The demonetization of gold: transactions and the change in control

Abstract

Three models of a monetary economy are considered, in order to show the effects of a gold demonetization: the first with a gold money, the second with demonetized gold but no central bank, and the third with demonetized gold, but with a central bank. The distinctions between ownership and control are discussed. Our results show a gain in efficiency (in the case of “enough money”) when a switch is made from a durable commodity money to a fiat money. This is due to players being able to enjoy both the full service value of gold and transactions value of money—something that cannot be done in the original model with gold money. When we further add in the central bank, there is a somewhat further efficiency gain in the case of “not enough money”. We close the paper with a discussion of the usefulness of central banks.

Appendices

Appendix 1: A buy–sell model of competitive money lending with gold

In this model there are the usual two continua of traders, plus a continuum of moneylenders. The trader types are endowed with $(a,0,m)$ and $(0,a,m)$ respectively, while the lenders have $(0,0,B)$. However, this time the money is gold. The gold can only be used for transactions or for jewelry (services) during the period, but not both. In the language of Quint–Shubik (2013, Chapter 6), the parameter values $(k_{1},k_{2},k_{3})$ are set equal to $(0,0,1)$.

The optimization for the traders of type 1 is

$$\begin{aligned}&\left. \max _{b,q,d}2\sqrt{(a-q)\frac{b}{\bar{p}}}+\left( m+\frac{d}{1+\rho }-b\right) +\Pi \left( m+\frac{d}{1+\rho }-b+pq-d\right) \right. \qquad \quad \,\, \end{aligned}$$

(41)

$$\begin{aligned}&\left. \text {s.t. }m+\frac{d}{1+\rho }-b\ge 0\qquad \qquad (\lambda ) \text { (cash flow constraint) }\right. \end{aligned}$$

(42)

$$\begin{aligned}&\qquad \left. m+\frac{d}{1+\rho }-b+pq-d\ge 0 \qquad \qquad (\mu ) \text { (budget constraint) }\right. \end{aligned}$$

(43)

$$\begin{aligned}&\qquad \left. b,d\ge 0,\quad 0\le q\le a\right. \end{aligned}$$

(44)

The first order conditions here are

$$\begin{aligned}&\left. \frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{a-q}{b}}=1+\Pi +\lambda +\mu \right. \end{aligned}$$

(45)

$$\begin{aligned}&\left. \frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{b}{a-q}}=(\mu +\Pi )p\right. \end{aligned}$$

(46)

$$\begin{aligned}&\left. \lambda =(\mu +\Pi )\rho -1\right. \end{aligned}$$

(47)

$$\begin{aligned}&\left. m+\frac{d}{1+\rho }-b=0\quad \text {or}\quad \lambda =0\right. \end{aligned}$$

(48)

$$\begin{aligned}&\left. m+\frac{d}{1+\rho }-b+pq-d=0\quad \text {or}\quad \mu =0\right. \end{aligned}$$

(49)

Similarly, the optimization for the Type 2 traders is

$$\begin{aligned}&\left. \max _{\bar{b},\bar{q},\bar{d}}2\sqrt{(a-\bar{q})\frac{\bar{b }}{p}}+\left( m+\frac{\bar{d}}{1+\rho }-\bar{b}\right) +\Pi \left( m+\frac{ \bar{d}}{1+\rho }-\bar{b}+\bar{p}\bar{q}-\bar{d}\right) \right. \\&\left. \text {s.t. }m+\frac{\bar{d}}{1+\rho }-\bar{b}\ge 0\qquad \qquad \qquad \qquad (\overline{\lambda })\right. \\&\qquad \left. m+\frac{\bar{d}}{1+\rho }-\bar{b}+\bar{p}\bar{q}- \bar{d}\ge 0 \qquad \qquad (\overline{\mu })\right. \\&\qquad \left. \bar{b},\bar{d}\ge 0,\quad 0\le \bar{q}\le a.\right. \end{aligned}$$

with first order conditions

$$\begin{aligned}&\left. \frac{1}{\sqrt{p}}\sqrt{\frac{a-\bar{q}}{\bar{b}}}=1+\Pi +\bar{\lambda }+\bar{\mu }\right. \end{aligned}$$

(50)

$$\begin{aligned}&\left. \frac{1}{\sqrt{p}}\sqrt{\frac{\bar{b}}{a-\bar{q}}}=(\bar{\mu }+\Pi ) \bar{p}\right. \end{aligned}$$

(51)

$$\begin{aligned}&\left. \overline{\lambda }=(\bar{\mu }+\Pi )\rho -1\right. \end{aligned}$$

(52)

$$\begin{aligned}&\left. m+\frac{\bar{d}}{1+\rho }-\bar{b}=0\quad \text {or}\quad \bar{\lambda }=0\right. \end{aligned}$$

(53)

$$\begin{aligned}&\left. m+\frac{\bar{d}}{1+\rho }-\bar{b}+\bar{p}\bar{q}-\bar{d}=0\quad \text {or}\quad \bar{\mu }=0.\right. \end{aligned}$$

(54)

For the continuum of moneylenders, the optimization is

$$\begin{aligned}&\left. \max _{_{b_{1}^{*},b_{2}^{*},g}}2\sqrt{\frac{b_{1}^{*}}{p}\frac{ b_{2}^{*}}{\overline{p}}}+B-b_{1}^{*}-b_{2}^{*}-g+\Pi \left( B-b_{1}^{*}-b_{2}^{*}+\rho g\right) \right. \end{aligned}$$

(55)

$$\begin{aligned}&\left. \text {s.t. }B-b_{1}^{*}-b_{2}^{*}-g\ge 0\qquad \qquad (\lambda ^{*})\right. \end{aligned}$$

(56)

$$\begin{aligned}&\qquad \left. b_{1}^{*},b_{2}^{*},g\ge 0\right. \end{aligned}$$

(57)

Note here that the money lenders have $g$ as a decision variable, and not $\rho$. The reason is that individually, each lender can decide how much to lend—but they cannot individually influence the interest rate. The first order conditions here are:

$$\begin{aligned}&\displaystyle \left. \sqrt{\frac{b_{2}^{*}}{p\overline{p}b_{1}^{*}}}=\Pi +\lambda ^{*}+1\right. \end{aligned}$$

(58)

$$\begin{aligned}&\displaystyle \left. \sqrt{\frac{b_{1}^{*}}{p\overline{p}b_{2}^{*}}}=\Pi +\lambda ^{*}+1\right. \end{aligned}$$

(59)

$$\begin{aligned}&\displaystyle \left. \Pi \rho -\lambda ^{*}-1=0\right. \end{aligned}$$

(60)

$$\begin{aligned}&\displaystyle \left. B-b_{1}^{*}-b_{2}^{*}-g=0\quad \text {or}\quad \lambda ^{*}=0\right. \end{aligned}$$

(61)

Finally, we have the following balance conditions: $p=\frac{\overline{b} +b_{1}^{*}}{q}, \overline{p}=\frac{b+b_{2}^{*}}{\overline{q}}$, and $1+\rho =\frac{d+\overline{d}}{g}$.

Case 1: We first analyze the case where the traders have little gold and the lenders have a lot. Thus the traders’ constraints are all tight and the lenders’ are loose, i.e. $\lambda >0, \mu >0, \overline{\lambda }>0, \overline{\mu }>0$, and $\lambda ^{*}=0$. We also assume a symmetric solution, i.e. $p=\overline{p}, b=\overline{b}, d=\overline{d}, q=\overline{q}$, and $b_{1}^{*}=b_{2}^{*}$.

Condition (60) gives $\rho =\frac{1}{\Pi }$. And (58) together with symmetry gives $\frac{1}{p}=\Pi +1$, which is $p=\frac{1}{\Pi +1}= \overline{p}$.

Next, we find an expression for the multiplier $\mu$. We begin with (46), which is $\sqrt{\frac{b}{a-q}}=(\mu +\Pi )p^{\frac{3}{2}}$, or $\sqrt{\frac{a-q}{b}}=\frac{1}{(\mu +\Pi )p^{\frac{3}{2}}}$. Substituting into (45), we have $\frac{1}{\sqrt{\overline{p}}}\frac{1}{(\mu +\Pi )p^{\frac{3}{ 2}}}=1+\Pi +\lambda +\mu =1+\Pi +(\mu +\Pi )\rho -1+\mu =(1+\rho )(\mu +\Pi )$. But $p=\frac{1}{\Pi +1}$; so we have $(\Pi +1)^{2}=(1+\rho )(\mu +\Pi )^{2}$, which is $(\Pi +1)=\sqrt{1+\rho }(\mu +\Pi )=\sqrt{1+\frac{1}{\Pi }} (\mu +\Pi )$. Solving for $\mu$ yields $\mu =\sqrt{\Pi (1+\Pi )}-\Pi$.

Next, we again start with (46), which is $\sqrt{\frac{b}{a-q}}=(\mu +\Pi )p^{\frac{3}{2}}$. Squaring both sides, we have $\frac{b}{a-q}=(\mu +\Pi )^{2}p^{3}$. Now substitute in $p=\frac{1}{\Pi +1}$ and $\mu =\sqrt{\Pi (1+\Pi )}-\Pi$. We end up with $b=(a-q)\frac{\Pi }{(1+\Pi )^{2}}$.

