt 20.\end{cases}$$

10. Write $\MU_y$ for $y\lt 80$ and for $y\gt 80$.

The same argument, with the reference level $80$: $$\MU_y=\begin{cases}\dfrac{1}{\sqrt{y}}, & y\lt 80,\\ \dfrac{3}{4\sqrt{y}}, & y\gt 80.\end{cases}$$

11. Compute $\mathrm{MRS}^{+}$, the $\mathrm{MRS}$ at $(20,80)$ using the expression of the region ${(x\gt 20 \text{ and } y\lt 80)}$. Then compute $\mathrm{MRS}^{-}$, using the expression of the region ${(x\lt 20 \text{ and } y\gt 80)}$.

With $X$ in gains and $Y$ in losses, $$\begin{aligned}\mathrm{MRS}^{+}&=\frac{3/(4\sqrt{x})}{1/\sqrt{y}}=\frac34\sqrt{\frac{y}{x}}\\ &=\frac34\cdot 2=\frac32.\end{aligned}$$ With $X$ in losses and $Y$ in gains, $$\begin{aligned}\mathrm{MRS}^{-}&=\frac{1/\sqrt{x}}{3/(4\sqrt{y})}=\frac43\sqrt{\frac{y}{x}}\\ &=\frac43\cdot 2=\frac83\approx 2.67.\end{aligned}$$

12. With $p_X=2$, does the price ratio lie between them? What does that say about $A$?

Yes: $\mathrm{MRS}^{+}=\tfrac32\le 2\le\tfrac83=\mathrm{MRS}^{-}$. Since $A$ is on the budget line and the price ratio lies between the two one-sided values, $A$ is Consumer $R$'s optimal basket, at the kink. Shifting a dollar from $Y$ to $X$ adds $\tfrac{3}{4\sqrt{20}}\cdot\tfrac12\approx 0.084$ but costs $\tfrac{1}{\sqrt{80}}\approx 0.112$; shifting it from $X$ to $Y$ adds $\tfrac{3}{4\sqrt{80}}\approx 0.084$ but costs $\tfrac{1}{\sqrt{20}}\cdot\tfrac12\approx 0.112$. Either way utility falls.

Consumer $R$'s indifference curve through $A$ kinks where it crosses the reference lines $x=\bar x$ and $y=\bar y$ (dash-dotted). The orange rays leave $A$ with slopes $-\mathrm{MRS}^{-}=-\tfrac83$ to the left and $-\mathrm{MRS}^{+}=-\tfrac32$ to the right. The budget line, with slope $-2$, runs between them, so it touches the indifference curve only at $A$.

The New Optimal Basket

Problem Set 4, Questions 13–16

13. Now $p_X=3$. Does it still lie between them? Which good must the consumer cut, and why?

No: $3\gt\tfrac83=\mathrm{MRS}^{-}$, so the price ratio lies above both one-sided values. Consumer $R$ must cut $X$. Since $3\cdot 20+80=140\gt 120$, $A$ is no longer affordable and something has to go. A dollar saved by cutting $X$ costs $\tfrac{1}{\sqrt{20}}\cdot\tfrac13\approx 0.075$ in utility, while a dollar saved by cutting $Y$ costs $\tfrac{1}{\sqrt{80}}\approx 0.112$. Keeping $X$ at $20$ and cutting $Y$ instead leads to ${C=(20,60)}$, which lies on a lower indifference curve than $B$ (see the figure below).

14. Suppose $Y$ stays pinned at its reference level $\bar y=80$. Solve the new budget constraint $3x+\bar y=120$ for $x$, and call that value $x_1$.

$3x+80=120$ gives $x_1=\tfrac{40}{3}\approx 13.3$.

15. Let ${B=(x_1,80)}$. Compute $\mathrm{MRS}^{+}$ using the expression of the region ${(x\lt 20 \text{ and } y\lt 80)}$, and $\mathrm{MRS}^{-}$ using the expression of the region ${(x\lt 20 \text{ and } y\gt 80)}$.

At $B$, $\sqrt{y/x}=\sqrt{80/(40/3)}=\sqrt{6}$. With both goods in losses, $$\mathrm{MRS}^{+}=\frac{1/\sqrt{x}}{1/\sqrt{y}}=\sqrt{\frac{y}{x}}=\sqrt{6}\approx 2.45.$$ With $X$ in losses and $Y$ in gains, $$\begin{aligned}\mathrm{MRS}^{-}&=\frac{1/\sqrt{x}}{3/(4\sqrt{y})}=\frac43\sqrt{\frac{y}{x}}\\ &=\frac43\sqrt{6}\approx 3.27.\end{aligned}$$

16. Does $p_X/p_Y=3$ lie between them? Since ${B=(x_1,80)}$ is on the budget line, what can we conclude? Is $B$ the optimal basket or not? Why?

Yes, $\sqrt{6}\approx 2.45\le 3\le 3.27\approx\tfrac43\sqrt{6}$, so $B$, which is on the budget line, is the optimal basket: no shift of a dollar between the goods raises utility. Consumer $R$ absorbs the whole price rise by cutting $X$ from $20$ to $\tfrac{40}{3}$ and keeps $Y$ at its reference level, whereas Consumer $T$ moves to $(10,90)$.

The price rise for Consumer $R$: budget lines at $p_X=2$ (solid) and $p_X=3$ (dashed), with the reference lines dash-dotted. The indifference curve through $B$ (light) kinks at $B$, where it crosses $y=\bar y$, and again near $(20,66)$, where it crosses $x=\bar x$. The hollow point ${C=(20,60)}$ keeps $X$ at its reference level but lies below that curve, so Consumer $R$ prefers $B$.