% paper/appendices/proofs.tex — UQO WP 27. Proofs of the propositions of Section "A stylized model".
\section{Proofs}
\label{app:proofs}

Throughout, A1--A4 hold and the notation is that of Section~\ref{sec:theory}:
$n_g=\phi_gN$, $\Phi=\sum_g\phi_g$, $H=\sum_g\phi_g^2$, $\sigma_s^2=\sigma_v^2+\sigma_\eta^2+\sigma_u^2$,
$\bar\varepsilon=N^{-1}\sum_i\varepsilon_i$ and $\bar Y=Y/(\beta N)$.

\subsection*{Two lemmas}

\begin{lemma}[Flow decomposition]
\label{lem:flow}
$Y=\beta N\,(v+\bar\varepsilon)+z$ and
$\operatorname{Var}(\bar\varepsilon)=\sigma_\eta^2\big[H+(1-\Phi)/N\big]+\sigma_u^2/N$. Consequently
$\operatorname{Var}(\bar Y-v)=\Omega$ as defined in \eqref{eq:lambda}.
\end{lemma}

\begin{proof}
By A2, $\sum_ix_i=\beta\sum_i(v+\varepsilon_i)=\beta N(v+\bar\varepsilon)$. Let $I$ be the set of
traders on private models, $|I|=N(1-\Phi)$. Then
$\sum_i\varepsilon_i=\sum_{g=1}^Gn_g\eta_g+\sum_{i\in I}\eta_i+\sum_{i=1}^Nu_i$, a sum of
$G+|I|+N$ independent centered Gaussians (A1), whence
$\operatorname{Var}(\sum_i\varepsilon_i)=\sigma_\eta^2\sum_gn_g^2+\sigma_\eta^2N(1-\Phi)+N\sigma_u^2$.
Dividing by $N^2$ and using $n_g^2/N^2=\phi_g^2$ gives the variance of $\bar\varepsilon$. Finally
$\bar Y-v=\bar\varepsilon+z/(\beta N)$ with $z$ independent of $\bar\varepsilon$, so
$\operatorname{Var}(\bar Y-v)=\operatorname{Var}(\bar\varepsilon)+\sigma_z^2/(\beta^2N^2)=\Omega$.
\end{proof}

\begin{lemma}[Projection]
\label{lem:proj}
$\lambda=\tau/(\beta N)$, $\tau=\sigma_v^2/(\sigma_v^2+\Omega)$, $P=\tau\bar Y$, and
$\operatorname{Var}(P-v)=\operatorname{Var}(v\mid Y)=\sigma_v^2(1-\tau)$.
\end{lemma}

\begin{proof}
$(v,Y)$ is jointly Gaussian, so $\mathbb E[v\mid Y]=\lambda Y$ with
$\lambda=\operatorname{Cov}(v,Y)/\operatorname{Var}(Y)$ and
$\operatorname{Var}(v\mid Y)=\sigma_v^2-\operatorname{Cov}(v,Y)^2/\operatorname{Var}(Y)$. By Lemma~\ref{lem:flow},
$\operatorname{Cov}(v,Y)=\beta N\sigma_v^2$ and $\operatorname{Var}(Y)=\beta^2N^2(\sigma_v^2+\Omega)$. Hence
$\tau=1-\operatorname{Var}(v\mid Y)/\sigma_v^2=\operatorname{Cov}(v,Y)^2/(\sigma_v^2\operatorname{Var}(Y))
=\sigma_v^2/(\sigma_v^2+\Omega)$ and $\lambda=\sigma_v^2/[\beta N(\sigma_v^2+\Omega)]=\tau/(\beta N)$, so
$P=\lambda Y=\tau\bar Y$. For the pricing error,
\begin{align*}
\operatorname{Var}(P-v)&=\lambda^2\operatorname{Var}(Y)-2\lambda\operatorname{Cov}(v,Y)+\sigma_v^2
=\frac{\operatorname{Cov}(v,Y)^2}{\operatorname{Var}(Y)}-2\frac{\operatorname{Cov}(v,Y)^2}{\operatorname{Var}(Y)}+\sigma_v^2\\
&=\sigma_v^2-\frac{\operatorname{Cov}(v,Y)^2}{\operatorname{Var}(Y)}=\operatorname{Var}(v\mid Y),
\end{align*}
which equals $\sigma_v^2(1-\tau)$ by definition of $\tau$.
\end{proof}

\subsection*{Proof of Proposition~\ref{prop:info}}

With a single model of $n=\phi N$ users, $H=\phi^2$ and $\Phi=\phi$, so Lemma~\ref{lem:flow} gives
\eqref{eq:tau_phi}. Since $\tau=\sigma_v^2/(\sigma_v^2+\Omega)$ is strictly decreasing in $\Omega$, every
monotonicity claim about $\tau$ is the reverse claim about $\Omega$.