Next, since the traders’ cash flow and budget constraints are both tight, we know $d=pq=\frac{q}{1+\Pi }$. But the tight cash flow constraint also means $\small d=(b-m)(1+\rho )=(b-m)(1+\frac{1}{\Pi })=\left( (a-q)\frac{\Pi }{(1+\Pi )^{2}}-m\right) (1+\frac{1}{\Pi })$. Hence $\small \left( (a-q)\frac{\Pi }{(1+\Pi )^{2}} -m\right) (1+\frac{1}{\Pi })=$.$\small \frac{q}{1+\Pi }$. Solving for $q$ yields

$$\begin{aligned} q=\frac{a}{2}-\frac{m(1+\Pi )^{2}}{2\Pi }=\overline{q}. \end{aligned}$$

(62)

At this point it becomes easy to solve for the other variables: $d=pq=\frac{q }{1+\Pi }=\frac{a}{2(1+\Pi )}-\frac{m(1+\Pi )}{2\Pi }, b=(a-q)\frac{\Pi }{ (1+\Pi )^{2}}=\frac{a\Pi }{2(1+\Pi )^{2}}+\frac{m}{2}$, and $b_{1}^{*}=b_{2}^{*}=pq-b=d-b=\frac{a}{2(1+\Pi )^{2}}-\frac{m}{2\Pi }-m$. In addition, $g=\frac{2d}{1+\rho }=\frac{a\Pi }{(1+\Pi )^{2}}-m$. For the multipliers, we already found $\mu =\sqrt{\Pi (1+\Pi )}-\Pi =\overline{\mu }$ and we are given $\lambda ^{*}=0$. Finally, we have $\lambda =\rho (\Pi +\mu )-1=\frac{1}{\Pi }(\Pi +\sqrt{\Pi (1+\Pi )}-\Pi )-1=\sqrt{\frac{1+\Pi }{\Pi }}-1$.

The results hold if $B-b_{1}^{*}-b_{2}^{*}-g\ge 0$, i.e. if $B-\frac{a}{(1+\Pi )^{2}}+\frac{m}{\Pi }+2m-\frac{a\Pi }{(1+\Pi )^{2}}+m\ge 0$. This yields a condition of $B\ge \frac{a}{1+\Pi }-\frac{m(1+\Pi )}{\Pi }$. In addition, we need the variables $q, d$, and $b_{1}^{*}=b_{2}^{*}$ to be nonnegative. This occurs if $m\le \frac{\Pi a}{(1+\Pi )^{2}}$.

Case 2: We analyze the case where $B$ is smaller. Hence the traders’ cash flow and budget constraints are tight as before, but the lenders’ cash flow constraints are tight, i.e. $\lambda >0, \mu >0$, and $\lambda ^{*}>0$. Again assume a symmetric solution, i.e. $p=\overline{p}, b=\overline{b}, d=\overline{d}, q=\overline{q}$, and $b_{1}^{*}=b_{2}^{*}$.

First, (60) implies $\lambda ^{*}=\Pi \rho -1$. But also, from (58) and symmetry, we have $\lambda ^{*}=\frac{1}{p}-\Pi -1$. Hence $\Pi \rho =\frac{1}{p}-\Pi$, which is

$$\begin{aligned} p=\frac{1}{\Pi (1+\rho )}=\overline{p}. \end{aligned}$$

(63)

Next, since the traders’ cash flow and budget constraints are tight, we have $d=pq$. But by the balance condition $pq$ is equal to $\overline{b} +b_{1}^{*}$, and since the lenders’ cash flow constraints are tight $b_{1}^{*}=b_{2}^{*}=\frac{B-g}{2}$. So we have $d=pq=\overline{b}+b_{1}^{*}= \overline{b}+\frac{B-g}{2}=\overline{b}+\frac{B-\frac{2d}{1+\rho }}{2}=\frac{B}{2}+\overline{b}-\frac{d}{1+\rho }=$.$\frac{B}{2}+m$, where the fourth equality follows from the balance constraint, the fifth is just algebra, and the last follows from symmetry $b=\overline{b}$ and the tight traders’ cash flow constraint. So we have

$$\begin{aligned} d=\frac{B}{2}+m=\overline{d}. \end{aligned}$$

(64)

At this point, we can obtain expressions for all of the variables in terms of $\rho$: $g=\frac{2d}{1+\rho }=\frac{B+2m}{1+\rho }, b=\frac{d}{1+\rho } +m=\frac{\frac{B}{2}+m}{1+\rho }+m=\overline{b}, b_{1}^{*}=b_{2}^{*}=\frac{1}{2}(B-g)=\frac{1}{2}(\frac{B+\rho B-B-2m}{1+\rho })=\frac{1}{2}(\frac{\rho B-2m}{1+\rho })$, Recalling $p=\frac{1}{\Pi (1+\rho )}$, we also have $q= \overline{q}=\frac{d}{p}=\frac{\frac{B}{2}+m}{\frac{1}{\Pi (1+\rho )}}=\Pi (1+\rho )(\frac{B}{2}+m)$.

So now all that remains to find an expression for $\rho$. To this end, we observe that (47) implies that

$$\begin{aligned} \lambda =\rho (\Pi +\mu )-1. \end{aligned}$$

(65)

Then

$$\begin{aligned} (45)&\Rightarrow \frac{1}{\sqrt{\bar{p}}}\sqrt{\frac{a-q}{b}}=1+\Pi +\lambda +\mu \end{aligned}$$

(66)

$$\begin{aligned}&\Rightarrow \frac{1}{\bar{p}}\left( \frac{a-q}{b}\right) =(1+\Pi +\lambda +\mu )^{2} \end{aligned}$$

(67)

$$\begin{aligned}&\Rightarrow \frac{1}{\bar{p}}\left( \frac{a-q}{b}\right) =(1+\rho )^{2}(\Pi +\mu )^{2}\quad \text {(using (65))} \end{aligned}$$

(68)

$$\begin{aligned}&\Rightarrow \frac{1}{\bar{p}}\left( \frac{a-q}{b}\right) =(1+\rho )^{2} \frac{1}{\overline{p}^{3}}\left( \frac{b}{a-q}\right) \quad \text {(using (46))} \end{aligned}$$

(69)

$$\begin{aligned}&\Rightarrow \left( \frac{a-q}{b}\right) ^{2}=\frac{(1+\rho )^{2}}{ \overline{p}^{2}}=\Pi ^{2}(1+\rho )^{4}\quad \text {(using 63)} \end{aligned}$$

(70)

$$\begin{aligned}&\Rightarrow \frac{a-q}{b}=\Pi (1+\rho )^{2} \end{aligned}$$

(71)

$$\begin{aligned}&\Rightarrow \frac{a-\Pi (1+\rho )\left( \frac{B}{2}+m\right) }{\frac{\frac{B}{2}+m}{ 1+\rho }+m}=\Pi (1+\rho )^{2} \end{aligned}$$

(72)

$$\begin{aligned}&\Rightarrow a-\Pi (1+\rho )\left( \frac{B}{2}+m\right) =\Pi (1+\rho )\left( \frac{B}{2} +m\right) +\Pi m(1+\rho )^{2} \end{aligned}$$

(73)

$$\begin{aligned}&\Rightarrow \Pi m(1+\rho )^{2}+2\Pi (1+\rho )\left( \frac{B}{2}+m\right) -a=0 \end{aligned}$$

(74)

The above quadratic expression for $1+\rho$ solves in two cases: first, if $m=0$, we have $1+\rho =\frac{a}{\Pi (B+2m)}$. Otherwise, if $m>0$, we have

$$\begin{aligned} 1+\rho =\frac{-\Pi (B+2m)+\sqrt{\Pi ^{2}(B+2m)^{2}+4a\Pi m}}{2\Pi m} \end{aligned}$$

(75)

Finally, we may solve for the multipliers. First, we have (46) $\Rightarrow \mu =\frac{\frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{b}{a-q}}-\Pi p}{p}=\frac{1}{p^{1.5}}\sqrt{\frac{1}{\Pi (1+\rho )^{2}}}-\Pi =\Pi ^{1.5}(1+\rho )^{1.5}\sqrt{\frac{1}{\Pi (1+\rho )^{2}}}=\Pi \sqrt{1+\rho }-\Pi$, where the second equality follows from (71) and the third from (63). Also, $\lambda =\rho (\Pi +\mu )-1=\Pi \rho \sqrt{1+\rho } -1$. Finally, condition (60) directly gives us $\lambda ^{*}=\Pi \rho -1$.

Note: In order for the above to be valid, we need $\rho$ and all the multipliers to be nonnegative. This requires $\rho \ge \frac{1}{\Pi }$. In the case where $m=0$, this means $B+2m\le \frac{a}{1+\Pi }$. In the case where $m>0$, the condition is not so easily stated, but does hold whenever $a$ is large compared to $B+2m$. We also need for $b_{1}^{*}$ and $b_{1}^{*}$ to be nonnegative, which means $\rho B\ge 2m$. Again, if $m=0$, this holds if $\rho \ge \frac{1}{\Pi }$, i.e. if $B+2m\le \frac{a}{1+\Pi }$. And again, if $m>0$ the condition is more complicated but holds if $a\gg B+2m$.

Appendix 2: A sell-all model of competitive money lending with gold

Both trader types are endowed with $m$ units of gold money. The continuum of money lenders is endowed with $B=M-2m$ units of money (where $0\le 2m\le M$). The lenders are utility maximizers, who (just like the traders) derive benefit from consumption of perishables, the service utility of the gold, and the salvage utility of the gold at the end of the game.

