Asset price bubbles, market liquidity, and systemic risk

Abstract

This paper studies an equilibrium model with heterogeneous agents, asset price bubbles, and trading constraints. Market liquidity is modeled as a stochastic quantity impact from trading on the price. Bubbles are larger in liquid markets and when trading constraints are more binding. Systemic risk is defined as an unanticipated shock that results in the nonexistence of an equilibrium in the economy. A realization of systemic risk results in a significant loss of wealth. Systemic risk increases as: (i) the fraction of agents seeing an asset price bubble increases, (ii) as the market becomes more illiquid, and (iii) as trading constraints are relaxed.

Appendix: Proofs

1.1 Trading constraint

The trading constraint set

$$\begin{aligned} K_{t}(\omega )=\left\{ x\in {\mathbb {R}}:Y_{t}-\varphi _{t}(X_{t+1}-X_{t})S_{t}\ge S_{t}x\left( \gamma 1_{x\ge 0}+(1+\gamma )1_{x\le 0}\right) \right\} \end{aligned}$$

where $0\le \gamma \le 1$ and $K_{t}(\omega )\subset {\mathbb {R}}$ is a ${\mathcal {F}}_{t}$ measurable, nonempty, closed, convex set.

Proof

It is nonempty because $0\in K_{t}(\omega )$. Using $Y_{t+1}=Y_{t}-\varphi _{t}(X_{t+1}-X_{t})S_{t}$, we can rewrite the constraint set as

$$\begin{aligned} K_{t}(\omega )=\left\{ X_{t+1}\ge 0:\;Y_{t+1}\ge -\gamma S_{t}X_{t+1}\right\} +\left\{ X_{t+1}\le 0:\;Y_{t+1}\ge -(1+\gamma )S_{t}X_{t+1}\right\} . \end{aligned}$$

This set is closed and since each subset in the sum is convex, the sum is convex, see Ruszczynski [58], p. 18. $\square$

1.2 Lemma 1

Proof

If $X_{t+1}\in int(K_{t})$, then $N_{K_{t}}(X_{t+1})=\{0\}$, see Tuy [64], p. 22. Hence, in this case if $\kappa \in N_{K_{t}}(X_{t+1})$, then $\kappa =0$.

(Case a) If $X_{t+1}\in bd(K_{t})$ and $X_{t+1}>0$, then by the convexity of $K_{t}$, $[0,X_{t+1}]\subset K_{t}$. This implies $N_{K_{t}}(X_{t+1})\ne \{0\}$. Hence, if $\kappa \ne 0$, then $\kappa (Z-X_{t+1})\le 0$ for all $0\le Z\le X_{t+1}$, implying $\kappa >0$.

(Case b) follows similarly.

(Case c) If $X_{t+1}\in bd(K_{t})$ and $X_{t+1}<0$, then by the convexity of $K_{t}$, $[X_{t+1},0]\subset K_{t}$. This implies $N_{K_{t}}(X_{t+1})\ne \{0\}$. Hence, if $\kappa \ne 0$, then $\kappa (Z-X_{t+1})\le 0$ for all $0\le X_{t+1}\le Z$, implying $\kappa <0$.

(Case d) follows similarly. This completes the proof. $\square$

1.3 Theorem 2

Proof

(Step 1) Existence and Uniqueness

The above assumptions in conjunction with Examples 2.1 and 5.2 of Pennanen [54] imply that the hypothesis of Theorem 5.1 in Pennanen [54] hold, which proves that an optimal trading strategy exists. The trading strategy is unique by the concavity of the utility function.

(Step 2) Characterization of the Solution

We use backward induction.

At time T, with share holdings $(X_{T}^{i},Y_{T}^{i})$, the optimal trading strategy is $\Delta X_{T+1}=-X_{T}$ since the portfolio must be liquidated at this date.

At time $t<T$ with share holdings $(X_{t}^{i},Y_{t}^{i})$, given the optimal $\{X_{j+1}^{i}:\,j\in \{t+1,\ldots ,T-1\}\}$, the optimal $Z\in K_{t}$ must maximize

$$\begin{aligned}&E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}\left( Y_{T+1}^{i}(Z)\right) \right] =E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}\left( y_{i}-\varphi _{T}\left( -\sum _{j=t+1}^{T-1}\Delta X_{j+1}^{i}-Z\right) S_{T}\right. \right. \\&\quad \left. \left. -\sum _{j=t+1}^{T-1}\varphi _{j}\left( \Delta X_{j+1}^{i}\right) S_{j}-\varphi _{t}\left( Z-X_{t}\right) S_{t}+\left( Y_{t}^{i}-y_{i}\right) \right) \right] . \end{aligned}$$

The first order condition, which is necessary and sufficient (see Tuy [64], p. 75) , is that

$$\begin{aligned} 0\in \partial _{X_{t+1}^{i}}E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}\left( Y_{T+1}^{i}\left( X_{t+1}\right) \right) \right] +N_{K_{t}}\left( X_{t+1}^{i}\right) . \end{aligned}$$

