1% paper/appendices/proofs.tex — UQO WP 27. Proofs of the propositions of Section "A stylized model".2\section{Proofs}3\label{app:proofs}45Throughout, A1--A4 hold and the notation is that of Section~\ref{sec:theory}:6$n_g=\phi_gN$, $\Phi=\sum_g\phi_g$, $H=\sum_g\phi_g^2$, $\sigma_s^2=\sigma_v^2+\sigma_\eta^2+\sigma_u^2$,7$\bar\varepsilon=N^{-1}\sum_i\varepsilon_i$ and $\bar Y=Y/(\beta N)$.89\subsection*{Two lemmas}1011\begin{lemma}[Flow decomposition]12\label{lem:flow}13$Y=\beta N\,(v+\bar\varepsilon)+z$ and14$\operatorname{Var}(\bar\varepsilon)=\sigma_\eta^2\big[H+(1-\Phi)/N\big]+\sigma_u^2/N$. Consequently15$\operatorname{Var}(\bar Y-v)=\Omega$ as defined in \eqref{eq:lambda}.16\end{lemma}1718\begin{proof}19By A2, $\sum_ix_i=\beta\sum_i(v+\varepsilon_i)=\beta N(v+\bar\varepsilon)$. Let $I$ be the set of20traders on private models, $|I|=N(1-\Phi)$. Then21$\sum_i\varepsilon_i=\sum_{g=1}^Gn_g\eta_g+\sum_{i\in I}\eta_i+\sum_{i=1}^Nu_i$, a sum of22$G+|I|+N$ independent centered Gaussians (A1), whence23$\operatorname{Var}(\sum_i\varepsilon_i)=\sigma_\eta^2\sum_gn_g^2+\sigma_\eta^2N(1-\Phi)+N\sigma_u^2$.24Dividing by $N^2$ and using $n_g^2/N^2=\phi_g^2$ gives the variance of $\bar\varepsilon$. Finally25$\bar Y-v=\bar\varepsilon+z/(\beta N)$ with $z$ independent of $\bar\varepsilon$, so26$\operatorname{Var}(\bar Y-v)=\operatorname{Var}(\bar\varepsilon)+\sigma_z^2/(\beta^2N^2)=\Omega$.27\end{proof}2829\begin{lemma}[Projection]30\label{lem:proj}31$\lambda=\tau/(\beta N)$, $\tau=\sigma_v^2/(\sigma_v^2+\Omega)$, $P=\tau\bar Y$, and32$\operatorname{Var}(P-v)=\operatorname{Var}(v\mid Y)=\sigma_v^2(1-\tau)$.33\end{lemma}3435\begin{proof}36$(v,Y)$ is jointly Gaussian, so $\mathbb E[v\mid Y]=\lambda Y$ with37$\lambda=\operatorname{Cov}(v,Y)/\operatorname{Var}(Y)$ and38$\operatorname{Var}(v\mid Y)=\sigma_v^2-\operatorname{Cov}(v,Y)^2/\operatorname{Var}(Y)$. By Lemma~\ref{lem:flow},39$\operatorname{Cov}(v,Y)=\beta N\sigma_v^2$ and $\operatorname{Var}(Y)=\beta^2N^2(\sigma_v^2+\Omega)$. Hence40$\tau=1-\operatorname{Var}(v\mid Y)/\sigma_v^2=\operatorname{Cov}(v,Y)^2/(\sigma_v^2\operatorname{Var}(Y))41=\sigma_v^2/(\sigma_v^2+\Omega)$ and $\lambda=\sigma_v^2/[\beta N(\sigma_v^2+\Omega)]=\tau/(\beta N)$, so42$P=\lambda Y=\tau\bar Y$. For the pricing error,43\begin{align*}44\operatorname{Var}(P-v)&=\lambda^2\operatorname{Var}(Y)-2\lambda\operatorname{Cov}(v,Y)+\sigma_v^245=\frac{\operatorname{Cov}(v,Y)^2}{\operatorname{Var}(Y)}-2\frac{\operatorname{Cov}(v,Y)^2}{\operatorname{Var}(Y)}+\sigma_v^2\\46&=\sigma_v^2-\frac{\operatorname{Cov}(v,Y)^2}{\operatorname{Var}(Y)}=\operatorname{Var}(v\mid Y),47\end{align*}48which equals $\sigma_v^2(1-\tau)$ by definition of $\tau$.49\end{proof}5051\subsection*{Proof of Proposition~\ref{prop:info}}5253With a single model of $n=\phi N$ users, $H=\phi^2$ and $\Phi=\phi$, so Lemma~\ref{lem:flow} gives54\eqref{eq:tau_phi}. Since $\tau=\sigma_v^2/(\sigma_v^2+\Omega)$ is strictly decreasing in $\Omega$, every55monotonicity claim about $\tau$ is the reverse claim about $\Omega$.5657\emph{(i) Discrete monotonicity.