The notation we will use is the following. For the Type 1 traders, $b_{1}$ and $b_{2}$ denote the amount of gold bid in the market for Good 1 and Good 2 respectively. The amount they borrow from the banks is $\frac{d}{1+\rho }$ at interest rate $\rho$, so the amount they must pay back is $d$. For the Type 2 traders, the notation $\bar{b}_{1}, \bar{b}_{2}$, and $\bar{d}$ is defined similarly. The prices for the perishables are $p$ and $\overline{p}$ , while the salvage value for the gold is $\Pi$ per unit. We assume that the service utility of the gold is one per unit-time,

The Type 1 traders face an optimization described by:

$$\begin{aligned}&\max _{b_{1},b_{2},d}2\sqrt{\frac{b_{1}}{p}\frac{b_{2}}{\bar{p}}}+m+ \frac{d}{1+\rho }-b_{1}-b_{2}\nonumber \\&\quad +pa+\Pi \left( m+\frac{d}{1+\rho } -b_{1}-b_{2}+pa-d\right) \end{aligned}$$

(76)

$$\begin{aligned}&\left. \text {s.t. }m+\frac{d}{1+\rho }-b_{1}-b_{2}\ge 0 \qquad \qquad (\lambda )\text { (cash flow constraint) }\right. \end{aligned}$$

(77)

$$\begin{aligned}&\qquad \left. m+\frac{d}{1+\rho }-b_{1}-b_{2}+pa-d\ge 0\qquad \qquad (\mu )\text { (budget constraint) }\qquad \quad \right. \end{aligned}$$

(78)

$$\begin{aligned}&\quad \left. b_{1},b_{2},d\ge 0\right. \end{aligned}$$

(79)

The first order conditions wrt $b_{1}, b_{2}$, and $d$ yield

$$\begin{aligned}&\displaystyle \left. \sqrt{\frac{b_{2}}{p\overline{p}b_{1}}}=1+\Pi +\lambda +\mu \right.&\end{aligned}$$

(80)

$$\begin{aligned}&\displaystyle \left. \sqrt{\frac{b_{1}}{p\overline{p}b_{2}}}=1+\Pi +\lambda +\mu \right.&\end{aligned}$$

(81)

$$\begin{aligned}&\displaystyle \left. \frac{1}{1+\rho }+\frac{\Pi }{1+\rho }-\Pi +\frac{\lambda }{1+\rho } +\frac{\mu }{1+\rho }-\mu =0\right.&\end{aligned}$$

(82)

$$\begin{aligned}&\displaystyle \left. m+\frac{d}{1+\rho }-b_{1}-b_{2}=0\quad \text {or}\quad \lambda =0\right.&\end{aligned}$$

(83)

$$\begin{aligned}&\displaystyle \left. m+\frac{d}{1+\rho }-b_{1}-b_{2}+pa-d=0\quad \text {or}\quad \mu =0\right.&\end{aligned}$$

(84)

We also point out that we require the feasibility conditions (77)–(79) for our solution.²⁵

Similarly, the Type 2 traders face the optimization below:

$$\begin{aligned}&\max _{\bar{b}_{1},\overline{b}_{2},\bar{d}}2\sqrt{\frac{\overline{b} _{1}}{p}\frac{\overline{b}_{2}}{\bar{p}}}+m+\frac{\bar{d}}{1+\rho }- \overline{b}_{1}-\overline{b}_{2}+\bar{p}a\nonumber \\&\quad +\Pi \left( m+\frac{\bar{d}}{ 1+\rho }-\overline{b}_{1}-\overline{b}_{2}+\bar{p}a-\bar{d}\right) \end{aligned}$$

(85)

$$\begin{aligned}&\left. \text {s.t. }m+\frac{\bar{d}}{1+\rho }-\overline{b}_{1}-\overline{b} _{2}\ge 0\qquad \qquad (\bar{\lambda })\;\text { (cash flow constraint)}\right. \end{aligned}$$

(86)

$$\begin{aligned}&\qquad \left. m+\frac{\bar{d}}{1+\rho }-\overline{b}_{1}-\overline{ b}_{2}+\bar{p}a-\bar{d}\ge 0 \qquad (\bar{\mu })\;\text {(budget constraint)}\qquad \right. \end{aligned}$$

(87)

$$\begin{aligned}&\qquad \left. \bar{b}_{1},\overline{b}_{2},\overline{d}\ge 0\right. \end{aligned}$$

(88)

The first order conditions here are

$$\begin{aligned}&\sqrt{\frac{\overline{b}_{2}}{p\overline{p}\overline{b}_{1}}}=1+\Pi +\bar{\lambda }+\bar{\mu } \end{aligned}$$

(89)

$$\begin{aligned}&\sqrt{\frac{\overline{b}_{1}}{p\overline{p}\overline{b}_{2}}}=1+\Pi +\bar{\lambda }+\bar{\mu } \end{aligned}$$

(90)

$$\begin{aligned} \frac{1}{1+\rho }&+\frac{\Pi }{1+\rho }-\Pi +\frac{\bar{\lambda }}{1+\rho }+ \frac{\bar{\mu }}{1+\rho }-\bar{\mu }=0 \end{aligned}$$

(91)

$$\begin{aligned}&m+\frac{\bar{d}}{1+\rho }-\overline{b}_{1}-\overline{b}_{2}=0\quad \text {or}\quad \bar{\lambda }=0 \end{aligned}$$

(92)

$$\begin{aligned}&m+\frac{\bar{d}}{1+\rho }-\overline{b}_{1}-\overline{b}_{2}+\bar{p}a-\bar{d}=0\quad \text {or}\quad \bar{\mu }=0 \end{aligned}$$

(93)

The money lender decision variables are $b_{1}^{*}$ (the amount they collectively bid for Good 1), $b_{2}^{*}$ (the amount they collectively bid for Good 2), and $g$ (the amount they offer for loan). Their optimization is expressed as:

$$\begin{aligned}&\left. \max _{b_{1}^{*}\text {,}b_{2}^{*}\text {,}g}2\sqrt{\frac{b_{1}^{*}}{p} \frac{b_{2}^{*}}{\bar{p}}}+B-g-b_{1}^{*}-b_{2}^{*}+\Pi (B-g-b_{1}^{*}-b_{2}^{*}+(1+\rho )g)\right. \qquad \quad \end{aligned}$$

(94)

$$\begin{aligned}&\left. \text {s.t. }B-g-b_{1}^{*}-b_{2}^{*}\ge 0\qquad \qquad (\lambda ^{*})\right. \end{aligned}$$

(95)

$$\begin{aligned}&\qquad \left. b_{1}^{*}, b_{2}^{*}, g\ge 0\right. \end{aligned}$$

(96)

The first order equations here are

$$\begin{aligned} \sqrt{\frac{b_{2}^{*}}{p\overline{p}b_{1}^{*}}}&= \Pi +\lambda ^{*}+1 \end{aligned}$$

(97)

$$\begin{aligned} \sqrt{\frac{b_{1}^{*}}{p\overline{p}b_{2}^{*}}}&= \Pi +\lambda ^{*}+1 \end{aligned}$$

(98)

$$\begin{aligned} \Pi \rho -\lambda ^{*}-1&= 0 \end{aligned}$$

(99)

$$\begin{aligned} B-b_{1}^{*}-b_{2}^{*}-g&= 0\quad \text {or}\quad \lambda ^{*}=0 \end{aligned}$$

(100)

Finally, the balance conditions are $p=\frac{b_{1}+\bar{b}_{1}+b_{1}^{*}}{a}$ , $\bar{p}=\frac{b_{2}+\bar{b}_{2}+b_{2}^{*}}{a}$ and $1+\rho =\frac{d+\bar{d }}{g}$. We also remark that here the problems for Types 1 and 2 are isomorphic and so we may assume a symmetric solution where $b_{1}=b_{2}=\bar{ b}_{1}=\bar{b}_{2}\equiv b, d=\bar{d}, b_{1}^{*}=b_{2}^{*}\equiv b^{*}$, and $p=\bar{p}$.

Case 1: $m$ low and $B$ high.

There are several cases, but as before we first consider for the case in which the traders have little money ($m$ small) and the bankers have a lot of money ($B$ large). In terms of our multipliers, we are assuming $\lambda$ , $\bar{\lambda }, \mu$, and $\bar{\mu }$ are positive, while $\lambda ^{*}=0$. The first observation is that condition (97) or (98), $\lambda ^{*}=0$, and symmetry together imply that $p=\overline{p}=\frac{1}{ 1+\Pi }$. Next, conditions $(\lambda )$ and $(\mu )$ holding tightly together imply that $pa=d$. Hence $d=\frac{a}{1+\Pi }=\bar{d}$. The cash flow constraint holding tightly is $m+\frac{d}{1+\rho }-2b=0$, which gives $b=\frac{m+\frac{d}{1+\rho }}{2}=\frac{m(1+\rho )(1+\Pi )+a}{2(1+\rho )(1+\Pi )}$; but then condition (99) and $\lambda ^{*}=0$ together give $\rho =\frac{1}{\Pi }$, so $b=\frac{m(1+\frac{1}{\Pi })(1+\Pi )+a}{2(1+\frac{1}{ \Pi })(1+\Pi )}=\frac{m(1+\Pi )^{2}+\Pi a}{2(1+\Pi )^{2}}$. So the traders consume $\frac{b}{p}=\frac{m(1+\Pi )^{2}+\Pi a}{2(1+\Pi )}$ of each good. Then, we can calculate $b^{*}$ via the balance condition for price: $b^{*}=pa-2b=\frac{a}{1+\Pi }-\frac{m(1+\Pi )^{2}+\Pi a}{(1+\Pi )^{2}}=\frac{a-m(1+\Pi )^{2}}{(1+\Pi )^{2}}$. So the lenders consume $\frac{b^{*}}{p}= \frac{a-m(1+\Pi )^{2}}{(1+\Pi )}$ of each good. They also lend an amount of $g=\frac{2d}{1+\rho }=\frac{2\Pi a}{(1+\Pi )^{2}}$.

Finally, for the multipliers, condition (80) with $p=\bar{p}=\frac{1}{ 1+\Pi }$ and $b_{1}=b_{2}$ gives $1+\Pi =1+\Pi +\lambda +\mu$; since $\lambda$ and $\mu$ are nonnegative, this gives $\lambda =\mu =0$.

We remark that the above results are only valid if: (a) the above expression for $b^{*}$ is nonnegative, and if (b) the lenders’ cash flow constraints $(\lambda ^{*})$ hold. These gives the conditions: (a) $m\le \frac{a}{(1+\Pi )^{2}}$ and (b) $B\ge \frac{2\Pi a}{(1+\Pi )^{2}}+2\frac{a-m(1+\Pi )^{2}}{(1+\Pi )^{2}}=\frac{2a}{1+\Pi }-2m$, which is $B+2m\ge \frac{2a}{1+\Pi }$.

Case 2: $m$ low and $B$ low.