\emph{(i) Discrete monotonicity.} Writing $\Omega$ as a function of the integer $n$,
$\Omega(n)=\sigma_\eta^2\,(n^2-n+N)/N^2+C$ with $C=\sigma_u^2/N+\sigma_z^2/(\beta^2N^2)$. Then
\[
\Omega(n+1)-\Omega(n)=\frac{\sigma_\eta^2}{N^2}\Big[(n+1)^2-(n+1)-n^2+n\Big]=\frac{2n\,\sigma_\eta^2}{N^2},
\]
which is zero for $n=0$ and strictly positive for every $n\ge1$ when $\sigma_\eta^2>0$. Hence
$\Omega(0)=\Omega(1)<\Omega(2)<\dots<\Omega(N)$ and $\tau$ is strictly decreasing on $\{1/N,\dots,1\}$,
with $\tau(0)=\tau(1/N)$. For $N=1$ the grid is $\{0,1\}$ and $\Omega(0)=\Omega(1)$. In the continuous
relaxation, $\Omega'(\phi)=\sigma_\eta^2(2\phi-1/N)$, which is strictly positive if and only if
$\phi>1/(2N)$; so $\partial\tau/\partial\phi=-\sigma_v^2\Omega'(\phi)/(\sigma_v^2+\Omega)^2<0$ exactly
there. (For $0<\phi<1/(2N)$ the relaxation is slightly increasing in $\phi$; no integer population lies
in this region, since it corresponds to fewer than one-half of a trader using the model.)

\emph{(ii) Uniform bound and limit.} For every $N$ and $\phi\in[0,1]$,
$\Omega(\phi)-\sigma_\eta^2\phi^2=\sigma_\eta^2(1-\phi)/N+\sigma_u^2/N+\sigma_z^2/(\beta^2N^2)\ge0$, so
$\Omega(\phi)\ge\sigma_\eta^2\phi^2$ and $\tau(\phi)\le\sigma_v^2/(\sigma_v^2+\phi^2\sigma_\eta^2)=\tau_\infty(\phi)$,
which is strictly below one when $\phi\sigma_\eta>0$. The difference $\Omega(\phi)-\sigma_\eta^2\phi^2$ is
$O(1/N)$ for fixed $\phi$ and fixed $\sigma_z^2$, so $\tau(\phi)\to\tau_\infty(\phi)$.

\emph{(iii) Private-model benchmark.} At $\phi=0$, $\Omega=(\sigma_\eta^2+\sigma_u^2)/N+\sigma_z^2/(\beta^2N^2)$,
which tends to zero whenever $\sigma_z^2/N^2\to0$; then $\tau\to1$. If instead noise trading scales
with the market, $\sigma_z^2=cN^2$, the limit is $\sigma_v^2/(\sigma_v^2+c/\beta^2)<1$ even at $\phi=0$;
this is the usual noisy-rational-expectations bound and is unrelated to model sharing. \qed

\subsection*{Proof of Proposition~\ref{prop:corr}}

\emph{(i)} By Lemma~\ref{lem:proj}, $\operatorname{Var}(P-v)=\sigma_v^2(1-\tau)=\sigma_v^2\Omega/(\sigma_v^2+\Omega)$,
an increasing function of $\Omega$; the monotonicity in $\phi$ and the uniform lower bound follow from
parts (i)--(ii) of Proposition~\ref{prop:info} since $y\mapsto\sigma_v^2y/(\sigma_v^2+y)$ is increasing.