} Writing $\Omega$ as a function of the integer $n$,58$\Omega(n)=\sigma_\eta^2\,(n^2-n+N)/N^2+C$ with $C=\sigma_u^2/N+\sigma_z^2/(\beta^2N^2)$. Then59\[60\Omega(n+1)-\Omega(n)=\frac{\sigma_\eta^2}{N^2}\Big[(n+1)^2-(n+1)-n^2+n\Big]=\frac{2n\,\sigma_\eta^2}{N^2},61\]62which is zero for $n=0$ and strictly positive for every $n\ge1$ when $\sigma_\eta^2>0$. Hence63$\Omega(0)=\Omega(1)<\Omega(2)<\dots<\Omega(N)$ and $\tau$ is strictly decreasing on $\{1/N,\dots,1\}$,64with $\tau(0)=\tau(1/N)$. For $N=1$ the grid is $\{0,1\}$ and $\Omega(0)=\Omega(1)$. In the continuous65relaxation, $\Omega'(\phi)=\sigma_\eta^2(2\phi-1/N)$, which is strictly positive if and only if66$\phi>1/(2N)$; so $\partial\tau/\partial\phi=-\sigma_v^2\Omega'(\phi)/(\sigma_v^2+\Omega)^2<0$ exactly67there. (For $0<\phi<1/(2N)$ the relaxation is slightly increasing in $\phi$; no integer population lies68in this region, since it corresponds to fewer than one-half of a trader using the model.)6970\emph{(ii) Uniform bound and limit.} For every $N$ and $\phi\in[0,1]$,71$\Omega(\phi)-\sigma_\eta^2\phi^2=\sigma_\eta^2(1-\phi)/N+\sigma_u^2/N+\sigma_z^2/(\beta^2N^2)\ge0$, so72$\Omega(\phi)\ge\sigma_\eta^2\phi^2$ and $\tau(\phi)\le\sigma_v^2/(\sigma_v^2+\phi^2\sigma_\eta^2)=\tau_\infty(\phi)$,73which is strictly below one when $\phi\sigma_\eta>0$. The difference $\Omega(\phi)-\sigma_\eta^2\phi^2$ is74$O(1/N)$ for fixed $\phi$ and fixed $\sigma_z^2$, so $\tau(\phi)\to\tau_\infty(\phi)$.7576\emph{(iii) Private-model benchmark.} At $\phi=0$, $\Omega=(\sigma_\eta^2+\sigma_u^2)/N+\sigma_z^2/(\beta^2N^2)$,77which tends to zero whenever $\sigma_z^2/N^2\to0$; then $\tau\to1$. If instead noise trading scales78with the market, $\sigma_z^2=cN^2$, the limit is $\sigma_v^2/(\sigma_v^2+c/\beta^2)<1$ even at $\phi=0$;79this is the usual noisy-rational-expectations bound and is unrelated to model sharing. \qed8081\subsection*{Proof of Proposition~\ref{prop:corr}}8283\emph{(i)} By Lemma~\ref{lem:proj}, $\operatorname{Var}(P-v)=\sigma_v^2(1-\tau)=\sigma_v^2\Omega/(\sigma_v^2+\Omega)$,84an increasing function of $\Omega$; the monotonicity in $\phi$ and the uniform lower bound follow from85parts (i)--(ii) of Proposition~\ref{prop:info} since $y\mapsto\sigma_v^2y/(\sigma_v^2+y)$ is increasing.8687\emph{(ii)} By A2, $\operatorname{Corr}(x_i,x_j)=\operatorname{Corr}(s_i,s_j)$ since $\beta>0$. By A1,88$\operatorname{Var}(s_i)=\sigma_s^2$ for every $i$, and for $i\neq j$,89$\operatorname{Cov}(s_i,s_j)=\sigma_v^2+\operatorname{Cov}(\eta_{g(i)},\eta_{g(j)})$, which equals90$\sigma_v^2+\sigma_\eta^2$ if $g(i)=g(j)\neq\varnothing$ and $\sigma_v^2$ otherwise (two private models, or91two different shared models, have independent $\eta$'s). This gives $\rho_1$ and $\rho_0$, and92$\rho_1-\rho_0=\sigma_\eta^2/\sigma_s^2>0$ if and only if $\sigma_\eta^2>0$.9394\emph{(iii)} Among the $N(N-1)$ ordered pairs $i\neq j$, exactly $n(n-1)$ have both members on the95shared model; all others have correlation $\rho_0$. Hence96$\bar\rho=\rho_0+w(\rho_1-\rho_0)$ with $w=n(n-1)/[N(N-1)]=\phi(N\phi-1)/(N-1)$. As a function of97real $\phi$, $w'(\phi)=(2N\phi-1)/(N-1)$ and $w''(\phi)=2N/(N-1)>0$: $w$ is convex and increasing98for $\phi>1/(2N)$. On the integer grid, $w(n+1)-w(n)=2n/[N(N-1)]>0$ for $n\ge1$. As $N\to\infty$,99$w\to\phi^2$.100101\emph{(iv)} Let $(x_i,x_j)$ be centered jointly Gaussian with correlation $\rho\in(-1,1)$. Write102$x_j/\sigma_j=\rho\,x_i/\sigma_i+\sqrt{1-\rho^2}\,\xi$ with $\xi$ standard normal independent of $x_i$.103The classical orthant formula gives $\Pr(x_i>0,x_j>0)=\tfrac14+\arcsin(\rho)/(2\pi)$; by symmetry104$\Pr(x_i<0,x_j<0)$ is the same, so $S(\rho)=\Pr(x_ix_j>0)=\tfrac12+\arcsin(\rho)/\pi$.\footnote{For105completeness: $\Pr(x_i>0,x_j>0)=\Pr(\vartheta\in(0,\pi/2)\text{ and }\vartheta\in(\alpha-\pi/2,\alpha+\pi/2))$106where $\vartheta$ is the uniformly distributed angle of a standard bivariate normal vector and107$\cos\alpha=\rho$; the two arcs overlap on a set of measure $\pi-\alpha$, so the probability is108$(\pi-\arccos\rho)/(2\pi)=\tfrac14+\arcsin(\rho)/(2\pi)$.} Since $\arcsin$ is strictly increasing on109$[-1,1]$, so is $S$. Averaging over ordered pairs with the same count as in (iii) gives \eqref{eq:sync}.110Because $\bar\rho$ is affine in $w$ with slope $\rho_1-\rho_0>0$ (when $\sigma_\eta^2>0$), we may111regard $\bar S$ as a function of $\bar\rho$ with112\[113\frac{d\bar S}{d\bar\rho}=\frac{1}{\pi}\cdot\frac{\arcsin\rho_1-\arcsin\rho_0}{\rho_1-\rho_0}>0 .114\]115For the Jensen bound, $(\arcsin)''(\rho)=\rho(1-\rho^2)^{-3/2}\ge0$ on $[0,1)$, and $\rho_0,\rho_1\ge0$,116so $(1-w)\arcsin\rho_0+w\arcsin\rho_1\ge\arcsin\big((1-w)\rho_0+w\rho_1\big)=\arcsin\bar\rho$, with117equality if and only if $\rho_0=\rho_1$ or $w\in\{0,1\}$; for $0<w<1$ equality thus requires118$\sigma_\eta^2=0$.119120\emph{The placebo.} If $\sigma_\eta^2=0$ then $\Omega$ does not depend on $\phi$ (the bracket in121\eqref{eq:tau_phi} is multiplied by zero), $\rho_1=\rho_0$, and $w$ drops out of \eqref{eq:rhobar} and122\eqref{eq:sync}. Hence none of the four objects varies with $\phi$. This shows that the restriction123$\sigma_\eta^2>0$ is necessary for every strict claim in the proposition. \qed124125\subsection*{Proof of Proposition~\ref{prop:diversity}}126127\emph{(i)} By Lemma~\ref{lem:flow}, $\Omega=\sigma_\eta^2[H+(1-\Phi)/N]+\sigma_u^2/N+\sigma_z^2/(\beta^2N^2)$;128for fixed $\Phi$ the shares enter only through $H$, and $\partial\Omega/\partial H=\sigma_\eta^2>0$, so129$\partial\tau/\partial H=-\sigma_v^2\sigma_\eta^2/(\sigma_v^2+\Omega)^2<0$. Setting $\Phi=1$ and $H=1/M_{\mathrm{eff}}$130gives the stated form.131132\emph{(ii)} By the Cauchy--Schwarz inequality,133$\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 if134and only if $(\phi_g)$ is proportional to $(1,\dots,1)$, i.e.