Now we cover the case in which $\lambda , \bar{\lambda }, \mu$, and $\bar{\mu }$ are again all positive, but $\lambda ^{*}>0$ also. We first note that condition (97) and symmetry together imply $\lambda ^{*}=\frac{1}{p}-1-\Pi$; substituting this expression for $\lambda ^{*}$ back into (99) gives $\Pi \rho -(\frac{1 }{p}-1-\Pi )-1=0$, which is

$$\begin{aligned} p=\frac{1}{\Pi (1+\rho )}=\overline{p}. \end{aligned}$$

(101)

Next, we note from symmetry, the balance conditions, and the tight $(\lambda ^{*})$ constraint that $d=\frac{(1+\rho )g}{2}, 2b+b^{*}=pa$, and $b^{*}= \frac{B-g}{2}$. Hence, starting with the tight cash flow constraint for the traders, we have $m+\frac{d}{1+\rho }-b-b=0\Rightarrow m+\frac{g}{2} -2b=0\Rightarrow m+\frac{g}{2}-pa+b^{*}=0\Rightarrow m+\frac{g}{2}-pa+\frac{ B-g}{2}=0\Rightarrow p=\frac{B+2m}{2a}=\overline{p}$. Not only does this give an expression for $p$, but this in combination with (101) gives $\frac{1}{\Pi (1+\rho )}=\frac{B+2m}{2a}$, or $\rho =\frac{2a}{\Pi (B+2m)}-1$.

It is now a simple exercise to calculate expressions for the other variables: $d=pa=\frac{B+2m}{2}=\overline{d}, g=\frac{2d}{1+\rho }=\frac{ \Pi (B+2m)^{2}}{2a}, b=b_{1}=b_{2}=\frac{m+\frac{g}{2}}{2}=\frac{4am+\Pi (B+2m)^{2}}{8a}=\overline{b}_{1}=\overline{b}_{2}, b^{*}=b_{1}^{*}=b_{2}^{*}=\frac{B-g}{2}=\frac{2aB-\Pi (B+2m)^{2}}{4a}, \lambda ^{*}=\frac{1}{p}-1-\Pi =\frac{2a}{B+2m}-\Pi -1, \mu =\frac{1}{p}-1-\Pi -\lambda =\frac{1}{p}-1-\Pi -(\Pi +\mu )\rho +1\Rightarrow \mu =0$, and $\lambda =\frac{2a}{B+2m}-\Pi -1$.

In order for these results to hold, we need $b^{*}\ge 0, \lambda ^{*}\ge 0$, and $\lambda \ge 0$. These hold if $\Pi (B+2m)^{2}\le 2aB$ and $(1+\Pi )B\le 2a$. If $m=0$, all that is needed is $(1+\Pi )B\le 2a$.

Appendix 3: Demonetized gold, no central bank, lenders also act as gold merchants

This model is similar to “Appendix 1”, except now the gold has been demonetized. All holders of $x$ units of gold are now endowed with $x$ units of demonetized gold, plus $x$ units of fiat (“gold strips”). Hence Trader Type 1 is now endowed with $a$ units of good #1, plus $m$ units of gold, plus $m$ units of fiat. Trader Type 2 is endowed with $a$ units of good #1, $m$ units of gold, and $m$ units of fiat. The lenders (now lender-merchants) are endowed with only $B$ units of gold and $B$ units of fiat.

We assume that the financial market (for fiat) operates first, over the time period $[0,k_{1}]$. This is followed by the goods markets for gold and for perishable, both of which operate over $[k_{2},k_{3}]$ ($0\le k_{1}\le k_{2}\le k_{3}$). However, while any money borrowed during the financial market is not obtained until the end of the market at time $k_{1}$, any gold or perishable bought during the second phase is credited at the beginning of that phase, at time $k_{2}$. This is an important assumption for the trade of gold, as the final owners of gold get to enjoy its service value over the period $[k_{2},k_{3}]$. As in the previous model, we assume $k_{1}=0, k_{2}=0$, and $k_{3}=1$.

With the new market for demonetized gold, we have to define some new variables. First, for the traders, the amount they bid for gold is $b_{3}$. The amount of gold they put up for sale is $q_{3}$. For the lenders, the amount they bid for gold is denoted $b_{3}^{*}$, while the amount they put up for sale is $q_{3}^{*}$. The price of gold is given by $p_{3}$ and the parameters $\Pi _{1}$ and $\Pi _{2}$ are the per unit salvage value of strips (fiat) and gold respectively, at the end of the game.

The traders of Type 1 attempt to solve

$$\begin{aligned}&\max _{b,q,d,b_{3},q_{3}}2\sqrt{(a-q)\frac{b}{\bar{p}}}+\left( m+ \frac{b_{3}}{p_{3}}-q_{3}\right) \nonumber \\&\quad +\Pi _{1}\left( m+\frac{d}{1+\rho } -b-b_{3}+pq-d+p_{3}q_{3}\right) \end{aligned}$$

(102)

$$\begin{aligned}&\quad \left. +\Pi _{2}\left( m+\frac{b_{3}}{p_{3}}-q_{3}\right) \right. \end{aligned}$$

(103)

$$\begin{aligned}&\quad \left. \text {s.t. }m+\frac{d}{1+\rho }-b-b_{3}\ge 0\qquad \qquad (\lambda )\text { (cash flow constraint) }\right. \end{aligned}$$

(104)

$$\begin{aligned}&\qquad \left. m+\frac{d}{1+\rho }-b-b_{3}+pq-d+p_{3}q_{3}\ge 0\qquad \qquad (\mu )\text { (budget constraint) }\right. \end{aligned}$$

(105)

$$\begin{aligned}&\qquad \left. b,b_{3},d\ge 0,\quad 0\le q\le a, 0\le q_{3}\le m\right. \end{aligned}$$

(106)

The first order conditions here, with respect to $b, q, d, b_{3}$, and $q_{3}$ are

$$\begin{aligned}&\displaystyle \left. \frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{a-q}{b}}=\Pi _{1}+\lambda +\mu \right. \end{aligned}$$

(107)

$$\begin{aligned}&\displaystyle \left. \frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{b}{a-q}}=(\mu +\Pi _{1})p\right. \end{aligned}$$

(108)

$$\begin{aligned}&\displaystyle \left. \lambda =(\mu +\Pi _{1})\rho \right. \end{aligned}$$

(109)

$$\begin{aligned}&\displaystyle \left. \frac{1+\Pi _{2}}{p_{3}}=\Pi _{1}+\lambda +\mu \right. \end{aligned}$$

(110)

$$\begin{aligned}&\displaystyle \left. \Pi _{1}p_{3}-1-\Pi _{2}+\mu p_{3}=0\right. \end{aligned}$$

(111)

In addition, we have the complementarity constraints

$$\begin{aligned} m+\frac{d}{1+\rho }-b-b_{3}&= 0\quad \text {or}\quad \lambda =0 \end{aligned}$$

(112)

$$\begin{aligned} m+\frac{d}{1+\rho }-b-b_{3}+pq-d+p_{3}q_{3}&= 0\quad \text {or}\quad \mu =0 \end{aligned}$$

(113)

Similarly, the optimization for the Type 2 traders is

$$\begin{aligned}&\max _{\bar{b},\bar{q},\bar{d},\overline{b}_{3},\overline{q}_{3}} 2\sqrt{(a-\bar{q})\frac{\bar{b}}{p}}+\left( m+\frac{\overline{b}_{3} }{p_{3}}-\overline{q}_{3}\right) \nonumber \\&\quad +\Pi _{1}\left( m+\frac{\bar{d}}{1+\rho }- \bar{b}-\overline{b}_{3}+\bar{p}\bar{q}-\bar{d}+p_{3}\overline{q}_{3}\right) \\&\quad \left. +\Pi _{2}\left( m+\frac{\overline{b}_{3}}{p_{3}}-\overline{q}_{3}\right) \right. \\&\left. \text {s.t. }m+\frac{\bar{d}}{1+\rho }-\bar{b}-\overline{b}_{3}\ge 0\qquad \qquad \qquad \qquad (\overline{\lambda })\right. \\&\qquad \left. m+\frac{\bar{d}}{1+\rho }-\bar{b}-\overline{b}_{3}+ \bar{p}\bar{q}-\bar{d}+p_{3}\overline{q}_{3}\ge 0\qquad \qquad (\overline{\mu })\right. \\&\qquad \left. \bar{b},\overline{b}_{3},\bar{d}\ge 0,0\le \bar{q}\le a, 0\le \overline{q}_{3}\le m.\right. \end{aligned}$$

with first order conditions

$$\begin{aligned}&\displaystyle \left. \frac{1}{\sqrt{p}}\sqrt{\frac{a-\bar{q}}{\bar{b}}}=\Pi _{1}+\bar{\lambda }+\bar{\mu }\right. \end{aligned}$$

(114)

$$\begin{aligned}&\displaystyle \left. \frac{1}{\sqrt{p}}\sqrt{\frac{\bar{b}}{a-\bar{q}}}=(\bar{\mu }+\Pi _{1})\bar{p}\right. \end{aligned}$$

(115)

$$\begin{aligned}&\displaystyle \left. \overline{\lambda }=(\bar{\mu }+\Pi _{1})\rho \right. \end{aligned}$$

(116)

$$\begin{aligned}&\displaystyle \left. \frac{1+\Pi _{2}}{p_{3}}=\Pi _{1}+\overline{\lambda }+\overline{\mu }\right. \end{aligned}$$

(117)

$$\begin{aligned}&\displaystyle \left. \Pi _{1}p_{3}-1-\Pi _{2}+\overline{\mu }p_{3}=0\right. \end{aligned}$$

(118)

$$\begin{aligned}&\displaystyle \left. m+\frac{\bar{d}}{1+\rho }-\bar{b}=0\quad \text {or}\quad \bar{\lambda }=0\right. \end{aligned}$$

(119)

$$\begin{aligned}&\displaystyle \left. m+\frac{\bar{d}}{1+\rho }-\bar{b}-\overline{b}_{3}+\bar{p}\bar{q}- \bar{d}+p_{3}\overline{q}_{3}=0\quad \text {or}\quad \bar{\mu }=0.\right. \end{aligned}$$