\emph{(ii)} By A2, $\operatorname{Corr}(x_i,x_j)=\operatorname{Corr}(s_i,s_j)$ since $\beta>0$. By A1,
$\operatorname{Var}(s_i)=\sigma_s^2$ for every $i$, and for $i\neq j$,
$\operatorname{Cov}(s_i,s_j)=\sigma_v^2+\operatorname{Cov}(\eta_{g(i)},\eta_{g(j)})$, which equals
$\sigma_v^2+\sigma_\eta^2$ if $g(i)=g(j)\neq\varnothing$ and $\sigma_v^2$ otherwise (two private models, or
two different shared models, have independent $\eta$'s). This gives $\rho_1$ and $\rho_0$, and
$\rho_1-\rho_0=\sigma_\eta^2/\sigma_s^2>0$ if and only if $\sigma_\eta^2>0$.

\emph{(iii)} Among the $N(N-1)$ ordered pairs $i\neq j$, exactly $n(n-1)$ have both members on the
shared model; all others have correlation $\rho_0$. Hence
$\bar\rho=\rho_0+w(\rho_1-\rho_0)$ with $w=n(n-1)/[N(N-1)]=\phi(N\phi-1)/(N-1)$. As a function of
real $\phi$, $w'(\phi)=(2N\phi-1)/(N-1)$ and $w''(\phi)=2N/(N-1)>0$: $w$ is convex and increasing
for $\phi>1/(2N)$. On the integer grid, $w(n+1)-w(n)=2n/[N(N-1)]>0$ for $n\ge1$. As $N\to\infty$,
$w\to\phi^2$.

\emph{(iv)} Let $(x_i,x_j)$ be centered jointly Gaussian with correlation $\rho\in(-1,1)$. Write
$x_j/\sigma_j=\rho\,x_i/\sigma_i+\sqrt{1-\rho^2}\,\xi$ with $\xi$ standard normal independent of $x_i$.
The classical orthant formula gives $\Pr(x_i>0,x_j>0)=\tfrac14+\arcsin(\rho)/(2\pi)$; by symmetry
$\Pr(x_i<0,x_j<0)$ is the same, so $S(\rho)=\Pr(x_ix_j>0)=\tfrac12+\arcsin(\rho)/\pi$.\footnote{For
completeness: $\Pr(x_i>0,x_j>0)=\Pr(\vartheta\in(0,\pi/2)\text{ and }\vartheta\in(\alpha-\pi/2,\alpha+\pi/2))$
where $\vartheta$ is the uniformly distributed angle of a standard bivariate normal vector and
$\cos\alpha=\rho$; the two arcs overlap on a set of measure $\pi-\alpha$, so the probability is
$(\pi-\arccos\rho)/(2\pi)=\tfrac14+\arcsin(\rho)/(2\pi)$.} Since $\arcsin$ is strictly increasing on
$[-1,1]$, so is $S$. Averaging over ordered pairs with the same count as in (iii) gives \eqref{eq:sync}.
Because $\bar\rho$ is affine in $w$ with slope $\rho_1-\rho_0>0$ (when $\sigma_\eta^2>0$), we may
regard $\bar S$ as a function of $\bar\rho$ with
\[
\frac{d\bar S}{d\bar\rho}=\frac{1}{\pi}\cdot\frac{\arcsin\rho_1-\arcsin\rho_0}{\rho_1-\rho_0}>0 .
\]
For the Jensen bound, $(\arcsin)''(\rho)=\rho(1-\rho^2)^{-3/2}\ge0$ on $[0,1)$, and $\rho_0,\rho_1\ge0$,
so $(1-w)\arcsin\rho_0+w\arcsin\rho_1\ge\arcsin\big((1-w)\rho_0+w\rho_1\big)=\arcsin\bar\rho$, with
equality if and only if $\rho_0=\rho_1$ or $w\in\{0,1\}$; for $0<w<1$ equality thus requires
$\sigma_\eta^2=0$.

\emph{The placebo.} If $\sigma_\eta^2=0$ then $\Omega$ does not depend on $\phi$ (the bracket in
\eqref{eq:tau_phi} is multiplied by zero), $\rho_1=\rho_0$, and $w$ drops out of \eqref{eq:rhobar} and
\eqref{eq:sync}. Hence none of the four objects varies with $\phi$. This shows that the restriction
$\sigma_\eta^2>0$ is necessary for every strict claim in the proposition. \qed

\subsection*{Proof of Proposition~\ref{prop:diversity}}

\emph{(i)} By Lemma~\ref{lem:flow}, $\Omega=\sigma_\eta^2[H+(1-\Phi)/N]+\sigma_u^2/N+\sigma_z^2/(\beta^2N^2)$;
for fixed $\Phi$ the shares enter only through $H$, and $\partial\Omega/\partial H=\sigma_\eta^2>0$, so
$\partial\tau/\partial H=-\sigma_v^2\sigma_\eta^2/(\sigma_v^2+\Omega)^2<0$. Setting $\Phi=1$ and $H=1/M_{\mathrm{eff}}$
gives the stated form.

