1% ============================================================================2% Appendix: proofs of the propositions in sections/theory.tex and sections/welfare.tex3% Author: Simon-Pierre Boucher — simon-pierre.boucher@uqo.ca4% ============================================================================5\providecommand{\E}{\mathbb{E}}6\providecommand{\Var}{\operatorname{Var}}7\providecommand{\Cov}{\operatorname{Cov}}8\providecommand{\MSE}{\operatorname{MSE}}9\providecommand{\one}{\mathbf{1}}10\providecommand{\med}{\operatorname{med}}1112\section{Proofs}13\label{app:proofs}1415Throughout, $v$ is treated as fixed, so that the error of a pool $w^{\prime}s$ with $w^{\prime}\one=1$ is $w^{\prime}b+w^{\prime}\eta$ and \eqref{eq:msew} holds.1617\begin{proof}[Proof of Proposition~\ref{prop:exchangeable}]18Write $\Sigma=\sigma^{2}[(1-\rho)I+\rho\one\one^{\prime}]$. Its eigenvalues are $\sigma^{2}(1-\rho)$, with multiplicity $K-1$ on the subspace orthogonal to $\one$, and $\sigma^{2}(1-\rho+\rho K)$ on $\one$; both are positive exactly when $-1/(K-1)<\rho<1$, and the lower endpoint gives a singular but positive semi-definite matrix. Since $\Sigma\one=\sigma^{2}(1-\rho+\rho K)\one$,19\[20\MSE_{\mathrm{eq}}=\frac{\one^{\prime}\Sigma\one}{K^{2}}=\frac{\sigma^{2}(1-\rho+\rho K)}{K}=\sigma^{2}\Big[\frac{1-\rho}{K}+\rho\Big],21\]22which is \eqref{eq:msemean}. For $\rho>-1/(K-1)$, $\Sigma^{-1}\one=\one/[\sigma^{2}(1-\rho+\rho K)]$, so the Bates--Granger weights $\Sigma^{-1}\one/(\one^{\prime}\Sigma^{-1}\one)$ equal $\one/K$ and the equal-weight pool is the minimum-variance pool (Proposition~\ref{prop:general} shows that this pool is optimal among all linear pools). Finally $(1-\rho)/K\ge0$ gives $\MSE_{\mathrm{eq}}\ge\rho\sigma^{2}$ for every $K$, and $(1-\rho)\sigma^{2}/K$ is the only term that depends on $K$.23\end{proof}2425\begin{proof}[Proof of Proposition~\ref{prop:general}]26Let $b=0$. Minimising $w^{\prime}\Sigma w$ subject to $w^{\prime}\one=1$ with multiplier $\mu$ gives $2\Sigma w=\mu\one$, hence $w^{\ast}=\Sigma^{-1}\one/(\one^{\prime}\Sigma^{-1}\one)$ and $\MSE^{\ast}=w^{\ast\prime}\Sigma w^{\ast}=1/(\one^{\prime}\Sigma^{-1}\one)$; the minimiser is unique because $\Sigma$ is positive definite. For the inequality, apply the Cauchy--Schwarz inequality to $\Sigma^{1/2}\one$ and $\Sigma^{-1/2}\one$:27\[28K^{2}=(\one^{\prime}\one)^{2}=\big((\Sigma^{1/2}\one)^{\prime}(\Sigma^{-1/2}\one)\big)^{2}\le(\one^{\prime}\Sigma\one)(\one^{\prime}\Sigma^{-1}\one),29\]30which rearranges to $1/(\one^{\prime}\Sigma^{-1}\one)\le\one^{\prime}\Sigma\one/K^{2}$. Equality holds if and only if $\Sigma^{1/2}\one$ and $\Sigma^{-1/2}\one$ are proportional, that is $\Sigma\one\propto\one$. If $b\neq0$, $\MSE(w)=w^{\prime}(\Sigma+bb^{\prime})w$ and the same Lagrangian argument gives $w\propto(\Sigma+bb^{\prime})^{-1}\one$. If $b=\beta\one$ then $w^{\prime}b=\beta$ for every $w$ with $w^{\prime}\one=1$, so the bias contributes $\beta^{2}$ to every pool.31\end{proof}3233\begin{proof}[Proof of Proposition~\ref{prop:nested}]34The error of the equal-weight pool is $K^{-1}\sum_k\eta_k=c+\sum_m(n_m/K)g_m+K^{-1}\sum_ke_k$. The three groups of terms are independent, so the variance is $\sigma_c^{2}+\sigma_g^{2}\sum_m(n_m/K)^{2}+\sigma_e^{2}/K$, which is \eqref{eq:nested}. With $\sum_mn_m=K$, the Cauchy--Schwarz inequality $(\sum_mn_m)^{2}\le M\sum_mn_m^{2}$ gives $\sum_m(n_m/K)^{2}\ge1/M$ with equality if and only if all $n_m$ are equal. Neither $\sigma_c^{2}$ nor the minimised provider term $\sigma_g^{2}/M$ depends on $K$, and $\sigma_c^{2}$ does not depend on $M$.35\end{proof}3637\begin{proof}[Proof of Proposition~\ref{prop:consensus}]38(i) Suppose $T(x^{(r)})=T_0:=T(x^{(0)})$. Using translation and scale equivariance with $a=1-\lambda>0$ when $\lambda<1$,39\[40T(x^{(r+1)})=T\big((1-\lambda)x^{(r)}+\lambda T_0\one\big)=(1-\lambda)T(x^{(r)})+\lambda T_0=T_0 ;41\]42when $\lambda=1$, $x^{(r+1)}=T_0\one$ and $T(T_0\one)=T_0$ by translation equivariance applied to $x=0$ together with $T(0)=0$, which follows from scale equivariance. By induction $T(x^{(r)})=T_0$ for all $r$. Substituting into \eqref{eq:revision}, $x^{(r+1)}-T_0\one=(1-\lambda)(x^{(r)}-T_0\one)$, hence $x^{(r)}-T_0\one=(1-\lambda)^{r}(x^{(0)}-T_0\one)$, which is \eqref{eq:trajectory}. Differences $x^{(r)}_k-x^{(r)}_l$ follow by subtraction. For a dispersion statistic $D$ with $D(x+b\one)=D(x)$ and $D(ax)=aD(x)$ for $a>0$, \eqref{eq:trajectory} gives $D(x^{(r)})=D((1-\lambda)^{r}x^{(0)})=(1-\lambda)^{r}D(x^{(0)})$ when $\lambda<1$, and $D(x^{(r)})=D(T_0\one)=D(0)=0$ when $\lambda=1$, consistent with the formula. Nothing in the argument involves $\Sigma$.4344(ii) Averaging \eqref{eq:trajectory} over $k$ with $T_0=\bar x^{(0)}$ gives $\bar x^{(r)}=(1-\lambda)^{r}\bar x^{(0)}+[1-(1-\lambda)^{r}]\bar x^{(0)}=\bar x^{(0)}$.4546(iii) For $0\le\rho<1$ the exchangeable vector can be written $\eta=c\one+e$ with $c\sim N(0,\rho\sigma^{2})$ independent of $e$, whose components are independent $N(0,(1-\rho)\sigma^{2})$. Then $\bar x^{(0)}=v+c+\bar e$ and $\med(x^{(0)})=v+c+\med(e)$. Averaging \eqref{eq:trajectory} with $T=\med$ and writing $\alpha=(1-\lambda)^{r}$,47\[48\bar x^{(r)}=\alpha\bar x^{(0)}+(1-\alpha)\med(x^{(0)})=v+c+\bar e+(1-\alpha)U,\qquad U:=\med(e)-\bar e .49\]50Since $e$ is symmetric about zero, $\E[\med(e)]=0$, so $\E[\bar x^{(r)}]=v$. The invariance $\med(x^{(r)})=\med(x^{(0)})$ is part (i). Consider the Gaussian location model in which $e_k=\mu+\epsilon_k$ with $\epsilon_k$ independent $N(0,(1-\rho)\sigma^{2})$ and the variance known; $\bar e$ is a complete sufficient statistic for $\mu$ and is therefore the uniformly minimum-variance unbiased estimator by the Lehmann--Scheff\'e theorem, while $\med(e)$ is unbiased for $\mu$ by symmetry. Hence $U$ is an unbiased estimator of zero with finite variance, and for every real $a$ the estimator $\bar e+aU$ is unbiased, so that51\[52\Var(\bar e+aU)=\Var(\bar e)+2a\Cov(\bar e,U)+a^{2}\Var(U)\ge\Var(\bar e)\quad\text{for all }a,53\]54which is possible only if $\Cov(\bar e,U)=0$. The identity holds at every $\mu$, in particular at $\mu=0$. Therefore, using the independence of $c$ and $e$,55\[56\MSE(\bar x^{(r)})=\rho\sigma^{2}+\Var(\bar e)+(1-\alpha)^{2}\Var(U)=\MSE(\bar x^{(0)})+(1-\alpha)^{2}\Var(U),57\]58and, because $\Cov(\bar e,\med(e))=\Var(\bar e)+\Cov(\bar e,U)=\Var(\bar e)$,59\[60\Var(U)=\Var(\med(e))-2\Cov(\bar e,\med(e))+\Var(\bar e)=\Var(\med(e))-\Var(\bar e).61\]62This is \eqref{eq:medianpath} with $\Delta_K=\Var(U)\ge0$. The factor $(1-\alpha)^{2}=[1-(1-\lambda)^{r}]^{2}$ is non-decreasing in $r$ and tends to one, so that $\MSE(\bar x^{(r)})$ converges to $\rho\sigma^{2}+\Var(\med(e))$, the mean-squared error of $\med(x^{(0)})$.63\end{proof}6465\begin{proof}[Proof of Proposition~\ref{prop:degroot}]66(i) Let $u=(W^{r})^{\prime}w$. Then $u^{\prime}x^{(0)}=w^{\prime}W^{r}x^{(0)}=w^{\prime}x^{(r)}$ and $u^{\prime}\one=w^{\prime}W^{r}\one=w^{\prime}\one=1$ because $W\one=\one$ implies $W^{r}\one=\one$. Hence every round-$r$ pool is a round-0 pool and its mean-squared error is at least $\MSE^{\ast}$. If $W^{r}$ is non-singular, then $(W^{r})^{-1}\one=\one$, and for any $u$ with $u^{\prime}\one=1$ the vector $w=((W^{r})^{\prime})^{-1}u$ satisfies $w^{\prime}\one=u^{\prime}(W^{r})^{-1}\one=u^{\prime}\one=1$ and $(W^{r})^{\prime}w=u$; taking $u=w^{\ast}$ shows that the optimum is attained at round $r$.6768(ii) A primitive row-stochastic matrix has the simple eigenvalue $1$ with right eigenvector $\one$ and a strictly positive left eigenvector $\pi$ normalised by $\pi^{\prime}\one=1$, all other eigenvalues having modulus strictly below one, so that $W^{r}\to\one\pi^{\prime}$ (Perron--Frobenius). Hence $x^{(r)}\to\one\pi^{\prime}x^{(0)}$ and, for any $w$ with $w^{\prime}\one=1$, $w^{\prime}x^{(r)}\to\pi^{\prime}x^{(0)}$. Since $x^{(0)}$ is Gaussian and $W^{r}$ converges, second moments converge and $\MSE(w^{\prime}x^{(r)})\to\pi^{\prime}\Sigma\pi$. As $\pi^{\prime}\one=1$, $\pi$ is a round-0 pool, so $\pi^{\prime}\Sigma\pi\ge\MSE^{\ast}$ with equality if and only if $\pi=w^{\ast}$, the unique minimiser of