(120)

For the continuum of lender-merchants, the new decision variables are $b_{3}^{*}$ (the amount the lenders bid for gold) and $q_{3}^{*}$ (the amount of gold they put up for sale). Their optimization is

$$\begin{aligned}&\left. \max _{_{b_{1}^{*},b_{2}^{*},b_{3}^{*},q_{3}^{*},g}}2\sqrt{\frac{ b_{1}^{*}}{p}\frac{b_{2}^{*}}{\overline{p}}}+\left( B+\frac{b_{3}^{*}}{p_{3}} -q_{3}^{*}\right) +\Pi _{1}\left( B-b_{1}^{*}-b_{2}^{*}+\rho g+p_{3}q_{3}^{*}-b_{3}^{*}\right) \right. \end{aligned}$$

(121)

$$\begin{aligned}&\qquad \left. +\Pi _{2}\left( B+\frac{b_{3}^{*}}{p_{3}}-q_{3}^{*}\right) \right. \end{aligned}$$

(122)

$$\begin{aligned}&\left. \text {s.t. }B-b_{1}^{*}-b_{2}^{*}-b_{3}^{*}-g\ge 0\qquad \qquad (\lambda ^{*})\right. \end{aligned}$$

(123)

$$\begin{aligned}&\qquad \left. b_{1}^{*},b_{2}^{*},b_{3}^{*},g\ge 0,\quad 0\le q_{3}^{*}\le B\right. \end{aligned}$$

(124)

The first order conditions here are:

$$\begin{aligned}&\displaystyle \left. \sqrt{\frac{b_{2}^{*}}{p\overline{p}b_{1}^{*}}}=\Pi _{1}+\lambda ^{*}\right.&\end{aligned}$$

(125)

$$\begin{aligned}&\displaystyle \left. \sqrt{\frac{b_{1}^{*}}{p\overline{p}b_{2}^{*}}}=\Pi _{1}+\lambda ^{*}\right.&\end{aligned}$$

(126)

$$\begin{aligned}&\displaystyle \left. \Pi _{1}\rho -\lambda ^{*}=0\right. \end{aligned}$$

(127)

$$\begin{aligned}&\displaystyle \left. -\Pi _{1}+\frac{1+\Pi _{2}}{p_{3}}-\lambda ^{*}=0\right.&\end{aligned}$$

(128)

$$\begin{aligned}&\displaystyle \left. \Pi _{1}p_{3}-(1+\Pi _{2})=0\right.&\end{aligned}$$

(129)

$$\begin{aligned}&\displaystyle \left. B-b_{1}^{*}-b_{2}^{*}-b_{3}^{*}-g=0\quad \text {or}\quad \lambda ^{*}=0\right.&\end{aligned}$$

(130)

Finally, we have the balance conditions $p=\frac{\overline{b}+b_{1}^{*}}{q}, \overline{p}=\frac{b+b_{2}^{*}}{\overline{q}}, p_{3}=\frac{b_{3}+ \overline{b}_{3}+b_{3}^{*}}{q_{3}+\overline{q}_{3}+q_{3}^{*}}$, and $1+\rho = \frac{d+\overline{d}}{g}$.

Case 1: We first consider the case where $m$ is small and $B$ is large, so we assume the traders’ cash flow and budget constraints are both tight, but the lender-merchants’ cash flow constraint is loose ($\lambda ^{*}=0$). In addition, we assume $m$ is much smaller than $B$, and so the lenders will be selling gold to the traders. Mathematically, this means $q_{3}=\overline{q}_{3}=0$ and $b_{3}^{*}=0$. However, by doing this we may no longer assume conditions (111), (118), and (128).

We also assume a symmetric solution, i.e. $p=\overline{p}, q=\overline{q}, b=\overline{b}, d=\overline{d}$, and $b_{1}^{*}=b_{2}^{*}$.

Our analysis is as follows. First, condition (129) implies that $p_{3}= \frac{1+\Pi _{2}}{\Pi _{1}}$. But then (110) is $\lambda +\mu =\frac{ 1+\Pi _{2}}{p_{3}}-\Pi _{1}=\Pi _{1}-\Pi _{1}=0$. Since $\lambda$ and $\mu$ are both nonnegative, this implies $\lambda =\mu =0=\overline{\lambda }= \overline{\mu }$. But then (109) implies $\rho =0$. And then (127 ) gives $\lambda ^{*}=0$. And then (125) gives $p=\frac{1}{\Pi _{1}}= \overline{p}$.

We can now get expressions for all of the other variables in terms of $q$. To begin, note that (107) along with $\lambda =\mu =0$ gives $\sqrt{\frac{a-q}{b}}=\Pi _{1}\sqrt{\overline{p}}$; plugging in our previous expression for $\overline{p}$ gives $\sqrt{\frac{a-q}{b}}=\sqrt{\Pi _{1}}$, or $\frac{a-q}{b}=\Pi _{1}.$ We can rewrite this as

$$\begin{aligned} b=\frac{a-q}{\Pi _{1}}. \end{aligned}$$

(131)

Next, Trader Type 1’s cash flow and budget constraints being tight together imply $pq-d+p_{3}q_{3}=0$; the assumption that $q_{3}=0$ further gives $d=pq= \frac{q}{\Pi _{1}}$. In addition, the balance condition for $p$ gives $pq=d= \overline{b}+b_{1}^{*}$. Hence $b_{1}^{*}=d-\overline{b}=d-b=\frac{q}{\Pi _{1}}-\frac{a-q}{\Pi _{1}}=\frac{2q-a}{\Pi _{1}}=b_{2}^{*}$. Also, the balance condition for interest $1+\rho =\frac{d+\overline{d}}{g}$ yields $g=2d$ (because of $\rho =0$ and symmetry), so $g=\frac{2q}{\Pi _{1}}$. Furthermore, $b_{3}=m+\frac{d}{1+\rho }-b=m+d-b=m+\frac{q}{\Pi _{1}}-\frac{ a-q}{\Pi _{1}}=m+\frac{2q-a}{\Pi _{1}}=\overline{b}_{3}$. This in turn implies $q_{3}^{*}=\frac{b_{3}+\overline{b}_{3}}{p_{3}}=\frac{2\left( m+ \frac{2q-a}{\Pi _{1}}\right) \Pi _{1}}{1+\Pi _{2}}=\frac{2(\Pi _{1}m+2q-a)}{1+\Pi _{2}}$.

There is no way to pin down an exact value for $q$, so we have a continuum of solutions parametrized by $q$. However, we can find upper and lower bounds for $q$, so that we end up with a “line segment” of solutions. First, since $b_{1}^{*}$ and $b_{2}^{*}$ are nonnegative, we must have $q\ge \frac{a}{2}$. But also the cash flow constraint for the lenders must be satisfied, i.e. $B-b_{1}^{*}-b_{2}^{*}-b_{3}^{*}-g\ge 0$. This reduces to $B-2\left( \frac{2q-a}{\Pi _{1}}\right) -\frac{2q}{\Pi _{1}}\ge 0$, or $q\le \frac{2a+\Pi _{1}B}{6}$. In addition, we must have $q_{3}^{*}\le B$, which is $\frac{2(\Pi _{1}m+2q-a)}{1+\Pi _{2}}\le B$ or $q\le \frac{(1+\Pi _{2})B-2\Pi _{1}m+2a}{4}$. So the range for $q$ is $\frac{a}{2}\le q\le \min (\frac{2a+\Pi _{1}B}{6},\frac{(1+\Pi _{2})B-2\Pi _{1}m+2a}{4},a)$. In order for this range to be nonempty we must have $\Pi _{1}B\ge a$ and $(1+\Pi _{2})B\ge 2\Pi _{1}m$.

If one also requires a 100 % reserve requirement for lending, i.e. $B-q_{3}^{*}\ge g$, this gives the further condition $B\ge \frac{2(\Pi _{1}m+2q-a)}{1+\Pi _{2}}+\frac{2q}{\Pi _{1}}$. There will be at least one $q$ to satisfy this (namely $q=\frac{a}{2}$) so long as $B\ge \frac{2\Pi _{1}m}{ 1+\Pi _{2}}+\frac{a}{\Pi _{1}}$.

Case 2: Now we consider the case where $B$ is smaller (but still significantly larger than $m$). In this case, we assume that the gold market goes inactive—so in addition to $q_{3}=\overline{q}_{3}=0$ and $b_{3}^{*}=0$, we also assume $b_{3}=\overline{b}_{3}=q_{3}^{*}=0$. This means that we cannot use condition (110), (111), (117), (118), (128), or (129) in our analysis. In addition, we assume all trader and lender-merchant constraints are tight, including now the lender-merchants’ cash flow constraints.