\emph{(ii)} By the Cauchy--Schwarz inequality,
$\Phi^2=\big(\sum_g\phi_g\cdot1\big)^2\le\big(\sum_g\phi_g^2\big)\big(\sum_g1^2\big)=GH$, with equality if
and only if $(\phi_g)$ is proportional to $(1,\dots,1)$, i.e.\ $\phi_g=\Phi/G$. For the upper bound,
$\Phi^2=\sum_g\phi_g^2+\sum_{g\neq g'}\phi_g\phi_{g'}\ge H$, with equality if and only if all cross
products vanish, i.e.\ at most one share is positive. In the integer version one minimizes $\sum_gn_g^2$
subject to $\sum_gn_g=\Phi N$: if some $n_a\ge n_b+2$, moving one trader from $a$ to $b$ changes the
sum by $(n_a-1)^2+(n_b+1)^2-n_a^2-n_b^2=-2(n_a-n_b-1)<0$, so at a minimum all $n_g$ differ by at most one.

\emph{(iii)} Combine (i) and (ii): $H_{\text{single}}=\Phi^2\ge H_{\text{mixed}}\ge\Phi^2/G=H_{\text{uniform}}$
and $\tau$ is decreasing in $H$. Strictness: $H_{\text{mixed}}<\Phi^2$ as soon as two shares are
positive, and $\partial\tau/\partial H<0$ requires $\sigma_\eta^2>0$.

\emph{(iv)} Ordered pairs with both members on the same shared model number
$\sum_gn_g(n_g-1)=N^2H-N\Phi$; dividing by $N(N-1)$ gives $w=(NH-\Phi)/(N-1)$. For fixed $\Phi$, $w$
is increasing in $H$, and $\bar\rho$ and $\bar S$ are increasing in $w$ by the proof of
Proposition~\ref{prop:corr}. \qed

\subsection*{Proof of Proposition~\ref{prop:feedback} and Corollary~\ref{cor:llmtrend}}

Under A5 the round's price change solves $\Delta P=\lambda(Y_1+\gamma\Delta P)$.

\emph{(i)} If $\lambda\gamma\neq1$ the unique solution is $\Delta P^\ast=\lambda Y_1/(1-\lambda\gamma)=\kappa\lambda Y_1$.
The iterated response satisfies $\Delta P^{(k+1)}-\Delta P^\ast=\lambda\gamma(\Delta P^{(k)}-\Delta P^\ast)$,
so $\Delta P^{(k)}-\Delta P^\ast=(\lambda\gamma)^k(\Delta P^{(0)}-\Delta P^\ast)$, which converges to zero
for every starting point if and only if $|\lambda\gamma|<1$; with $\gamma\ge0$ this is $\lambda\gamma<1$.
Then $\operatorname{Var}(\Delta P)=\kappa^2\lambda^2\operatorname{Var}(Y_1)$, i.e.\ amplification by
$\kappa^2=1/(1-\lambda\gamma)^2$ relative to the no-feedback price change $\lambda Y_1$.

\emph{(ii)} If $\lambda\gamma=1$ the equation reads $\Delta P=\lambda Y_1+\Delta P$, impossible for
$Y_1\neq0$, and the iteration gives $\Delta P^{(k)}=\Delta P^{(0)}+k\lambda Y_1$, unbounded. If
$\lambda\gamma>1$ the algebraic solution $\Delta P^\ast=\lambda Y_1/(1-\lambda\gamma)$ exists but has the
sign of $-Y_1$ (a price that falls when the flow is a net purchase), and since $(\lambda\gamma)^k\to\infty$
the iteration diverges from every $\Delta P^{(0)}\neq\Delta P^\ast$; in particular from the natural start
$\Delta P^{(0)}=\lambda Y_1$. This is the counterexample announced in the text: the closed form
$1/(1-\lambda\gamma)^2$ is meaningless outside $\lambda\gamma<1$.