Proposition~\ref{prop:general}.6970(iii) $W=(1-\lambda)I+\lambda\one\one^{\prime}/K$ has every entry strictly positive for $\lambda\in(0,1]$ and is therefore primitive; $W\one=\one$; and $(\one/K)^{\prime}W=(1-\lambda)\one^{\prime}/K+\lambda\one^{\prime}/K=(\one/K)^{\prime}$, so $\pi=\one/K$ and the consensus is $\bar x^{(0)}$.71\end{proof}7273\begin{proof}[Proof of Lemma~\ref{lem:additive}]74The pooled error after the revision is $w^{\prime}\eta+w^{\prime}\zeta$ (biases are unchanged since $\E[\zeta]=0$). Expanding the square and taking expectations,75$\E[(w^{\prime}\eta+w^{\prime}\zeta)^{2}]=w^{\prime}\Sigma w+w^{\prime}\Gamma w+2\,w^{\prime}\E[\zeta\eta^{\prime}]w$, which is \eqref{eq:additive}. If $\Cov(\zeta,\eta)=0$ the change equals $w^{\prime}\Gamma w\ge0$, strictly positive for every $w\neq0$ when $\Gamma$ is positive definite. A negative change requires $2w^{\prime}\Cov(\zeta,\eta)w<-w^{\prime}\Gamma w<0$.76\end{proof}7778\begin{proof}[Proof of Proposition~\ref{prop:exchange}]79Because $a_k^{\prime}\one=1$, the signal \eqref{eq:evidencesignal} equals $v+c+a_k^{\prime}u+g_{m(k)}+e_k$, and the equal-weight pool has error $c+\omega^{\prime}u+\sum_m(n_m/K)g_m+K^{-1}\sum_ke_k$ with $\omega=K^{-1}\sum_ka_k$. The four groups are independent and $\Var(\omega^{\prime}u)=\sigma_u^{2}\|\omega\|^{2}$, which gives \eqref{eq:evidencemse}. An exchange changes only the $a_k$, hence only $\omega$, so the change in the pooled error is $\sigma_u^{2}(\|\omega^{\prime}\|^{2}-\|\omega\|^{2})$.8081(i) An item in $\bigcap_kA_k$ already belongs to every view, so adding it changes no $A_k$ and no $a_k$. Any exchange with $\omega^{\prime}=\omega$ leaves the pooled error unchanged by the formula just derived, and $\sigma_c^{2}$, $\sigma_g^{2}$ and the $\tau_k^{2}$ never enter the exchange term.8283(ii) Let $U=\bigcup_kA_k$ and $J_U=|U|$. Each $a_k$ is supported on $U$ and sums to one, so $\omega$ is supported on $U$ with $\sum_{j\in U}\omega_j=1$. By the Cauchy--Schwarz inequality, $1=(\sum_{j\in U}\omega_j)^{2}\le J_U\sum_{j\in U}\omega_j^{2}=J_U\|\omega\|^{2}$, with equality if and only if $\omega_j=1/J_U$ for all $j\in U$. The full exchange sets every $a_k$ equal to the uniform vector on $U$, hence $\omega^{\prime}$ is uniform on $U$ and the bound is attained. A single agent shown $U$ has evidence error $c+J_U^{-1}\sum_{j\in U}u_j$ with variance $\sigma_c^{2}+\sigma_u^{2}/J_U$, the same quantity.8485(iii) Before the exchange $a_1=(1,0)$, $a_2=a_3=(0,1)$, so $\omega=(1/3,2/3)$ and $\|\omega\|^{2}=1/9+4/9=5/9$. After giving item~2 to agent~1, $a_1=(1/2,1/2)$, $\omega=(1/6,5/6)$ and $\|\omega\|^{2}=1/36+25/36=13/18>5/9$.86\end{proof}8788\begin{proof}[Proof of Proposition~\ref{prop:overshrink}]89Let $S=\sum_j\sigma_j^{2}$ and $\alpha=(1-\lambda)^{r}\in(0,1)$ for $r\ge1$. By Proposition~\ref{prop:consensus}(ii) with the mean as anchor, $x^{(r)}=\alpha x^{(0)}+(1-\alpha)\bar x^{(0)}\one$, so the error of agent $k$ at round $r$ is $\alpha\eta_k+(1-\alpha)\bar\eta$. With independent round-0 errors, $\Cov(\eta_k,\bar\eta)=\sigma_k^{2}/K$ and $\Var(\bar\eta)=S/K^{2}$, hence90\[91\Var(x^{(r)}_k)=\beta\sigma_k^{2}+\gamma,\qquad \beta=\alpha^{2}+\frac{2\alpha(1-\alpha)}{K}>0,\qquad \gamma=\frac{(1-\alpha)^{2}S}{K^{2}}>0 .92\]93The inverse-variance weights are $w_k=(\beta\sigma_k^{2}+\gamma)^{-1}/C$ with $C=\sum_j(\beta\sigma_j^{2}+\gamma)^{-1}$, and the pooled estimate is94\[95\sum_kw_kx^{(r)}_k=v+\alpha\sum_kw_k\eta_k+(1-\alpha)\bar\eta=v+\sum_ku_k\eta_k,\qquad u_k=\alpha w_k+\frac{1-\alpha}{K},96\]97with $\sum_ku_k=1$. Its mean-squared error is $\sum_ku_k^{2}\sigma_k^{2}$, which by Proposition~\ref{prop:general} is at least $\MSE^{\ast}=(\sum_k\sigma_k^{-2})^{-1}$, with equality if and only if $u=w^{\ast}$, $w^{\ast}_k=\tau\sigma_k^{-2}$, $\tau=(\sum_j\sigma_j^{-2})^{-1}$. At round 0, $\alpha=1$, $u=w$ and $w_k\propto\sigma_k^{-2}$, so the pool is optimal. Suppose that $u=w^{\ast}$ for some $r\ge1$. Then for every $k$, $\alpha w_k=\tau\sigma_k^{-2}-(1-\alpha)/K$, and multiplying by $\beta\sigma_k^{2}+\gamma$,98\[99\frac{\alpha}{C}=\Big(\frac{\tau}{\sigma_k^{2}}-\frac{1-\alpha}{K}\Big)(\beta\sigma_k^{2}+\gamma)=\tau\beta-\frac{(1-\alpha)\gamma}{K}+\frac{\tau\gamma}{\sigma_k^{2}}-\frac{(1-\alpha)\beta}{K}\,\sigma_k^{2}.100\]101The right-hand side is $\phi(\sigma_k^{2})$ for the function $\phi(y)=\text{const}+\tau\gamma/y-(1-\alpha)\beta y/K$, which is strictly decreasing on $(0,\infty)$ because $\tau\gamma>0$ and $(1-\alpha)\beta/K>0$. A strictly monotone function takes the same value at all $\sigma_k^{2}$ only if they are all equal, which contradicts the hypothesis. Hence the inverse-variance pool at any round $r\ge1$ has mean-squared error strictly above $\MSE^{\ast}$, the error of the round-0 inverse-variance pool. As $r\to\infty$, $\alpha\to0$, so $\beta\to0$, $\gamma\to S/K^{2}$, $w_k\to1/K$ and $u_k\to1/K$; the error converges to $S/K^{2}=\MSE_{\mathrm{eq}}$, which exceeds $\MSE^{\ast}$ by Proposition~\ref{prop:general} because a diagonal $\Sigma$ with unequal entries does not satisfy $\Sigma\one\propto\one$.102\end{proof}103104\begin{proof}[Proof of Proposition~\ref{prop:placebo}]105(i) Let $P_\sigma$ be the permutation matrix with $(P_\sigma x)_k=x_{\sigma(k)}$. With provider assignment randomised independently of the slot, as in the design, the signal vector of arm~I has the distribution of $P_\sigma x^{G}$, where $x^{G}$ is the signal vector