We begin with the tight traders’ cash flow constraints: $m+\frac{d}{1+\rho } -b-b_{3}=0$. Assuming $b_{3}=0$, the balance condition $1+\rho =\frac{2d}{g}$, and simplifying, this yields $m+\frac{g}{2}-b=0$. But the tight lender-merchants’ cash flow constraints (plus symmetry $b_{1}^{*}=b_{2}^{*}$ ) imply $g=B-2b_{2}^{*}$; hence we have $m+\frac{B}{2}-b_{2}^{*}-b=0$. Next, the balance constraint for price implies $b+b_{2}^{*}=\overline{p}\overline{q }=pq$, and furthermore the tight cash flow and budget constraints (plus assuming $q_{3}=0$) imply $pq=d$. So $m+\frac{B}{2}-b_{2}^{*}-b=0$ implies $m+\frac{B}{2}-d=0$, or

$$\begin{aligned} d=m+\frac{B}{2}=\overline{d}. \end{aligned}$$

(132)

Next, (125) together with symmetry implies that $\frac{1}{p}=\Pi _{1}+\lambda ^{*}=\Pi _{1}+\Pi _{1}\rho =\Pi _{1}(1+\rho )$, where the second equality follows from (127). This is

$$\begin{aligned} 1+\rho =\frac{1}{\Pi _{1}p}. \end{aligned}$$

(133)

Next, substituting (132), (133), and $b_{3}=0$ into the tight traders’ cash flow constraint gives $m+\frac{m+\frac{B}{2}}{\frac{1}{\Pi _{1}p}}-b=0$, which is

$$\begin{aligned} b=m+\Pi _{1}p\left( m+\frac{B}{2}\right) =\overline{b}. \end{aligned}$$

(134)

Next, condition (107) is $\frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{a-q}{b}}=\Pi _{1}+\lambda +\mu$. Substituting in for $\lambda$ using (109) gives $\frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{a-q}{b}}=\Pi _{1}+\mu +\rho (\Pi _{1}+\mu )=(1+\rho )(\Pi _{1}+\mu )$. Now (108) implies $(\Pi _{1}+\mu )=$ $\frac{1}{p^{\frac{3}{2}}}\sqrt{\frac{b}{a-q}}$, so we have $\frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{a-q}{b}}=(1+\rho )\frac{1}{p^{\frac{ 3}{2}}}\sqrt{\frac{b}{a-q}}$, which is $p\left( \frac{a-q}{b}\right) =1+\rho$. Substituting in for $1+\rho$ using (133) gives $p\left( \frac{a-q}{ b}\right) =\frac{1}{\Pi _{1}p}$, which is $b=\Pi _{1}p^{2}(a-q)$. Now the traders’ cash flow and budget constraints being tight, together with the assumption $q_{3}=0$ imply $pq=d$; hence we can substitute $q=\frac{d}{p}= \frac{m+\frac{B}{2}}{p}=\frac{B+2m}{2p}$ into the last equation for $b$, obtaining

$$\begin{aligned} b=\Pi _{1}p^{2}\left( a-\frac{B+2m}{2p}\right) =\overline{b}. \end{aligned}$$

(135)

Equations (134) and (135) give two expressions for $b$; setting them equal to each other gives the following equation, which can be solved for $p$ using computational methods:

$$\begin{aligned} m+\Pi _{1}p\left( m+\frac{B}{2}\right) =\Pi _{1}p^{2}\left( a-\frac{B+2m}{2p} \right) . \end{aligned}$$

(136)

Once we know $p$($=\overline{p}$), it is then a simple matter to compute $q$ (and $\overline{q}$) via $q=\overline{q}=\frac{m+\frac{B}{2}}{p}, b= \overline{b}$ via (134) or (135), $\rho$ via (133), $g$ from $1+\rho =\frac{2d}{g}=\frac{B+2m}{g}$, and $b_{1}^{*}=b_{2}^{*}$ from $g=B-2b_{2}^{*}$. Consumption levels and final utilities can then be computed accordingly. Finally, for the multipliers, we may calculate $\lambda ^{*}$ from $\frac{1}{p}=\Pi _{1}+\lambda ^{*}, \mu$ from (108), and then $\lambda$ from (109).

In the special case of $m=0$, Eq. (136) solves easily, with $p= \overline{p}=\frac{B}{a}$. We also get $q=\overline{q}=\frac{a}{2}, b= \overline{b}=\frac{\Pi _{1}B^{2}}{2a}, \rho =\frac{a}{\Pi _{1}B}-1, g= \frac{\Pi _{1}B^{2}}{a}$, and $b_{1}^{*}=b_{2}^{*}=\frac{B}{2}(1-\frac{\Pi _{1}B}{a})$. For the multipliers, $\lambda ^{*}=\frac{a}{B}-\Pi _{1}, \mu = \sqrt{\Pi _{1}}(\sqrt{\frac{a}{B}}-\sqrt{\Pi _{1}})=\overline{\mu }$, and $\lambda =\sqrt{\frac{a}{\Pi _{1}B}}(\frac{a}{B}-\Pi _{1})$. We note that in order for these quantities to be nonnegative, we need for $\Pi _{1}B\le a$.

Appendix 4: Demonetized gold with a central bank

Now we assume the demonetization of gold in the presence of a central bank. When the demonetization occurs, the holders of $x$ units of gold are now endowed just with $x$ units of demonetized gold. The accompanying $x$ units of “gold strips” (fiat) go to the central bank. For the moneylenders of the previous models, this means that they are endowed only with ($B$ units of) demonetized gold, making them “gold merchants”.

The central bank is a strategic dummy, lending an amount $G$ of fiat no matter what. It is essentially owned by the individuals of the society (the traders and the merchants). Thus, at the end of the game, its profits are divided up between the two types of trader and the merchants, in the ratio of $m$ to $m$ to $B$, reflecting the original ownership of the gold which backs up the bank in the first place.

The traders of Type 1 are endowed with $m$ units of gold, $0$ units of fiat, and $a$ units of perishable good #1. Their optimization problem looks like this:

$$\begin{aligned}&\max _{b,q,d,b_{3},q_{3}}2\sqrt{(a-q)\frac{b}{\bar{p}}}+\left( m+ \frac{b_{3}}{p_{3}}-q_{3}\right) \nonumber \\&\qquad +\Pi _{1}\left( \frac{d}{1+\rho } -b-b_{3}+pq-d+p_{3}q_{3}+D\right) \end{aligned}$$

(137)

$$\begin{aligned}&\qquad \left. +\Pi _{2}\left( m+\frac{b_{3}}{p_{3}}-q_{3}\right) \right. \end{aligned}$$

(138)

$$\begin{aligned}&\left. \text {s.t. }\frac{d}{1+\rho }-b-b_{3}\ge 0\qquad \qquad (\lambda )\; \text { (cash flow constraint) }\right. \end{aligned}$$

(139)

$$\begin{aligned}&\qquad \left. \frac{d}{1+\rho }-b- b_{3}+pq-d+p_{3}q_{3} +D\ge 0\qquad (\mu )\;\text { (budget constraint) }\right. \end{aligned}$$

(140)

$$\begin{aligned}&\qquad \left. b,b_{3},d\ge 0,\quad 0\le q\le a, 0\le q_{3}\le m\right. \end{aligned}$$

(141)

Here the new quantity “$D$” is the amount of fiat that comes back to the Type 1 traders, as a result of their ownership of $\frac{m}{B+2m}$ of the profits of the central bank. Also, note that here the cash flow and budget constraints lack the quantity “$m$” on the left hand side, reflecting that the traders now are assumed to begin with gold but no fiat as compensation for their gold’s demonetization.

The first order conditions in the above, with respect to $b, q, d, b_{3}$, and $q_{3}$ are

$$\begin{aligned}&\displaystyle \left. \frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{a-q}{b}}=\Pi _{1}+\lambda +\mu \right.&\end{aligned}$$

(142)

$$\begin{aligned}&\displaystyle \left. \frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{b}{a-q}}=(\mu +\Pi _{1})p\right.&\end{aligned}$$

(143)

$$\begin{aligned}&\displaystyle \left. \lambda =(\mu +\Pi _{1})\rho \right.&\end{aligned}$$

(144)

$$\begin{aligned}&\displaystyle \left. \frac{1+\Pi _{2}}{p_{3}}=\Pi _{1}+\lambda +\mu \right.&\end{aligned}$$

(145)

$$\begin{aligned}&\displaystyle \left. \Pi _{1}p_{3}-1-\Pi _{2}+\mu p_{3}=0\right.&\end{aligned}$$

(146)

In addition, we have the complementarity constraints

$$\begin{aligned}&\left. \frac{d}{1+\rho }-b-b_{3}=0\quad \text {or}\quad \lambda =0\right. \end{aligned}$$

(147)

$$\begin{aligned}&\left. \frac{d}{1+\rho }-b-b_{3}+pq-d+p_{3}q_{3}+D=0\quad \text {or}\quad \mu =0\right. \end{aligned}$$

(148)

Similarly, the optimization for the Type 2 traders is

$$\begin{aligned}&\max _{\bar{b},\bar{q},\bar{d},\overline{b}_{3},\overline{q}_{3}} 2\sqrt{(a-\bar{q})\frac{\bar{b}}{p}}+\left( m+\frac{\overline{b}_{3} }{p_{3}}-\overline{q}_{3}\right) \nonumber \\&\quad +\Pi _{1}\left( \frac{\bar{d}}{1+\rho }- \bar{b}-\overline{b}_{3}+\bar{p}\bar{q}-\bar{d}+p_{3}\overline{q} _{3}+D\right) \\&\quad \left. +\Pi _{2}\left( m+\frac{\overline{b}_{3}}{p_{3}}-\overline{q }_{3}\right) \right. \\&\left. \text {s.t. }\frac{\bar{d}}{1+\rho }-\bar{b}-\overline{b}_{3}\ge 0\qquad \qquad \qquad (\overline{\lambda })\right. \\&\qquad \left. \frac{\bar{d}}{1+\rho }-\bar{b}-\overline{b}_{3}+ \bar{p}\bar{q}-\bar{d}+p_{3}\overline{q}_{3}+D\ge 0 \qquad \qquad (\overline{\mu })\right. \\&\qquad \left. \bar{b},\overline{b}_{3},\bar{d}\ge 0,\quad 0\le \bar{q}\le a, 0\le \overline{q}_{3}\le m.\right. \end{aligned}$$

with first order conditions

$$\begin{aligned}&\displaystyle \left. \frac{1}{\sqrt{p}}\sqrt{\frac{a-\bar{q}}{\bar{b}}}=\Pi _{1}+\bar{\lambda }+\bar{\mu }\right.&\end{aligned}$$

(149)

$$\begin{aligned}&\displaystyle \left. \frac{1}{\sqrt{p}}\sqrt{\frac{\bar{b}}{a-\bar{q}}}=(\bar{\mu }+\Pi _{1})\bar{p}\right.&\end{aligned}$$

(150)

$$\begin{aligned}&\displaystyle \left. \overline{\lambda }=(\bar{\mu }+\Pi _{1})\rho \right.&\end{aligned}$$