But,

$$\begin{aligned}&\partial _{X_{t+1}^{i}}E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}\left( Y_{T+1}^{i}\left( X_{t+1}\right) \right) \right] =\frac{dE_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}\left( Y_{T+1}^{i}\left( X_{t+1}\right) \right) \right] }{dX_{t+1}^{i}}\\&\quad =E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'\left( Y_{T+1}^{i}\right) \left( \varphi '_{T}\left( \Delta X_{T+1}^{i}\right) S_{T}-\varphi '_{t}\left( \Delta X_{t+1}^{i}\right) S_{t}\right) \right] . \end{aligned}$$

Hence, there exists a unique $\kappa _{t}^{i}\in N_{K_{t}}(X_{t+1}^{i})$ such that

$$\begin{aligned} E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'\left( Y_{T+1}^{i}\right) \left( \varphi '_{T}\left( \Delta X_{T+1}^{i}\right) S_{T}-\varphi '_{t}\left( \Delta X_{t+1}^{i}\right) S_{t}\right) \right] +\kappa _{t}=0. \end{aligned}$$

(20)

This iimplies the following expressions.

$$\begin{aligned}&\frac{E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'\left( Y_{T+1}^{i}\right) \left( \varphi '_{T}(\Delta X_{T+1}^{i})S_{T}-\varphi '_{t}(\Delta X_{t+1}^{i})S_{t}\right) \right] }{E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'(Y_{T+1}^{i})\right] }+\frac{\kappa _{t}}{E^{{\mathbb {P}}_{i}}\left[ U_{i}'(Y_{T+1}^{i})\right] }=0.\\&E_{t}^{\mathbb {{\mathbb {Q}}}_{i}}\left[ \varphi '_{T}(\Delta X_{T+1}^{i})S_{T}-\varphi '_{t}(\Delta X_{t+1}^{i})S_{t}\right] +\frac{\kappa _{t}}{E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'(Y_{T+1}^{i})\right] }\frac{\varphi '_{t}(\Delta X_{t+1}^{i})}{\varphi '_{t}(\Delta X_{t+1}^{i})}=0 \end{aligned}$$

where $E_{t}^{{\mathbb {Q}}_{i}}\left[ \cdot \right] =E^{{\mathbb {Q}}_{i}}\left[ \cdot \left| {\mathcal {F}}_{t}\right. \right]$ is conditional expectation under ${\mathbb {Q}}_{i}$. Then,

$$\begin{aligned} E_{t}^{\mathbb {{\mathbb {Q}}}_{i}}\left[ \varphi '_{T}(\Delta X_{T+1}^{i})S_{T}-\varphi '_{t}(\Delta X_{t+1}^{i})S_{t}+\varphi '_{t}(\Delta X_{t+1}^{i})\frac{\kappa _{t}}{E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'(Y_{T+1}^{i})\right] \varphi '_{t}(\Delta X_{t+1}^{i})}\right] =0. \end{aligned}$$

Define $v_{t}^{i}=\frac{\kappa _{t}^{i}}{E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'(Y_{T+1}^{i})\right] \varphi '_{t}(\Delta X_{t+1}^{i})}$. Note that $\kappa _{T}^{i}=0$ since $N_{K_{T}}(X)=\{0\}$ for all $X\in K_{T}$. Then,

$$\begin{aligned} E_{t}^{\mathbb {{\mathbb {Q}}}_{i}}\left[ \varphi '_{T}(\Delta X_{T+1}^{i})(S_{T}-v_{T}^{i})-\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})\right] =0. \end{aligned}$$

This implies that $\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-\nu _{t}^{i})$ is a martingale for $t\in \{0,\ldots ,T\}$ under $\frac{d{\mathbb {Q}}_{i}}{d{\mathbb {P}}_{i}}=\frac{U_{i}'(Y_{T+1}^{i})}{E^{{\mathbb {P}}_{i}}\left[ U_{i}'(Y_{T+1}^{i})\right] }>0.$ This completes the proof. $\square$

1.4 Corollary 3

Proof

Given $E_{t}^{\mathbb {{\mathbb {P}}}_{i}}\left[ \varphi '_{T}(\Delta X_{T+1}^{i})(S_{T}-v_{T}^{i})-\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})\right] =0$.

Algebra yields

$$\begin{aligned}E_{t}^{{\mathbb {P}}_{i}}\left[ \varphi '_{T}(\Delta X_{T+1}^{i})S_{T}\right] -\varphi '_{t}(\Delta X_{t+1}^{i})S_{t}+\varphi '_{t}(\Delta X_{t+1}^{i})v_{t}^{i}=0.\end{aligned}$$

This implies

$$\begin{aligned}&\varphi '_{t}(\Delta X_{t+1}^{i})S_{t}-E_{t}^{{\mathbb {P}}_{i}}\left[ \varphi '_{T}(\Delta X_{T+1}^{i})S_{T}\right] =-\varphi '_{t}(\Delta X_{t+1}^{i})v_{t}^{i}\hbox {, i.e.}\\&\beta _{t}^{i}=-\varphi '_{t}(\Delta X_{t+1}^{i})v_{t}^{i}\hbox {. But,} v_{t}^{i}=\frac{\kappa _{t}^{i}}{E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'(Y_{T+1}^{i})\right] \varphi '_{t}(\Delta X_{t+1}^{i})}.\end{aligned}$$

Substitution yields ${\beta _{t}^{i}=\frac{-\kappa _{t}^{i}}{E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'\left( Y_{T+1}^{i}\right) \right] }}$.