of arm~G. An aggregator $\mathcal{A}$ that is invariant to permutations satisfies $\mathcal{A}(P_\sigma x)=\mathcal{A}(x)$ for every $x$, so $\mathcal{A}(x^{I})$ and $\mathcal{A}(x^{G})$ have the same distribution, and in particular the same error distribution.106107(ii) With views fixed to the slots, $\omega$ and the provider term in \eqref{eq:evidencemse} are unchanged, and the reasoning term is $K^{-2}\sum_k\tau^{2}(\pi_{\sigma(k)})=K^{-2}\sum_k\tau^{2}(\pi_k)$ because a permutation does not change a sum. Hence $\MSE_{\mathrm{eq}}$ is the same in both arms.108109(iii) Substituting $\tau_k^{2}=\tau^{2}(\pi_{\sigma(k)},A_k)$ and $\tau_k^{2}=\tau^{2}(\pi_k,A_k)$ into \eqref{eq:evidencemse} and subtracting gives the stated difference; the other three terms cancel because they do not involve the prompts.110\end{proof}111112\begin{proof}[Proof of Proposition~\ref{prop:rationing}]113Conditional on the seller's draw, the match fails for lack of financing when $W+L<S$, that is when $\ell(1-h)(1+b+sZ)<(1+u_S)-(1-\ell)\kappa$, or $Z<a(u_S)/(\ell(1-h)s)$, which has probability $F=\Phi\big(a(u_S)/(\ell(1-h)s)\big)$ with $a(u_S)=(1+u_S)-(1-\ell)\kappa-\ell(1-h)(1+b)$. Write the argument as114\[115q(b,h,s)=\frac{(1+u_S)-(1-\ell)\kappa}{\ell(1-h)s}-\frac{1+b}{s}.116\]117Since $\Phi$ is strictly increasing, the monotonicity of $F$ is that of $q$. (i) $\partial q/\partial b=-1/s<0$. When financing is needed, $(1+u_S)>(1-\ell)\kappa$, the first term is positive and strictly increasing in $h$, so $\partial q/\partial h>0$. (ii) $\partial q/\partial s=-a(u_S)/(\ell(1-h)s^{2})$, which is positive when $a(u_S)<0$ and negative when $a(u_S)>0$. (iii) The failure rate among beneficial matches is $\E[F\mid B\ge S]$; if $a(u_S)<0$ almost surely on $\{B\ge S\}$, each term is increasing in $s$ by (ii) and so is the expectation. In general the expectation splits into the integral over $\{a<0\}$, increasing in $s$, and the integral over $\{a>0\}$, decreasing in $s$. (iv) Under $h=h_0+h_1s$ with $h_1>0$, $dq/ds=\partial q/\partial s+h_1\,\partial q/\partial h$, and the second term is positive by (i), so on $\{a<0\}$ the derivative is larger than without the rule.118\end{proof}119120\begin{proof}[Proof of Proposition~\ref{prop:lenderloss}]121Let $Y=(1-c_\ell)V_1$, a Gaussian variable with positive variance, and $\varphi(L)=\E[\max(L-Y,0)]$. Then $\varphi^{\prime}(L)=\Pr(Y<L)\in(0,1)$ and $\varphi^{\prime\prime}(L)=f_Y(L)>0$ for every $L$, so $\varphi$ is strictly increasing and strictly convex on $\mathbb{R}$. With $L=\ell(1-h)(1+b+sZ)$ and $h$ fixed, $\mathrm{EL}(b,s)=p_d\E[\varphi(\ell(1-h)(1+b+sZ))]$. (i) $L$ is strictly increasing in $b$ for every $Z$, and so is $\varphi(L)$; hence $\mathrm{EL}$ is strictly increasing in $b$. (ii) Let $s^{\prime}>s$ and $\delta=(s^{\prime2}-s^{2})^{1/2}$. If $Z^{\prime}$ is standard normal and independent of $Z$, then $sZ+\delta Z^{\prime}$ has the distribution of $s^{\prime}Z$. Writing $m=\ell(1-h)$,122\[123\E\big[\varphi(m(1+b+s^{\prime}Z))\big]=\E\Big[\E\big[\varphi(m(1+b+sZ)+m\delta Z^{\prime})\mid Z\big]\Big]>\E\big[\varphi(m(1+b+sZ))\big],124\]125where the inequality is Jensen's inequality applied conditionally on $Z$ to the strictly convex $\varphi$ and the non-degenerate variable $m\delta Z^{\prime}$. (iii) Under $h=h_0+h_1s$ the loan is $L=\ell(1-h_0-h_1s)(1+b+sZ)$, whose mean $\ell(1-h_0-h_1s)(1+b)$ is strictly decreasing in $s$; the effect of $s$ on $\mathrm{EL}$ combines the increasing spread effect of (ii) with a decreasing location effect, and its sign is not determined without further restrictions. For any $s$ there is a haircut $h(s)$ that holds $\mathrm{EL}$ at its value at $s=0$, because $\mathrm{EL}$ is continuous and strictly increasing in $L$ and tends to zero as $h\to1$; by Proposition~\ref{prop:rationing}(i) this haircut raises the failure probability of every financeable match.126\end{proof}127128\begin{proof}[Proof of Proposition~\ref{prop:systemic}]129Let $g(z)=p_d\,\varphi(\ell(1-h)(1+b+sz))$ with $\varphi$ as above; $g$ is continuous, non-decreasing and grows at most linearly, hence square-integrable against the standard normal density. (i) By linearity of expectation, $\E[\sum_n\ell_n]=N\E[g(Z_1)]$, and $Z_1$ is standard normal for every $\rho$. (ii) By exchangeability, $\Var(\sum_n\ell_n)=N\Var(g(Z_1))+N(N-1)\Cov(g(Z_1),g(Z_2))$, and $(Z_1,Z_2)$ is standard bivariate normal with correlation $\rho$. Let $\{h_n\}_{n\ge0}$ be the orthonormal Hermite polynomials with respect to the standard normal density and $g=\sum_{n\ge0}a_nh_n$ in $L^{2}$. Mehler's formula gives $\E[h_m(Z_1)h_n(Z_2)]=\rho^{n}\mathbf{1}\{m=n\}$, hence130\[131\Cov\big(g(Z_1),g(Z_2)\big)=\sum_{n\ge1}a_n^{2}\rho^{n},132\]133a power series with non-negative coefficients, which is non-decreasing in $\rho$ on $[0,1]$ and strictly increasing because $g$ is not almost surely constant, so that some $a_n$ with $n\ge1$ is non-zero. At $\rho=0$ the covariance is zero and at $\rho=1$ it equals $\sum_{n\ge1}a_n^{2}=\Var(g(Z_1))$, so the system variance rises from $N\Var(g)$ to $N^{2}\Var(g)$. (iii) With $N$ lenders using distinct models, $\Var(\sum_n\ell_n)=Nv_0[1+(N-1)\tilde\rho_0]$ where $v_0=\Var(g_0(Z))$ and $\tilde\rho_0=\Cov(g_0(Z_1),g_0(Z_2))/v_0\in[0,1]$; with a single common model of loss variance $v_1$ it equals $N^{2}v_1$. The latter exceeds the former if and only if $v_1/v_0>[1+(N-1)\tilde\rho_0]/N$.134\end{proof}135