(151)

$$\begin{aligned}&\displaystyle \left. \frac{1+\Pi _{2}}{p_{3}}=\Pi _{1}+\overline{\lambda }+\overline{\mu }\right.&\end{aligned}$$

(152)

$$\begin{aligned}&\displaystyle \left. \Pi _{1}p_{3}-1-\Pi _{2}+\overline{\mu }p_{3}=0\right.&\end{aligned}$$

(153)

$$\begin{aligned}&\displaystyle \left. m+\frac{\bar{d}}{1+\rho }-\bar{b}-\overline{b}_{3}=0\quad \text {or}\quad \bar{\lambda }=0\right.&\end{aligned}$$

(154)

$$\begin{aligned}&\displaystyle \left. m+\frac{\bar{d}}{1+\rho }-\bar{b}-\overline{b}_{3}+\bar{p}\bar{q}-\bar{d}+p_{3}\overline{q}_{3}=0\quad \text {or}\quad \bar{\mu }=0.\right.&\end{aligned}$$

(155)

The gold merchants begin with $B$ units of gold; but like the traders, they begin with no fiat. They have been stripped of their former money lending function. Instead, again like the traders, they have become money borrowers—the new variable $d^{*}$stands for the amount of loan (from the central bank) that they must repay. And as a result, they also now have a budget constraint, again just like the traders. Finally, they too have an ownership stake in the central bank, namely $\frac{B}{B+2m}$ of its profits, which we denote by $D^{*}$. The merchants’ optimization is as below

$$\begin{aligned}&\max _{_{b_{1}^{*},b_{2}^{*},b_{3}^{*},q_{3}^{*},d^{*}}}2\sqrt{\frac{ b_{1}^{*}}{p}\frac{b_{2}^{*}}{\overline{p}}}+\left( B+\frac{b_{3}^{*}}{p_{3}} -q_{3}^{*}\right) \nonumber \\&\qquad +\Pi _{1}\left( \frac{d^{*}}{1+\rho } -b_{1}^{*}-b_{2}^{*}-b_{3}^{*}-d^{*}+p_{3}q_{3}^{*}+D^{*}\right) \end{aligned}$$

(156)

$$\begin{aligned}&\qquad \left. +\Pi _{2}\left( B+\frac{b_{3}^{*}}{p_{3}}-q_{3}^{*}\right) \right. \end{aligned}$$

(157)

$$\begin{aligned}&\left. \text {s.t. }\frac{d^{*}}{1+\rho }-b_{1}^{*}-b_{2}^{*}-b_{3}^{*}\ge 0\qquad \qquad (\lambda ^{*})\right. \end{aligned}$$

(158)

$$\begin{aligned}&\qquad \left. \frac{d^{*}}{1+\rho } -b_{1}^{*}-b_{2}^{*}-b_{3}^{*}-d^{*}+p_{3}q_{3}^{*}+D^{*}\ge 0\qquad \qquad (\mu ^{*})\right. \end{aligned}$$

(159)

$$\begin{aligned}&\qquad \left. b_{1}^{*},b_{2}^{*},b_{3}^{*},d^{*}\ge 0,\quad 0\le q_{3}^{*}\le B\right. \end{aligned}$$

(160)

The first order conditions here are:

$$\begin{aligned}&\displaystyle \left. \sqrt{\frac{b_{2}^{*}}{p\overline{p}b_{1}^{*}}}=\Pi _{1}+\lambda ^{*}+\mu ^{*}\right. \end{aligned}$$

(161)

$$\begin{aligned}&\displaystyle \left. \sqrt{\frac{b_{1}^{*}}{p\overline{p}b_{2}^{*}}}=\Pi _{1}+\lambda ^{*}+\mu ^{*}\right. \end{aligned}$$

(162)

$$\begin{aligned}&\displaystyle \left. (\Pi _{1}+\mu ^{*})\rho -\lambda ^{*}=0\right. \end{aligned}$$

(163)

$$\begin{aligned}&\displaystyle \left. -\Pi _{1}+\frac{1+\Pi _{2}}{p_{3}}-\lambda ^{*}-\mu ^{*}=0\right. \end{aligned}$$

(164)

$$\begin{aligned}&\displaystyle \left. \Pi _{1}p_{3}+\mu ^{*}p_{3}-(1+\Pi _{2})=0\right. \end{aligned}$$

(165)

$$\begin{aligned}&\displaystyle \left. \frac{d^{*}}{1+\rho }-b_{1}^{*}-b_{2}^{*}-b_{3}^{*}=0\quad \text {or}\quad \lambda ^{*}=0\right. \end{aligned}$$

(166)

$$\begin{aligned}&\displaystyle \left. \frac{d^{*}}{1+\rho } -b_{1}^{*}-b_{2}^{*}-b_{3}^{*}-d^{*}+p_{3}q_{3}^{*}+D^{*}=0\quad \text {or}\quad \mu ^{*}=0\right. \end{aligned}$$

(167)

The central bank is endowed with only $B+2m$ units of fiat. It is a strategic dummy—it always just lends a total of $G$ units of fiat to the traders and merchants, where $G$ is a given quantity with $0\le G\le B+2m$ . Its profits are $\rho G$, and so each trader type’s share is $D=\frac{m}{B+2m}\rho G$, while the merchants’ share is $D^{*}=\frac{B}{B+2m}\rho G$.

Next, we have the balance conditions (a) $p=\frac{\overline{b}+b_{1}^{*}}{q}$, (b) $\overline{p}=\frac{b+b_{2}^{*}}{\overline{q}}$, (c) $p_{3}=\frac{b_{3}+ \overline{b}_{3}+b_{3}^{*}}{q_{3}+\overline{q}_{3}+q_{3}^{*}}$, and d) $1+\rho =\frac{d+\overline{d}+d^{*}}{G}$.

Finally, we make the same symmetry assumptions as usual: $p=\overline{p}, q=\overline{q}, b=\overline{b}, d=\overline{d}$, and $b_{1}^{*}=b_{2}^{*}$.

We will consider two cases. In both cases, $m$ is relatively small compared to $B$ and $G$. Since the traders are starting with much less gold than the merchants, we assume that the traders will not be selling gold to the merchants, i.e. $q_{3}= \overline{q}_{3}=0$ and $b_{3}^{*}=0$. Also, since the traders and merchants both start with no fiat, we assume the cash flow constraints are tight for both. In Case 1, the central bank lends out a relatively large amount of money ($G$ large). Since $B$ is large compared to $m$, we assume that when all of the money is recycled back to the merchants they will have loose budget constraints, so $\mu ^{*}=0$. In Case 2 ($G$ small) there is a lot less money in the economy, so all constraints, for both traders and merchants, are tight.

Case 1: $m$ small, $G$ large

We first analyze the case in which the cash flow constraints (for both types of traders and for merchants) are tight, and the budget constraints for the merchants are loose (so $\mu ^{*}$ is equal to zero), $q_{3}= \overline{q} _{3}=0$, and $b_{3}^{*}=0$.

The analysis of this case goes as follows. First, $\mu ^{*}=0$ and (165) together imply that $p_{3}=\frac{1+\Pi _{2}}{\Pi _{1}}$. Then (145) is $\frac{1+\Pi _{2}}{p_{3}}=\Pi _{1}+\lambda +\mu$. But $p_{3}=\frac{1+\Pi _{2}}{\Pi _{1}}$, so this reduces to $\lambda +\mu =0$. Since the multipliers $\lambda$ and $\mu$ are both constrained to be nonnegative, we have $\lambda =\mu =0$. Similarly, $\overline{\lambda }=\overline{\mu }=0$.

Next, $\lambda =0$ and (144) together imply $\rho =0$. But this in turn implies $D=D^{*}=0$. Also, $\rho =0$ together with (163) implies $\lambda ^{*}=0$. And then (161) plus symmetry imply $\frac{1}{p}=\Pi _{1}+\lambda ^{*}+\mu ^{*}=\Pi _{1}$, so we have $p=\frac{1}{\Pi _{1}}= \overline{p}$. And then (142) (with $\lambda =\mu =0$) gives $\sqrt{\frac{a-q}{b}}=\Pi _{1}\sqrt{\overline{p}}=\sqrt{\Pi _{1}}$, which in turn gives

$$\begin{aligned} q=a-\Pi _{1}b. \end{aligned}$$

(168)

The next piece is to obtain an expression for $b$. This will be a somewhat long, tedious process. We begin by working with the budget constraint for Trader Type 1: $\frac{d}{1+\rho }-b-b_{3}+pq-d+p_{3}q_{3}+D\ge 0$. Substituting $\rho =0, q_{3}=0$, and $D=0$ yields $pq-b-b_{3}\ge 0$. But by balancing condition a) $pq-b=pq-\overline{b}=b_{1}^{*}$, so we have

$$\begin{aligned} b_{1}^{*}-b_{3}\ge 0. \end{aligned}$$

(169)

But also, the budget constraint for the merchants is $\frac{d^{*}}{1+\rho } -b_{1}^{*}-b_{2}^{*}-b_{3}^{*}-d^{*}+p_{3}q_{3}^{*}+D^{*}\ge 0$. Substituting $\rho =0, b_{3}^{*}=0$, and $D^{*}=0$ yields $p_{3}q_{3}^{*}-b_{1}^{*}-b_{2}^{*}\ge 0$, which (by symmetry) is $p_{3}q_{3}^{*}-2b_{1}^{*}\ge 0$. But by balancing condition c) (plus the assumptions $q_{3}=\overline{q}_{3}=b_{3}^{*}=0$ and symmetry) we have $p_{3}q_{3}^{*}=2b_{3}$, and so $2b_{3}-2b_{1}^{*}\ge 0$. But this and (169) together imply

$$\begin{aligned} b_{1}^{*}=b_{3}. \end{aligned}$$

(170)

Now look at the cash flow constraint for the merchants. We assumed it was tight, i.e. $\frac{d^{*}}{1+\rho }-b_{1}^{*}-b_{2}^{*}-b_{3}^{*}=0$. Substituting in $\rho =0$ and $b_{3}^{*}=0$, and using symmetry gives $d^{*}-2b_{1}^{*}=0$, or $b_{1}^{*}=\frac{d^{*}}{2}$. But balancing condition d) (with $\rho =0$) gives $d^{*}=G-d-\overline{d}=G-2d$; hence $b_{1}^{*}= \frac{G-2d}{2}=\frac{G}{2}-d=\frac{G}{2}-b-b_{3}=\frac{G}{2}-b-b_{1}^{*}$. [In the previous, the penultimate equality follows from the tight Trader 1 cash flow constraint (together with $\rho =0$), while the last equality follows from (170).] Combining the “$b_{1}^{*}$” terms, we have $2b_{1}^{*}= \frac{G}{2}-b$, or

$$\begin{aligned} b_{1}^{*}=\frac{G}{4}-\frac{b}{2}. \end{aligned}$$

(171)

Now we are finally in a position to obtain our expression for $b$. We start with balancing condition a), which is $pq=b+b_{1}^{*}$. Substituting in our expressions $p=\frac{1}{\Pi _{1}}$, Eq. (168) for $q$, and (171) for $b_{1}^{*}$, we have $\frac{a-\Pi _{1}b}{\Pi _{1}}=b+\frac{G}{4}- \frac{b}{2}$, which solves with $b=\frac{4a-\Pi _{1}G}{6\Pi _{1}}=\overline{b}$.