Using Lemma 1 completes the proof. $\square$

1.5 Lemma 4

Proof

Using the fundamental theorem of calculus and the assumption that $\varphi _{t}^{k}(0)=0$, we have $\varphi _{t}^{k}(x)=\int _{0}^{x}\frac{d\varphi _{t}^{k}(u)}{du}du$ for $k\in \{l,nl\}$ for $x>0$ and $\varphi _{t}^{k}(x)=-\int _{0}^{x}\frac{d\varphi _{t}^{k}(u)}{du}du$ for $k\in \{l,nl\}$ for $x<0$. The result follows because the integral is a positive linear operator. $\square$

1.6 Theorem 5

Proof

From expression (9),

$$\begin{aligned} {{{\beta _{t}^{i}=\frac{-\kappa _{t}^{i}}{E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'\left( Y_{T+1}^{i}\right) \right] }}}}. \end{aligned}$$

Assume that $\kappa _{t}^{i}$, $X_{t+1}$ and $\Delta X_{t+1}^{i}$ are fixed. There are two cases to consider.

(Case 1) If $X_{t+1}^{i}>0$, show that $\beta _{t}^{i}(\varphi _{t}^{l})<\beta _{t}^{i}(\varphi _{t}^{nl})<0$.

Assume that $X_{t+1}^{i}>0$, then $\kappa _{t}^{i}>0$ and $\Delta X_{t+1}^{i}>0$. Thus, $\beta _{t}^{i}<0$ since $\varphi '_{t}(\Delta X_{t+1}^{i})>0$ for all $\Delta X_{t+1}^{i}$.

Given $Y_{T+1}^{i}=y_{i}-\varphi _{T}\left( -x_{i}-\sum _{t=0}^{T-1}\Delta X_{t+1}^{i}\right) S_{T}-\sum _{t=0}^{T-1}\varphi _{t}(\Delta X_{t+1}^{i})S_{t}$, from Lemma (4) we have that $\varphi _{t}^{l}(x)<\varphi _{t}^{nl}(x)$ for all $x\not =0$ a.e. ${\mathbb {P}}$.

Since $U_{i}$ is concave, a larger $\varphi _{t}$ (corresponding to $\varphi _{t}^{nl}(x)$) implies a smaller $Y_{T+1}^{i}$. This results in a larger $E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'\left( Y_{T+1}^{i}\right) \right]$. Hence, the denominator is larger for $\varphi _{t}^{nl}$ than for $\varphi _{t}^{l}$, and given $\beta _{t}^{i}<0$ this means that the size of the bubble has decreased, i.e. $\beta _{t}^{i}(\varphi _{t}^{l})<\beta _{t}^{i}(\varphi _{t}^{nl})<0$.

(Case 2) If $X_{t+1}^{i}<0$, show that $\beta _{t}^{i}(\varphi _{t}^{l})>\beta _{t}^{i}(\varphi _{t}^{nl})>0$.

Assume that $X_{t+1}^{i}<0$, then $\kappa _{t}^{i}<0$ and $\Delta X_{t+1}^{i}<0$. Thus, $\beta _{t}^{i}>0$ since $\varphi '_{t}(\Delta X_{t+1}^{i})>0$ for all $\Delta X_{t+1}^{i}$.

Using same reasoning as above, but for $\kappa _{t}^{i}<0$ and given $\beta _{t}^{i}>0$, the denominator is larger for $\varphi _{t}^{nl}$ as $\varphi _{t}^{l}(x)<\varphi _{t}^{nl}(x)$ holds for all $x\not =0$ a.e. ${\mathbb {P}}$. This implies that a positive bubble $\beta _{t}^{i}>0$ is smaller for a larger denominator corresponding to $\varphi _{t}^{nl}$. Thus, $\beta _{t}^{i}(\varphi _{t}^{nl})<\beta _{t}^{i}(\varphi _{t}^{l})$.