\emph{(iii)} $\Delta P^\ast=\kappa\lambda Y_1$ is a strictly monotone function of $Y_1$ (for $\kappa>0$), so
$\sigma(\Delta P^\ast)=\sigma(Y_1)$ and $\operatorname{Var}(v\mid P_1)=\operatorname{Var}(v\mid Y_1)$: $\tau$ is unchanged.
Using $\lambda^2\operatorname{Var}(Y_1)=\lambda\operatorname{Cov}(v,Y_1)=\sigma_v^2\tau$ (Lemma~\ref{lem:proj}),
\[
\operatorname{Var}(\kappa\lambda Y_1-v)=\kappa^2\lambda^2\operatorname{Var}(Y_1)-2\kappa\lambda\operatorname{Cov}(v,Y_1)+\sigma_v^2
=\sigma_v^2\big[\kappa^2\tau-2\kappa\tau+1\big]=\sigma_v^2\big[(1-\tau)+\tau(\kappa-1)^2\big].
\]
On $\gamma\in[0,1/\lambda)$, $\kappa$ increases from $1$ to $+\infty$, so $(\kappa-1)^2$ is increasing in $\gamma$.

\emph{Corollary.} Under A6 the total flow is $Y_1+\gamma_T\Delta P+\sum_i\kappa_{g(i)}\Delta P
=Y_1+\big(\gamma_T+\sum_gn_g\kappa_g\big)\Delta P=Y_1+\big(\gamma_T+N\sum_g\phi_g\kappa_g\big)\Delta P$,
with $\kappa_\varnothing\equiv0$ for private-model traders (any such loading can be folded into $\gamma_T$).
Applying (i)--(ii) with $\gamma=\gamma_T+N\sum_g\phi_g\kappa_g$ gives the statements; with a single
model, $\gamma(\phi)=\gamma_T+N\phi\kappa$ is increasing in $\phi$ when $\kappa>0$, $\kappa^2(\phi)$ is
increasing in $\gamma$ hence in $\phi$, and $\lambda\gamma(\phi)<1$ if and only if
$\phi<\phi^\ast=(1/\lambda-\gamma_T)/(N\kappa)$. If $\kappa=0$, $\gamma$ does not depend on $\phi$. With
several models, $\sum_g\phi_g\kappa_g$ is a share-weighted average that lies between $\Phi\min_g\kappa_g$
and $\Phi\max_g\kappa_g$; if the $\kappa_g$ differ in sign it can be smaller in absolute value than
$\Phi\kappa_g$ for every $g$. \qed

\subsection*{Proof of Proposition~\ref{prop:spiral}}

Let $R\equiv\{\Delta P:\Delta P<-b/X\}$ be the region in which $F>0$. On $R$,
$F(\Delta P)=(-X\Delta P-b)/m$ and the fixed-point equation $\Delta P=\Delta P_0-\lambda F(\Delta P)$ reads
\begin{equation}
  \Delta P=\Delta P_0+a\,\Delta P+\frac{\lambda b}{m},\qquad a=\frac{\lambda X}{m}.
  \label{eq:fp_affine}
\end{equation}
Off $R$, $F=0$ and the equation reads $\Delta P=\Delta P_0$.

\emph{(i)} If $\Delta P_0\ge-b/X$, then $\Delta P=\Delta P_0\notin R$ solves the equation, and the
cascade defined below stops at its first step since $F(\Delta P_0)=0$. For $a<1$ it is the only fixed
point: a solution in $R$ would satisfy \eqref{eq:fp_affine}, whose candidate
$(\Delta P_0+\lambda b/m)/(1-a)$ lies in $R$ if and only if $\Delta P_0+\lambda b/m<-(b/X)(1-a)=-b/X+\lambda b/m$,
i.e.\ $\Delta P_0<-b/X$, which fails. For $a>1$ and $b>0$ the same algebra shows that the candidate
\emph{does} lie in $R$ whenever $\Delta P_0>-b/X$ (dividing by $1-a<0$ reverses the inequality): a
second, self-fulfilling fixed point exists in which the cohort is forced to sell because the price has
fallen enough to exhaust the buffer, and the price has fallen because the cohort sells. This is the
multiplicity familiar from \citet{brunnermeier2009}. We select the cascade limit, which is
$\Delta P_0$; the simulator's forced-liquidation rule is likewise sequential and cannot jump to the
self-fulfilling point.