The rest of the variables can now be easily obtained. First, using (171 ), we have $b_{1}^{*}=\frac{G}{4}-\frac{b}{2}=\frac{G}{4}-\frac{a}{3\Pi _{1}} +\frac{G}{12}=\frac{\Pi _{1}G-a}{3\Pi _{1}}=b_{2}^{*}$. Because of (170 ), we also immediately have $b_{3}=\frac{\Pi _{1}G-a}{3\Pi _{1}}=\overline{b} _{3}$. Next, $q=a-\Pi _{1}b=a-\Pi _{1}\left( \frac{4a-\Pi _{1}G}{6\Pi _{1}} \right) =\frac{2a+\Pi _{1}G}{6}=\overline{q}$. Then $d^{*}=2b_{1}^{*}=\frac{ 2(\Pi _{1}G-a)}{3\Pi _{1}}$. And $d=\frac{G-d^{*}}{2}=\frac{\Pi _{1}G+2a}{ 6\Pi _{1}}=\overline{d}$. And finally $q_{3}^{*}=\frac{2b_{3}}{p_{3}}=\frac{ 2(\Pi _{1}G-a)}{3(1+\Pi _{2})}$. For completeness, we also list all of the other values previously found: $p=\frac{1}{\Pi _{1}}=\overline{p}, p_{3}= \frac{1+\Pi _{2}}{\Pi _{1}}$, and $\rho =D=D^{*}=\lambda =\overline{\lambda } =\mu =\overline{\mu }=\lambda ^{*}=\mu ^{*}=0$.

Last, we should state the values over which our calculations are valid. First, all of the calculated expressions for variables above must be nonnegative; this requires $a\le \Pi _{1}G\le 4a$. Second, the above value for $q_{3}^{*}$ must be less than or equal to $B$, i.e. $\frac{2(\Pi _{1}G-a) }{3(1+\Pi _{2})}\le B$.

Case 2: $m$ small, $G$ small

Now we consider what happens when $\Pi _{1}G<a$. In this case we assume cash flow and budget constraints for both traders and merchants are all tight, i.e. the multipliers $\lambda , \overline{\lambda }, \mu , \overline{ \mu }, \lambda ^{*}$, and $\mu ^{*}$ are all positive. The values of $q_{3}^{*}, b_{3}$, and $\overline{b}_{3}$ from Case 1 (from the boundary case $\Pi _{1}G=a$) lead us to assume the gold trade market shuts down, i.e. in addition to our previous assumptions of $q_{3}=$ $\overline{q}_{3}=0$ and $b_{3}^{*}=0$, we also have $q_{3}^{*}=b_{3}=\overline{b}_{3}=0$.

Our analysis begins with Eq. (142), namely $\frac{1}{\sqrt{ \overline{p}}}\sqrt{\frac{a-q}{b}}=\Pi _{1}+\lambda +\mu$. Substituting in for $\lambda$ using (144) yields $\frac{1}{\sqrt{\overline{p}}}\sqrt{ \frac{a-q}{b}}=(1+\rho )(\Pi _{1}+\mu )$. But then substituting in for $\Pi _{1}+\mu$ using (143) gives $\frac{1}{\sqrt{\overline{p}}}\sqrt{\frac{ a-q}{b}}=(1+\rho )\frac{1}{p\sqrt{\overline{p}}}\sqrt{\frac{b}{a-q}}$. Using symmetry $\overline{p}=p$ results in

$$\begin{aligned} p\left( \frac{a-q}{b}\right) =1+\rho . \end{aligned}$$

(172)

Next, the balancing constraint for price is $b=pq-b_{1}^{*}$. But the tight ($\lambda ^{*}$) constraint, together with $b_{3}^{*}=0$ and symmetry $b_{1}^{*}=b_{2}^{*}$ yields $b_{1}^{*}=\frac{d^{*}}{2(1+\rho )}$; substituting this in the previous expression for $b$ gives $b=pq-\frac{d^{*} }{2(1+\rho )}$. But the tight ($\lambda$) constraint implies $b=\frac{d}{ 1+\rho }$; setting the last two expressions for $b$ equal gives $pq-\frac{ d^{*}}{2(1+\rho )}=\frac{d}{1+\rho }$. This can be rearranged to read

$$\begin{aligned} q=\frac{d^{*}+2d}{2p(1+\rho )}. \end{aligned}$$

(173)

In addition, we may rewrite (172) as $\frac{p(a-q)}{1+\rho }=b=\frac{d}{1+\rho }$, so $d=ap-pq=ap-\frac{d^{*}+2d}{2(1+\rho )}$, which gives

$$\begin{aligned} ap=d+\frac{d^{*}+2d}{2(1+\rho )}. \end{aligned}$$

(174)

Next, the tight cash flow and budget constraints for the merchants together imply that $-d^{*}+p_{3}q_{3}^{*}+D^{*}=0$. Now $q_{3}^{*}$ is assumed to be zero, and $D^{*}$ is equal to $\frac{B}{B+2m}\rho G$, so this reduces to $d^{*}=\frac{B}{B+2m}\rho G$. Next, the balance constraint for the interest rate implies that $d=\frac{(1+\rho )G-d^{*}}{2}$. Substituting in our previous expression for $d^{*}$, this simplifies to $d=\frac{(1+\rho )G- \frac{B}{B+2m}\rho G}{2}=\frac{G}{2}+\frac{m}{B+2m}\rho G$. Hence, $d^{*}+2d=(1+\rho )G$, and our previous expressions (174) and (173) for $p$ and $q$ reduce to $p=\frac{2d+G}{2a}$ and $q=\frac{G}{2p}=\frac{aG}{2d+G}$ respectively. In addition, substituting in $d=\frac{G}{2}+\frac{m}{ B+2m}\rho G$ yields $p=\frac{1}{a}( G+\frac{m}{B+2m}\rho G) = \overline{p}$ and $q=\frac{a}{2}( \frac{B+2m}{B+2m+m\rho }) = \overline{q}$. We also have $b=\overline{b}=\frac{d}{1+\rho }=\frac{1}{ 1+\rho }\left( \frac{G}{2}+\frac{m}{B+2m}\rho G\right)$ and $b_{1}^{*}=b_{2}^{*}=\frac{d^{*}}{2(1+\rho )}=\frac{B}{B+2m}\frac{\rho G}{ 2(1+\rho )}$.

Note that we have a degree of freedom in our solutions here, as all of our variables are in terms of $\rho$.

Finally, we obtain expressions for the multipliers. First, from (143) we have $\mu =\frac{1}{p\sqrt{p}}\sqrt{\frac{b}{a-q}}-\Pi _{1}=\left( \frac{ 2a}{2d+G}\right) ^{\frac{3}{2}}\sqrt{\frac{d/(1+\rho )}{a-\frac{aG}{2d+G}}}= \frac{a}{2d+G}\frac{2}{\sqrt{1+\rho }}-\Pi _{1}=\frac{a}{\sqrt{1+\rho }(G+ \frac{m}{B+2m}\rho G)}-\Pi _{1}=\overline{\mu }$. We also have $\lambda =(\mu +\Pi _{1})\rho =\frac{a\rho }{\sqrt{1+\rho }(G+\frac{m}{B+2m}\rho G)}= \overline{\lambda }$. To find $\mu ^{*}$and $\lambda ^{*}$, we need to solve the two equations in two unknowns given by (162) and (163). First, (163) gives us $\lambda ^{*}=(\Pi _{1}+\mu ^{*})\rho$. Substituting this expression for $\lambda ^{*}$ into (162) and using symmetry, we have $\frac{1}{p}=(1+\rho )(\Pi _{1}+\mu ^{*})$, or $\mu ^{*}= \frac{1}{(1+\rho )p}-\Pi _{1}=\frac{a}{(1+\rho )\left( G+\frac{m}{B+2m}\rho G\right) }-\Pi _{1}$. Then $\lambda ^{*}=(\Pi _{1}+\mu ^{*})\rho =\frac{a\rho }{(1+\rho )\left( G+\frac{m}{B+2m}\rho G\right) }$.

In order for the results to be valid, $\rho$ must be nonnegative, and the above expressions for the multipliers $\mu$ and $\mu ^{*}$ must also be nonnegative. Looking closely at these, we see that there will be a range for $\rho$, with zero as the lower bound and the upper bound as the value of $\rho$ for which $(1+\rho )\left( G+\frac{m}{B+2m}\rho G\right)$ is equal to $\frac{a}{\Pi _{1}}$. This range will be nonempty so long as $\Pi _{1}G\le a$.