This completes the proof. $\square$

1.7 Corollary 6

Proof

$\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})$ for $t\in \{0,\ldots ,T\}$ a martingale under ${\mathbb {Q}}_{i}$

implies that $\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})\rho _{t}^{i}$ is a martingale under ${\mathbb {P}}_{i}$. Thus,

$E_{t}^{i}\left[ \frac{\varphi '_{t}(\Delta X_{t+2}^{i})(S_{t+1}-v_{t+1}^{i})}{\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})}\frac{\rho _{t+1}^{i}}{\rho _{t}^{i}}\right] =1$. Algebra yields

$$\begin{aligned}&E_{t}^{i}\left[ \frac{\varphi '_{t}(\Delta X_{t+2}^{i})(S_{t+1}-v_{t+1}^{i})}{\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})}\right] E_{t}^{i}\left[ \frac{\rho _{t+1}^{i}}{\rho _{t}^{i}}\right] \\&\quad +cov_{t}\left( \frac{\varphi '_{t}(\Delta X_{t+2}^{i})(S_{t+1}-v_{t+1}^{i})}{\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})},\frac{\rho _{t+1}^{i}}{\rho _{t}^{i}}\right) =1.\end{aligned}$$

But, $E_{t}^{i}\left[ \frac{\rho _{t+1}^{i}}{\rho _{t}^{i}}\right] =1$. Hence,

$$\begin{aligned}E_{t}^{i}\left[ \frac{\varphi '_{t}(\Delta X_{t+2}^{i})(S_{t+1}-v_{t+1}^{i})}{\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})}-1\right] =-cov_{t}^{i}\left( \frac{\varphi '_{t}(\Delta X_{t+2}^{i})(S_{t+1}-v_{t+1}^{i})}{\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})},\frac{\rho _{t+1}^{i}}{\rho _{t}^{i}}\right) .\end{aligned}$$

We can rewrite this as

$$\begin{aligned} E_{t}\left[ R_{t+1}^{v}\right] =-cov_{t}\left( R_{t+1}^{v},\frac{\rho _{t+1}}{\rho _{t}}\right) \hbox { where }R_{t+1}^{v}=\frac{\varphi '_{t}(\Delta X_{t+2}^{i})(S_{t+1}-v_{t+1}^{i})}{\varphi '_{t}(\Delta X_{t+1}^{i})(S_{t}-v_{t}^{i})}-1.\end{aligned}$$

Simple algebra gives $R_{t+1}^{v}=wR_{t+1}+(1-w)\theta _{t+1}$ where $w=\frac{S_{t}}{S_{t}-v_{t}^{i}}$.

Substitution yields

$$\begin{aligned} E_{t}\left[ wR_{t+1}+(1-w)\theta _{t+1}\right] =-cov_{t}\left( wR_{t+1}+(1-w)\theta _{t+1},\frac{\rho _{t+1}}{\rho _{t}}\right) . \end{aligned}$$

Or, $E_{t}\left[ R_{t+1}\right] =-\frac{(1-w)}{w}E_{t}\left[ \theta _{t+1}\right] -cov_{t}\left( R_{t+1},\frac{\rho _{t+1}}{\rho _{t}}\right) -\frac{(1-w)}{w}cov_{t}\left( \theta _{t+1},\frac{\rho _{t+1}}{\rho _{t}}\right)$.

But, $\frac{(1-w)}{w}=\frac{-v_{t}^{i}}{S_{t}}$. Substitution yields

$$\begin{aligned} E_{t}^{i}\left[ R_{t+1}\right] =-cov_{t}^{i}\left( R_{t+1},\frac{\rho _{t+1}}{\rho _{t}}\right) +\frac{v_{t}^{i}}{S_{t}}\left( E_{t}^{i}\left[ \theta _{t+1}\right] +cov_{t}^{i}\left( \theta _{t+1},\frac{\rho _{t+1}}{\rho _{t}}\right) \right) . \end{aligned}$$

Next, given $\beta _{t}^{i}=-\varphi '_{t}(\Delta X_{t+1}^{i})v_{t}^{i}$, solving for $v_{t}^{i}$ and substitution completes the proof. $\square$

1.8 Theorem 8

Proof

At time t, the optimal share holdings are either $X_{t+1}^{i}>0$ or $X_{t+1}^{i}<0$.

When $X_{t+1}^{i}>0$, the constraint being binding implies that $Y_{t+1}^{i}<0$. Similarly, when $X_{t+1}^{i}<0$ this implies that $Y_{t+1}^{i}>0$.

These implications follow by the definition of the constraint

$$\begin{aligned} K_{t}(\omega )=\left\{ X_{t+1}(\omega )\in {\mathbb {R}}:\;Y_{t+1}\ge -S_{t}X_{t+1}\left( \gamma 1_{X_{t+1}\ge 0}+(1+\gamma )1_{X_{t+1}\le 0}\right) \right\} \end{aligned}$$

Hence, in the time t optimization problem, the original constraint can be replaced by

$$\begin{aligned} K_{t}(\omega )=\left\{ X_{t+1}\in {\mathbb {R}}_{++}:-\gamma S_{t}X_{t+1}\le Y_{t+1}\right\} =\left\{ X_{t+1}\in {\mathbb {R}}_{++}:-\frac{Y_{t+1}}{S_{t}X_{t+1}}\le \gamma \right\} \end{aligned}$$

for $X_{t+1}>0$, and

$$\begin{aligned} K_{t}(\omega )= & {} \left\{ X_{t+1}\in {\mathbb {R}}_{--}:-\left( 1+\gamma \right) S_{t}X_{t+1}\le Y_{t+1}\right\} \\= & {} \left\{ X_{t+1}\in {{\mathbb {R}}}_{--}:\frac{Y_{t+1}}{S_{t}X_{t+1}}\le -\left( 1+\gamma \right) \right\} \end{aligned}$$

for $X_{t+1}<0$.