\emph{(ii)} Suppose $\Delta P_0<-b/X$ and $a<1$. The candidate $\Delta P=\Delta P_0$ is in $R$, where
$F>0$, so it is not a solution; the affine equation \eqref{eq:fp_affine} has the unique solution
$\Delta P^\ast=(\Delta P_0+\lambda b/m)/(1-a)$, which lies in $R$ by the equivalence just shown. With $b=0$
this is $\Delta P_0/(1-\lambda X/m)$ and the multiplier $1/(1-a)$ exceeds one. For the cascade, let
$\Delta P^{(0)}=\Delta P_0$ and $\Delta P^{(k)}=\Delta P_0-\lambda F(\Delta P^{(k-1)})$. Since
$\Delta P^{(0)}\in R$, $\Delta P^{(1)}-\Delta P^{(0)}=-\lambda F(\Delta P_0)<0$, so $\Delta P^{(1)}\in R$; by
induction all iterates are in $R$ and satisfy $\Delta P^{(k)}-\Delta P^\ast=a(\Delta P^{(k-1)}-\Delta P^\ast)$,
hence
\begin{equation}
  \Delta P^{(k)}-\Delta P^\ast=a^k(\Delta P_0-\Delta P^\ast)=-a^{k+1}\,\frac{|\Delta P_0|-b/X}{1-a},
  \label{eq:cascade_path}
\end{equation}
where we used $\Delta P_0-\Delta P^\ast=-\frac{\lambda}{m}\frac{X\Delta P_0+b}{1-a}=a\frac{|\Delta P_0|-b/X}{1-a}$.
The sequence is decreasing and converges to $\Delta P^\ast$ since $a<1$.

\emph{(iii)} Suppose $\Delta P_0<-b/X$ and $a\ge1$. No solution exists off $R$ (it would have to be
$\Delta P_0\in R$). On $R$, if $a=1$ then $\lambda b/m=(b/X)(\lambda X/m)=b/X$ and \eqref{eq:fp_affine} reduces to
$0=\Delta P_0+b/X$, which is false because $\Delta P_0<-b/X$; if $a>1$ the unique affine solution $(\Delta P_0+\lambda b/m)/(1-a)$ is the ratio of a
negative numerator and a negative denominator, hence positive, so it is not in $R$. Thus there is no
fixed point. Along the cascade, the increments obey
$\Delta P^{(k+1)}-\Delta P^{(k)}=-\lambda[F(\Delta P^{(k)})-F(\Delta P^{(k-1)})]=a(\Delta P^{(k)}-\Delta P^{(k-1)})$
as long as the iterates remain in $R$, which they do because the first increment is negative and the
ratio $a$ is positive; with $a\ge1$ the increments do not shrink, so $\Delta P^{(k)}\to-\infty$. In the
simulator the sequence stops at the price band.

\emph{(iv)} $\sum_{i\in L}x_i=\beta\sum_{i\in L}s_i$ is a centered Gaussian (A1--A2) with
\[
\operatorname{Var}\Big(\sum_{i\in L}s_i\Big)=\sum_{i\in L}\operatorname{Var}(s_i)+\sum_{i\neq j\in L}\operatorname{Cov}(s_i,s_j)
=\sigma_s^2\big[N_L+N_L(N_L-1)\bar\rho_L\big],
\]
by definition of the cohort-average correlation $\bar\rho_L$. Hence $X=|Z|$ with
$Z\sim\mathcal N(0,\sigma_X^2)$, $\sigma_X=\beta\sigma_s\sqrt{N_L[1+(N_L-1)\bar\rho_L]}$. Standard facts give
$\mathbb E|Z|=\sigma_X\sqrt{2/\pi}$ and $\Pr(|Z|\ge c)=2\bar\Phi(c/\sigma_X)$ for $c>0$, both strictly
increasing in $\sigma_X$, which is strictly increasing in $\bar\rho_L$ when $N_L>1$. Finally, applying
Proposition~\ref{prop:corr}(iii) to the cohort (with $N_L$ in place of $N$ and the cohort's own
shared-model share), $\bar\rho_L$ is increasing in that share on its admissible grid. The spiral
probability is $\Pr(a\ge1)=\Pr(X\ge m/\lambda)$, which gives the stated expression with $c=m/\lambda$.