\ $\phi_g=\Phi/G$. For the upper bound,135$\Phi^2=\sum_g\phi_g^2+\sum_{g\neq g'}\phi_g\phi_{g'}\ge H$, with equality if and only if all cross136products vanish, i.e.\ at most one share is positive. In the integer version one minimizes $\sum_gn_g^2$137subject to $\sum_gn_g=\Phi N$: if some $n_a\ge n_b+2$, moving one trader from $a$ to $b$ changes the138sum 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.139140\emph{(iii)} Combine (i) and (ii): $H_{\text{single}}=\Phi^2\ge H_{\text{mixed}}\ge\Phi^2/G=H_{\text{uniform}}$141and $\tau$ is decreasing in $H$. Strictness: $H_{\text{mixed}}<\Phi^2$ as soon as two shares are142positive, and $\partial\tau/\partial H<0$ requires $\sigma_\eta^2>0$.143144\emph{(iv)} Ordered pairs with both members on the same shared model number145$\sum_gn_g(n_g-1)=N^2H-N\Phi$; dividing by $N(N-1)$ gives $w=(NH-\Phi)/(N-1)$. For fixed $\Phi$, $w$146is increasing in $H$, and $\bar\rho$ and $\bar S$ are increasing in $w$ by the proof of147Proposition~\ref{prop:corr}. \qed148149\subsection*{Proof of Proposition~\ref{prop:feedback} and Corollary~\ref{cor:llmtrend}}150151Under A5 the round's price change solves $\Delta P=\lambda(Y_1+\gamma\Delta P)$.152153\emph{(i)} If $\lambda\gamma\neq1$ the unique solution is $\Delta P^\ast=\lambda Y_1/(1-\lambda\gamma)=\kappa\lambda Y_1$.154The iterated response satisfies $\Delta P^{(k+1)}-\Delta P^\ast=\lambda\gamma(\Delta P^{(k)}-\Delta P^\ast)$,155so $\Delta P^{(k)}-\Delta P^\ast=(\lambda\gamma)^k(\Delta P^{(0)}-\Delta P^\ast)$, which converges to zero156for every starting point if and only if $|\lambda\gamma|<1$; with $\gamma\ge0$ this is $\lambda\gamma<1$.157Then $\operatorname{Var}(\Delta P)=\kappa^2\lambda^2\operatorname{Var}(Y_1)$, i.e.\ amplification by158$\kappa^2=1/(1-\lambda\gamma)^2$ relative to the no-feedback price change $\lambda Y_1$.159160\emph{(ii)} If $\lambda\gamma=1$ the equation reads $\Delta P=\lambda Y_1+\Delta P$, impossible for161$Y_1\neq0$, and the iteration gives $\Delta P^{(k)}=\Delta P^{(0)}+k\lambda Y_1$, unbounded. If162$\lambda\gamma>1$ the algebraic solution $\Delta P^\ast=\lambda Y_1/(1-\lambda\gamma)$ exists but has the163sign of $-Y_1$ (a price that falls when the flow is a net purchase), and since $(\lambda\gamma)^k\to\infty$164the iteration diverges from every $\Delta P^{(0)}\neq\Delta P^\ast$; in particular from the natural start165$\Delta P^{(0)}=\lambda Y_1$. This is the counterexample announced in the text: the closed form166$1/(1-\lambda\gamma)^2$ is meaningless outside $\lambda\gamma<1$.167168\emph{(iii)} $\Delta P^\ast=\kappa\lambda Y_1$ is a strictly monotone function of $Y_1$ (for $\kappa>0$), so169$\sigma(\Delta P^\ast)=\sigma(Y_1)$ and $\operatorname{Var}(v\mid P_1)=\operatorname{Var}(v\mid Y_1)$: $\tau$ is unchanged.170Using $\lambda^2\operatorname{Var}(Y_1)=\lambda\operatorname{Cov}(v,Y_1)=\sigma_v^2\tau$ (Lemma~\ref{lem:proj}),171\[172\operatorname{Var}(\kappa\lambda Y_1-v)=\kappa^2\lambda^2\operatorname{Var}(Y_1)-2\kappa\lambda\operatorname{Cov}(v,Y_1)+\sigma_v^2173=\sigma_v^2\big[\kappa^2\tau-2\kappa\tau+1\big]=\sigma_v^2\big[(1-\tau)+\tau(\kappa-1)^2\big].174\]175On $\gamma\in[0,1/\lambda)$, $\kappa$ increases from $1$ to $+\infty$, so $(\kappa-1)^2$ is increasing in $\gamma$.176177\emph{Corollary.} Under A6 the total flow is $Y_1+\gamma_T\Delta P+\sum_i\kappa_{g(i)}\Delta P178=Y_1+\big(\gamma_T+\sum_gn_g\kappa_g\big)\Delta P=Y_1+\big(\gamma_T+N\sum_g\phi_g\kappa_g\big)\Delta P$,179with $\kappa_\varnothing\equiv0$ for private-model traders (any such loading can be folded into $\gamma_T$).180Applying (i)--(ii) with $\gamma=\gamma_T+N\sum_g\phi_g\kappa_g$ gives the statements; with a single181model, $\gamma(\phi)=\gamma_T+N\phi\kappa$ is increasing in $\phi$ when $\kappa>0$, $\kappa^2(\phi)$ is182increasing in $\gamma$ hence in $\phi$, and $\lambda\gamma(\phi)<1$ if and only if183$\phi<\phi^\ast=(1/\lambda-\gamma_T)/(N\kappa)$. If $\kappa=0$, $\gamma$ does not depend on $\phi$. With184several models, $\sum_g\phi_g\kappa_g$ is a share-weighted average that lies between $\Phi\min_g\kappa_g$185and $\Phi\max_g\kappa_g$; if the $\kappa_g$ differ in sign it can be smaller in absolute value than186$\Phi\kappa_g$ for every $g$. \qed187188\subsection*{Proof of Proposition~\ref{prop:spiral}}189190Let $R\equiv\{\Delta P:\Delta P<-b/X\}$ be the region in which $F>0$. On $R$,191$F(\Delta P)=(-X\Delta P-b)/m$ and the fixed-point equation $\Delta P=\Delta P_0-\lambda F(\Delta P)$ reads192\begin{equation}193 \Delta P=\Delta P_0+a\,\Delta P+\frac{\lambda b}{m},\qquad a=\frac{\lambda X}{m}.194 \label{eq:fp_affine}195\end{equation}196Off $R$, $F=0$ and the equation reads $\Delta P=\Delta P_0$.197198\emph{(i)} If $\Delta P_0\ge-b/X$, then $\Delta P=\Delta P_0\notin R$ solves the equation, and the199cascade defined below stops at its first step since $F(\Delta P_0)=0$. For $a<1$ it is the only fixed200point: a solution in $R$ would satisfy \eqref{eq:fp_affine}, whose candidate201$(\Delta P_0+\lambda b/m)/(1-a)$ lies in $R$ if and only if $\Delta P_0+\lambda b/m<-(b/X)(1-a)=-b/X+\lambda b/m$,202i.e.\ $\Delta P_0<-b/X$, which fails. For $a>1$ and $b>0$ the same algebra shows that the candidate203\emph{does} lie in $R$ whenever $\Delta P_0>-b/X$ (dividing by $1-a<0$ reverses the inequality): a204second, self-fulfilling fixed point exists in which the cohort is forced to sell because the price has205fallen enough to exhaust the buffer, and the price has fallen because the cohort sells. This is the206multiplicity familiar from \citet{brunnermeier2009}. We select the cascade limit, which is207$\Delta P_0$; the simulator's forced-liquidation rule is likewise sequential and cannot jump to the208self-fulfilling point.209210\emph{(ii)} Suppose $\Delta P_0<-b/X$ and $a<1$. The candidate $\Delta P=\Delta P_0$ is in $R$, where211$F>0$, so it is not a solution; the affine equation \eqref{eq:fp_affine} has the unique solution212$\Delta P^\ast=(\Delta P_0+\lambda b/m)/(1-a)$, which lies in $R$ by the equivalence just shown. With $b=0$213this is $\Delta P_0/(1-\lambda X/m)$ and the multiplier $1/(1-a)$ exceeds one. For the cascade, let214$\Delta P^{(0)}=\Delta P_0$ and $\Delta P^{(k)}=\Delta P_0-\lambda F(\Delta P^{(k-1)})$. Since215$\Delta P^{(0)}\in R$, $\Delta P^{(1)}-\Delta P^{(0)}=-\lambda F(\Delta P_0)<0$, so $\Delta P^{(1)}\in R$; by216induction all iterates are in $R$ and satisfy $\Delta P^{(k)}-\Delta P^\ast=a(\Delta P^{(k-1)}-\Delta P^\ast)$,217hence218\begin{equation}219 \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},220 \label{eq:cascade_path}221\end{equation}222where we used $\Delta P_0-\Delta P^\ast=-\frac{\lambda}{m}\frac{X\Delta P_0+b}{1-a}=a\frac{|\Delta P_0|-b/X}{1-a}$.223The sequence is decreasing and converges to $\Delta P^\ast$ since $a<1$.224225\emph{(iii)} Suppose $\Delta P_0<-b/X$ and $a\ge1$. No solution exists off $R$ (it would have to be226$\Delta P_0\in R$). On $R$, if $a=1$ then $\lambda b/m=(b/X)(\lambda X/m)=b/X$ and \eqref{eq:fp_affine} reduces to227$0=\Delta P_0+b/X$, 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 a228negative numerator and a negative denominator, hence positive, so it is not in $R$. Thus there is no229fixed point. Along the cascade, the increments obey230$\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)})$231as long as the iterates remain in $R$, which they do because the first increment is negative and the232ratio $a$ is positive; with $a\ge1$ the increments do not shrink, so $\Delta P^{(k)}\to-\infty$. In the233simulator the sequence stops at the price band.234235\emph{(iv)} $\sum_{i\in L}x_i=\beta\sum_{i\in L}s_i$ is a centered Gaussian (A1--A2) with236\[237\operatorname{Var}\Big(\sum_{i\in L}s_i\Big)=\sum_{i\in L}\operatorname{Var}(s_i)+\sum_{i\neq j\in L}\operatorname{Cov}(s_i,s_j)238=\sigma_s^2\big[N_L+N_L(N_L-1)\bar\rho_L\big],239\]240by definition of the cohort-average correlation $\bar\rho_L$. Hence $X=|Z|$ with241$Z\sim\mathcal N(0,\sigma_X^2)$, $\sigma_X=\beta\sigma_s\sqrt{N_L[1+(N_L-1)\bar\rho_L]}$. Standard facts give242$\mathbb E|Z|=\sigma_X\sqrt{2/\pi}$ and $\Pr(|Z|\ge c)=2\bar\Phi(c/\sigma_X)$ for $c>0$, both strictly243increasing in $\sigma_X$, which is strictly increasing in $\bar\rho_L$ when $N_L>1$. Finally, applying244Proposition~\ref{prop:corr}(iii) to the cohort (with $N_L$ in place of $N$ and the cohort's own245shared-model share), $\bar\rho_L$ is increasing in that share on its admissible grid. The spiral246probability is $\Pr(a\ge1)=\Pr(X\ge m/\lambda)$, which gives the stated expression with $c=m/\lambda$.247248\emph{Remark on the net position.