Note that $Y_{t+1}=Y_{t}-\varphi _{t}(X_{t+1}-X_{t})S_{t}$ from expression (2).

Define the function $f(X_{t+1})=-\left( \frac{Y_{t}-\varphi _{t}(X_{t+1}-X_{t})S_{t}}{S_{t}X_{t+1}}\right)$ on $X_{t+1}\in {\mathbb {R}}$.

We note that

$$\begin{aligned} f'(X_{t+1})= & {} \frac{Y_{t}-\varphi _{t}(X_{t+1}-X_{t})S_{t}}{S_{t}X_{t+1}^{2}}+\frac{\varphi _{t}'(X_{t+1}-X_{t})S_{t}}{S_{t}X_{t+1}}\\= & {} \frac{Y_{t+1}+\varphi _{t}'(X_{t+1}-X_{t})S_{t}X_{t+1}}{S_{t}X_{t+1}^{2}}. \end{aligned}$$

We first show that $f'(X_{t+1})>0$ when $X_{t+1}>0$.

Indeed, $\varphi _{t}'(X_{t+1}-X_{t})S_{t}X_{t+1}>\gamma S_{t}X_{t+1}$ when $X_{t+1}>0$ because $\varphi '_{t}(x)>1>\gamma$ when $x>0$.

Thus, $f'(X_{t+1})=\frac{Y_{t+1}+\varphi _{t}'(X_{t+1}-X_{t})S_{t}X_{t+1}}{S_{t}X_{t+1}^{2}}>\frac{Y_{t+1}+\gamma S_{t}X_{t+1}}{S_{t}X_{t+1}^{2}}\ge 0$ because $Y_{t+1}+\gamma S_{t}X_{t+1}\ge 0$ when $X_{t+1}>0$.

We now show that $f'(X_{t+1})>0$ when $X_{t+1}<0$.

Indeed, $\left( 1+\gamma \right) S_{t}X_{t+1}<\varphi _{t}'(X_{t+1}-X_{t})S_{t}X_{t+1}$ when $X_{t+1}<0$ because $\varphi '_{t}(x)<1<1+\gamma$ when $x<0$.

Thus, $f'(X_{t+1})=\frac{Y_{t+1}+\varphi _{t}'(X_{t+1}-X_{t})S_{t}X_{t+1}}{S_{t}X_{t+1}^{2}}>\frac{Y_{t+1}+\left( 1+\gamma \right) S_{t}X_{t+1}}{S_{t}X_{t+1}^{2}}\ge 0$ because $Y_{t+1}+\left( 1+\gamma \right) S_{t}X_{t+1}\ge 0$ when $X_{t+1}<0$.

We can rewrite the time t constraint as:

• (i) $f(X_{t+1})\le \gamma$ when the optimum is $X_{t+1}>0$, and

• (ii) $g(X_{t+1})\le \delta$ when the optimum is $X_{t+1}<0$ with $g(X_{t+1})=-f(X_{t+1})$ and $\delta =-(1+\gamma )$.

(Case 1) At time t the optimum is $X_{t+1}^{i}>0$ with a binding constraint and $f'(X_{t+1})>0$.

Looking at the proof of Theorem 2, we see that the time t the optimization problem is

$$\begin{aligned} \upsilon (\gamma )=\sup _{\{X_{t+1}\in {\mathbb {R}}\}}E^{{\mathbb {P}}_{i}}[U_{i}(Y_{T+1}(X_{t+1})]\quad \text {subject to}\quad f(X_{t+1})=\gamma \end{aligned}$$

when the constraint is binding. As written, the Lagrangian is

$$\begin{aligned} {\mathscr {L}}=E^{{\mathbb {P}}_{i}}\left[ U_{i}\left( Y_{T+1}^{i}(X_{t+1})\right) \right] +\lambda _{t}\left( f(X_{t+1})-\gamma \right) . \end{aligned}$$

The first order (necessary and sufficient) condition is

$$\begin{aligned} \frac{dE^{{\mathbb {P}}_{i}}\left[ U_{i}\left( Y_{T+1}^{i}(X_{t+1})\right) \right] }{dX_{t+1}}+\lambda _{t}f'(X_{t+1})=0. \end{aligned}$$

From the proof of Theorem 2, expression (20), we have that $\kappa _{t}=\lambda _{t}f'(X_{t+1})$, i.e. $\lambda _{t}=\frac{\kappa _{t}}{f'(X_{t+1})}$.

But, standard results yield $\upsilon '(\gamma )=\lambda _{t}$, see Holmes [31], p. 39. Hence, $\upsilon '(\gamma )=\lambda _{t}=\frac{\kappa _{t}}{f'(X_{t+1})}$. By Lemma 1, $\kappa _{t}>0$ for $X_{t+1}>0$, and $f'(X_{t+1})>0$ implies that $\lambda _{t}>0$.