\emph{Remark on the net position.} If the cohort contains both longs and shorts, a negative shock forces
sales only from the longs, whose gross exposure $\sum_{i\in L}\max\{x_i,0\}$ is at least
$\max\{\sum_{i\in L}x_i,0\}$; using the net position in the spiral coefficient therefore understates
$a$ for a shock against the cohort's net side, and the spiral condition stated is sufficient, not
necessary. \qed

\subsection*{Proof of Proposition~\ref{prop:safeguards}}

\emph{(i)} If $X\le L<m/\lambda$ then $a=\lambda X/m\le\lambda L/m<1$ for every realization, so case (iii)
of Proposition~\ref{prop:spiral} cannot occur and, by case (ii), the multiplier is at most
$1/(1-\lambda L/m)$. For the informational cost, let $g(s)=\min\{\max\{\beta s,-\ell\},\ell\}$, which is
Lipschitz with $g'(s)=\beta\,\mathbf 1\{|\beta s|<\ell\}$ almost everywhere. Since $(v,s_i)$ is jointly
Gaussian, write $v=\theta s_i+e$ with $\theta=\operatorname{Cov}(v,s_i)/\operatorname{Var}(s_i)$ and $e$ independent of
$s_i$. Then $\operatorname{Cov}(v,g(s_i))=\theta\operatorname{Cov}(s_i,g(s_i))=\theta\operatorname{Var}(s_i)\mathbb E[g'(s_i)]$ by
Stein's lemma, i.e.\ $\operatorname{Cov}(v,\tilde x_i)=\operatorname{Cov}(v,s_i)\,\beta\Pr(|\beta s_i|<\ell)=\Pr(|\beta s_i|<\ell)\operatorname{Cov}(v,x_i)$,
and $\Pr(|\beta s_i|<\ell)<1$ for any finite $\ell$. We do not claim a closed form for $\tau$ under
clipping, since the clipped flow is non-Gaussian and the market maker's pricing rule is then nonlinear.

\emph{(ii)} Under A7, after the halt the cascade is that of Proposition~\ref{prop:spiral} with $a$
replaced by $a_h\equiv\psi(h)a$ and the same starting point $\Delta P_0$ (nothing trades during the
halt). If $a_h<1$, part (ii) of that proposition gives convergence to
$\Delta P^\ast_h=(\Delta P_0+\lambda b/m)/(1-a_h)$, finite regardless of whether $a<1$. If also $a<1$,
then $1-a_h\ge1-a>0$ and the common numerator is negative, so $|\Delta P^\ast_h|\le|\Delta P^\ast|$, with
strict inequality if $\psi(h)<1$.

\emph{(iii)} From \eqref{eq:cascade_path}, $|\Delta P^{(k)}-\Delta P^\ast|=a^{k+1}(|\Delta P_0|-b/X)/(1-a)$,
which is at most $\epsilon$ if and only if $k+1\ge\log\big(\epsilon(1-a)/(|\Delta P_0|-b/X)\big)/\log a$
(the division by $\log a<0$ reverses the inequality). Hence
$T_\epsilon(a)=\max\{0,\lceil\log(\epsilon(1-a)/(|\Delta P_0|-b/X))/\log a\rceil-1\}$, which is the stated
formula when $b=0$. Under the halt, rounds $1,\dots,h$ leave the price at $\Delta P_0$ and the cascade
with coefficient $a_h$ then starts from $\Delta P_0$, so the first round within $\epsilon$ of
$\Delta P^\ast_h$ is $h+T_\epsilon(a_h)$. Since $T_\epsilon\ge0$, $T^h_\epsilon\ge h$; if $\psi(h)=1$ then
$a_h=a$, $\Delta P^\ast_h=\Delta P^\ast$ and $T^h_\epsilon=T_\epsilon(a)+h$. When $a\ge1$ and $a_h<1$ the
comparison is $T^h_\epsilon<\infty=T_\epsilon(a)$, where the latter is infinite because the no-halt cascade
has no finite limit (Proposition~\ref{prop:spiral}(iii)). \qed