} If the cohort contains both longs and shorts, a negative shock forces249sales only from the longs, whose gross exposure $\sum_{i\in L}\max\{x_i,0\}$ is at least250$\max\{\sum_{i\in L}x_i,0\}$; using the net position in the spiral coefficient therefore understates251$a$ for a shock against the cohort's net side, and the spiral condition stated is sufficient, not252necessary. \qed253254\subsection*{Proof of Proposition~\ref{prop:safeguards}}255256\emph{(i)} If $X\le L<m/\lambda$ then $a=\lambda X/m\le\lambda L/m<1$ for every realization, so case (iii)257of Proposition~\ref{prop:spiral} cannot occur and, by case (ii), the multiplier is at most258$1/(1-\lambda L/m)$. For the informational cost, let $g(s)=\min\{\max\{\beta s,-\ell\},\ell\}$, which is259Lipschitz with $g'(s)=\beta\,\mathbf 1\{|\beta s|<\ell\}$ almost everywhere. Since $(v,s_i)$ is jointly260Gaussian, write $v=\theta s_i+e$ with $\theta=\operatorname{Cov}(v,s_i)/\operatorname{Var}(s_i)$ and $e$ independent of261$s_i$. Then $\operatorname{Cov}(v,g(s_i))=\theta\operatorname{Cov}(s_i,g(s_i))=\theta\operatorname{Var}(s_i)\mathbb E[g'(s_i)]$ by262Stein's lemma, i.e.\ $\operatorname{Cov}(v,\tilde x_i)=\operatorname{Cov}(v,s_i)\,\beta\Pr(|\beta s_i|<\ell)=\Pr(|\beta s_i|<\ell)\operatorname{Cov}(v,x_i)$,263and $\Pr(|\beta s_i|<\ell)<1$ for any finite $\ell$. We do not claim a closed form for $\tau$ under264clipping, since the clipped flow is non-Gaussian and the market maker's pricing rule is then nonlinear.265266\emph{(ii)} Under A7, after the halt the cascade is that of Proposition~\ref{prop:spiral} with $a$267replaced by $a_h\equiv\psi(h)a$ and the same starting point $\Delta P_0$ (nothing trades during the268halt). If $a_h<1$, part (ii) of that proposition gives convergence to269$\Delta P^\ast_h=(\Delta P_0+\lambda b/m)/(1-a_h)$, finite regardless of whether $a<1$. If also $a<1$,270then $1-a_h\ge1-a>0$ and the common numerator is negative, so $|\Delta P^\ast_h|\le|\Delta P^\ast|$, with271strict inequality if $\psi(h)<1$.272273\emph{(iii)} From \eqref{eq:cascade_path}, $|\Delta P^{(k)}-\Delta P^\ast|=a^{k+1}(|\Delta P_0|-b/X)/(1-a)$,274which is at most $\epsilon$ if and only if $k+1\ge\log\big(\epsilon(1-a)/(|\Delta P_0|-b/X)\big)/\log a$275(the division by $\log a<0$ reverses the inequality). Hence276$T_\epsilon(a)=\max\{0,\lceil\log(\epsilon(1-a)/(|\Delta P_0|-b/X))/\log a\rceil-1\}$, which is the stated277formula when $b=0$. Under the halt, rounds $1,\dots,h$ leave the price at $\Delta P_0$ and the cascade278with coefficient $a_h$ then starts from $\Delta P_0$, so the first round within $\epsilon$ of279$\Delta P^\ast_h$ is $h+T_\epsilon(a_h)$. Since $T_\epsilon\ge0$, $T^h_\epsilon\ge h$; if $\psi(h)=1$ then280$a_h=a$, $\Delta P^\ast_h=\Delta P^\ast$ and $T^h_\epsilon=T_\epsilon(a)+h$. When $a\ge1$ and $a_h<1$ the281comparison is $T^h_\epsilon<\infty=T_\epsilon(a)$, where the latter is infinite because the no-halt cascade282has no finite limit (Proposition~\ref{prop:spiral}(iii)). \qed283