Also, it is well known that the value function is concave (in the parameter $\gamma$), see Holmes [31], p. 37. Hence, $\upsilon ''(\gamma )<0$, which implies that as $\gamma$ increases, $\upsilon '(\gamma )=$ $\lambda _{t}=\frac{\kappa _{t}}{f'(X_{t+1})}$ decreases. Since $f'(X_{t+1})>0$, all else equal, this implies that $\kappa _{t}>0$ decreases.

(Case 2) At time t the optimum is $X_{t+1}^{i}<0$ with a binding constraint and $g'(X_{t+1})=-f'(X_{t+1})<0$.

A similar argument to Case 1 yields that $\lambda _{t}=\frac{\kappa _{t}}{g'(X_{t+1})}$. By Lemma 1, $\kappa _{t}<0$ for ${X_{t+1}<0}$. So, ${g'}{(X_{t+1})<0}$ implies that $\lambda _{t}>0$. By standard results $\upsilon '(\delta )=\lambda _{t}$, hence $\upsilon '(\delta )=\frac{\kappa _{t}}{g'(X_{t+1})}>0$. By the concavity of $v(\delta )$, $\upsilon ''(\delta )<0$, which implies that as $\delta$ increases, ${\upsilon '(\delta )}$ $=\lambda _{t}=\frac{\kappa _{t}}{g'(X_{t+1})}$ decreases. Since ${g'(X_{t+1})<0}$, this implies that $\kappa _{t}<0$ becomes less negative, i.e. decreases in absolute value. But, $\delta =-\left( 1+\gamma \right)$ increasing implies that $\gamma$ decreases. Hence, as $\gamma$ increases, $\kappa _{t}<0$ becomes more negative, i.e. increases in absolute value.

This completes the proof. $\square$

1.9 Corollary 9

Proof

Given ${{\beta _{t}^{i}=\frac{-\kappa _{t}^{i}}{E_{t}^{{\mathbb {P}}_{i}}\left[ U_{i}'\left( Y_{T+1}^{i}\right) \right] }}}$, we have that $\gamma$ increasing (decreasing) implies:

• (i) $\kappa _{t}^{i}>0$ decreases (increases) when $X_{t+1}^{i}>0$, which implies $\beta _{t}^{i}<0$ decreases (increases) in absolute value, and

• (ii) $\kappa _{t}^{i}<0$ increases in absolute value when $X_{t+1}^{i}<0$, which means $\beta _{t}^{i}>0$ increases (note the negative sign in the above expression).

We note the asymmetry in the increase of $\gamma$.

In case (i) when $X_{t+1}^{i}>0$, $\gamma$ decreasing means the constraint is becoming more binding/restrictive (less borrowing possible).

In case (ii) when $X_{t+1}^{i}<0$, $\gamma$ increasing means the constraint is becoming more binding/restrictive (less short selling possible).

Thus, as $\gamma$ becomes more binding, in both cases the bubble increases in absolute value. $\square$

1.10 Theorem 12

Proof

Two cases to consider.

(Case 1) $X_{t+1}^{i}>0$ and $\beta _{t}^{i}<0$ with $\Delta S<0$ at time t.

Before the shock the borrowings are $Y_{t+1}^{i}=-\gamma S_{t}X_{t+1}^{i}<0$.

After the shock, the maximum borrowings are $-\gamma \left( S_{t}+\Delta S\right) X_{t+1}^{i}>-\gamma S_{t}X_{t+1}^{i}$. Hence, the constraint is violated and the change in wealth is negative.

Indeed,

$$\begin{aligned} \left( X_{t+1}^{i}S_{t}-\gamma S_{t}X_{t+1}^{i}\right) -\left( X_{t+1}^{i}S_{t}-\gamma \left( S_{t}+\Delta S\right) X_{t+1}^{i}\right) =\gamma X_{t+1}^{i}\Delta S<0. \end{aligned}$$

This implies that borrowers must sell shares $\Delta X^{i}<0$ to obtain cash to reduce their borrowings so that the constraint is not violated. Next, we determine the minimum shares they must sell to stay on the constraint.

After selling shares, the constraint is $-\gamma \left( S_{t}+\Delta S\right) \left( X_{t+1}^{i}+\Delta X^{i}\right) <0$.

The cash needed is $-\gamma \left( S_{t}+\Delta S\right) \left( X_{t+1}^{i}+\Delta X^{i}\right) +\gamma S_{t}X_{t+1}^{i}>0$.

From the liquidity cost of trading, the cash obtained from selling shares is

$$\begin{aligned} -\varphi {}_{t}(\Delta X^{i})\left( S_{t}+\Delta S\right) >0. \end{aligned}$$

From the above cash needed/obtained conditions, the solution is $\Delta X^{i}$ such that

$$\begin{aligned}&\gamma \left( S_{t}+\Delta S\right) \left( X_{t+1}^{i}+\Delta X^{i}\right) -\gamma S_{t}X_{t+1}^{i}=\varphi {}_{t}(\Delta X^{i})\left( S_{t}+\Delta S\right) \hbox {. Or,}\\&\gamma \left( S_{t}+\Delta S\right) X_{t+1}^{i}+\gamma \left( S_{t}+\Delta S\right) \Delta X^{i}-\gamma S_{t}X_{t+1}^{i}=\varphi {}_{t}(\Delta X^{i})\left( S_{t}+\Delta S\right) . \end{aligned}$$

This is equivalent to

$$\begin{aligned}\gamma \Delta SX_{t+1}^{i}+\gamma \left( S_{t}+\Delta S\right) \Delta X^{i}=\varphi {}_{t}(\Delta X^{i})\left( S_{t}+\Delta S\right) . \end{aligned}$$

Algebra yields

$$\begin{aligned}\gamma \frac{\Delta S}{S_{t}+\Delta S}X_{t+1}^{i}=\left( \varphi {}_{t}(\Delta X^{i})-\gamma \Delta X^{i}\right) . \end{aligned}$$

The left side is negative as $\Delta S<0$ and $X_{t+1}^{i}>0$.

A solution exists with $\Delta X^{i}<0$ if and only if $\varphi {}_{t}(\Delta X^{i})-\gamma \Delta X^{i}<0$.

(Case 2) $X_{t+1}^{i}<0$ and $\beta _{t}^{i}>0$ with $\Delta S>0$ at time t.

Before the shock the margin is $Y_{t+1}^{i}=-\left( 1+\gamma \right) S_{t}X_{t+1}^{i}>0$.

After the shock, the margin is $-\left( 1+\gamma \right) \left( S_{t}+\Delta S\right) X_{t+1}^{i}>-\left( 1+\gamma \right) S_{t}X_{t+1}^{i}$. Hence, the constraint is violated and the change in wealth is negative as in case 1 above.

This implies that short sellers must buy shares $\Delta X^{i}>0$ to reduce the short position so that the margin constraint is not violated. Next, we determine the minimum shares to buy to stay on the constraint.

After buying shares, the constraint is $-\left( 1+\gamma \right) \left( S_{t}+\Delta S\right) \left( X_{t+1}^{i}+\Delta X^{i}\right) >0$.

The cash reduction in the margin from purchase is

$$\begin{aligned}\left( 1+\gamma \right) \left( S_{t}+\Delta S\right) \left( X_{t+1}^{i}+\Delta X^{i}\right) -\left( 1+\gamma \right) S_{t}X_{t+1}^{i}>0. \end{aligned}$$

The cash needed to buy shares is $\varphi {}_{t}(\Delta X^{i})\left( S_{t}+\Delta S\right) >0$.

From the above cash reduction/needed conditions, the solution is $\Delta X^{i}$ such that

$$\begin{aligned}\left( 1+\gamma \right) \left( S_{t}+\Delta S\right) \left( X_{t+1}^{i}+\Delta X^{i}\right) -\left( 1+\gamma \right) S_{t}X_{t+1}^{i}=\varphi {}_{t}(\Delta X^{i})\left( S_{t}+\Delta S\right) . \end{aligned}$$

Or,

$$\begin{aligned}&\left( 1+\gamma \right) \left( S_{t}+\Delta S\right) X_{t+1}^{i}+\left( 1+\gamma \right) \left( S_{t}+\Delta S\right) \Delta X^{i}-\left( 1+\gamma \right) S_{t}X_{t+1}^{i}\\&\quad =\varphi {}_{t}(\Delta X^{i})\left( S_{t}+\Delta S\right) . \end{aligned}$$

Algebra yields

$$\begin{aligned}\left( 1+\gamma \right) \frac{\Delta S}{S_{t}+\Delta S}X_{t+1}^{i}=\left( \varphi {}_{t}(\Delta X^{i})-\left( 1+\gamma \right) \Delta X^{i}\right) . \end{aligned}$$

The left side is negative as $\Delta S>0$ and $X_{t+1}^{i}<0$.

A solution exists with $\Delta X^{i}>0$ if and only if $\varphi {}_{t}(\Delta X^{i})-\left( 1+\gamma \right) \Delta X^{i}<0$. $\square$

1.11 Corollary 13

Proof

(Step 1)

Consider the first case where the agent is borrowing to buy the stock. Here, before the shock, the agent’s wealth is

$$\begin{aligned} W_{t}^{i}=S_{t}X_{t+1}^{i}+Y_{t+1}^{i}=\left( 1-\gamma \right) S_{t}X_{t+1}^{i}>0 \end{aligned}$$

because $Y_{t+1}^{i}=-\gamma S_{t}X_{t+1}^{i}<0$. After the price shock of $\Delta S<0$ , the agent cannot satisfy the trading constraint. Given that the entire stock position is liquidated $(-X_{t+1}^{i}<0)$, the wealth after liquidation is

$$\begin{aligned} W_{t}^{i}(after)= & {} -\varphi {}_{t}(-X_{t+1}^{i})\left[ S_{t}+\Delta S\right] +Y_{t+1}^{i} \\= & {} -\varphi {}_{t}(-X_{t+1}^{i})\left[ S_{t}+\Delta S\right] -\gamma S_{t}X_{t+1}^{i}. \end{aligned}$$

Note here that the position in the mma is fixed. This new wealth is negative if $\varphi {}_{t}(-X_{t+1}^{i})\left[ S_{t}+\Delta S\right] >-\gamma S_{t}X_{t+1}^{i}$.

(Step 2)

Consider the second case where the agent has a margin account to short the stock. Here, before the shock, the agent’s wealth is

$$\begin{aligned} W_{t}^{i}=S_{t}X_{t+1}^{i}+Y_{t+1}^{i}=-\gamma S_{t}X_{t+1}^{i}>0 \end{aligned}$$

because $Y_{t+1}^{i}=-\left( 1+\gamma \right) S_{t}X_{t+1}^{i}>0$. After the price shock of $\Delta S>0$, the agent cannot satisfy the trading constraint. Given the entire mma position is liquidated to buy the stock to cover the short position, the wealth after buying back the stock is

$$\begin{aligned} W_{t}^{i}(after)=\left[ S_{t}+\Delta S\right] (X_{t+1}^{i}+Z) \end{aligned}$$

where $Z>0$ is the solution to $\varphi {}_{t}(Z)\left[ S_{t}+\Delta S\right] =-\left( 1+\gamma \right) S_{t}X_{t+1}^{i}$. Note in this case the position in the shorted shocks is fixed. This new wealth is negative if $Z<-X_{t+1}^{i}$. This completes the proof. $\square$

1.12 Theorem 14

Proof

By expression (19), we have three cases.

• (i) As ${\mathbb {I}}_{\beta _{t}}$ increases, $\left| \beta _{t}^{i}\right|$ is increasing, which makes $\beta _{t}^{i}<0$ more likely for $\Delta X^{i}<0$ and it makes $\beta _{t}^{i}>0$ more likely for $\Delta X^{i}>0$. This gives the result.

• (ii) As markets become more illiquid, $\varphi {}_{t}(\Delta X^{i})$ increases for all $\Delta X^{i}\ne 0$. See Lemma (4). This makes $\varphi {}_{t}(\Delta X^{i})>\gamma \Delta X^{i}$ more likely for $\Delta X^{i}<0$ and it makes $\varphi {}_{t}(\Delta X^{i})>(1+\gamma )\Delta X^{i}$ more likely for $\Delta X^{i}>0$. This gives the desired result.

• (iii) As the constraints become more binding, this means that $\gamma$ decreases for $X_{t+1}^{i}>0$ and that $1+\gamma$ increases for $X_{t+1}^{i}<0$. This makes $0>\varphi {}_{t}(\Delta X^{i})>\gamma \Delta X^{i}$ less likely for $\Delta X^{i}<0$ and it makes $\varphi {}_{t}(\Delta X^{i})>(1+\gamma )\Delta X^{i}>0$ less likely for $\Delta X^{i}>0$. This gives the desired result.

$\square$

1.13 Corollary 15

Proof

(Step 1)

Given$X_{t+1}^{i}>0$ and $\beta _{t}^{i}<0$, let the price shock be $\Delta S<0$ at time t. Then, by the previous Theorem (12) the minimum sale $\Delta X^{i}<0$ necessary to satisfy the borrowing constraint satisfies $\Delta X^{i}=\frac{1}{\left( \frac{\varphi {}_{t}(\Delta X^{i})}{\Delta X^{i}}-\gamma \right) }\gamma \frac{\Delta S}{S_{t}+\Delta S}X_{t+1}^{i}>\frac{1}{\left( 1-\gamma \right) }\gamma \frac{\Delta S}{S_{t}+\Delta S}X_{t+1}^{i}.$ The right side of this inequality is the trade size in a market with no illiquidities where $\varphi {}_{t}(\Delta X^{i})=\Delta X^{i}$.

(Step 2)

Given $X_{t+1}^{i}<0$ and $\beta _{t}^{i}>0$, let the price shock be $\Delta S>0$. Then, by the previous theorem the minimum purchase $\Delta X^{i}>0$ necessary to satisfy the margin constraint satisfies $\Delta X^{i}=\frac{1}{\left( \frac{\varphi {}_{t}(\Delta X^{i})}{\Delta X^{i}}-\left( 1+\gamma \right) \right) }\left( 1+\gamma \right) \frac{\Delta S}{S_{t}+\Delta S}X_{t+1}^{i}>\frac{1}{\gamma }\left( 1+\gamma \right) \frac{\Delta S}{S_{t}+\Delta S}X_{t+1}^{i}.$ The right side of this inequality is the trade size in a market with no illiquidities where $\varphi {}_{t}(\Delta X^{i})=\Delta X^{i}$.

In both steps above, the loss in wealth is larger than in the economy with liquidity costs.

(Step 3)

The change is wealth is computed by recognizing that the constraint is satisfied both before and after the trades. $\square$