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% WP18 — Appendix A: lemmas and proofs of the propositions of Section 6.
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\section{Proofs}
\label{app:proofs}
\setcounter{table}{0}\setcounter{figure}{0}

Throughout, $R = 1+r$, $d$, $\delta_2$, $\delta_3$ and $e_3$ are as in \eqref{eq:deltas}, the identity \eqref{eq:fisher} holds, and the hand-to-mouth consumption levels $c^o_j$ and terminal resources $W^o_3$ are those of Section~\ref{subsec:benchmark}. An owner who bought one unit at price $p$ pays $D_2 = \delta_2p$ at stage~2 and $D_3 = \delta_3p$ at stage~3, and sells at stage~3 for $(1-\kappa_s)p$ in a stationary state.

\subsection{Preliminary lemmas}

\begin{lemma}[Stage-3 split]
\label{lem:split}
A stage-3 household with resources $W_3$ chooses $c_3 = W_3/(1+\nu)$ and $b' = \nu W_3/(1+\nu)$.
\end{lemma}
\begin{proof}
Maximise $\ln c_3 + \nu\ln b'$ subject to $c_3 + b' = W_3$: the first-order condition $1/c_3 = \nu/b'$ with the constraint gives the result; the objective is strictly concave.
\end{proof}

\begin{lemma}[Hand-to-mouth values]
\label{lem:htm}
Under Assumptions~\ref{ass:pref}--\ref{ass:htm}, for $b\in\mathcal B$ the optimal plan of a renter is $n_1 = n_2 = 0$ and that of an owner is $B = (1-\theta)p$, $n_1 = n_2 = 0$. The value functions are
\begin{align}
V^O(b) &= \ln c^O_1(b) + \chi + \beta\ln c^O_2 + \beta^2(1+\nu)\,\mathbb E\ln W^O_3(y_3) + K, \label{eq:app_VO}\\
V^R(b) &= \ln c^R_1(b) + \beta\ln c^R_2 + \beta^2(1+\nu)\,\mathbb E\ln W^R_3(y_3) + K, \label{eq:app_VR}
\end{align}
with $K = \beta^2[\nu\ln\nu - (1+\nu)\ln(1+\nu)]$, and $V^O(b) - V^R(b) = \ln c^O_1(b) - \ln c^R_1(b) + \Delta$.
\end{lemma}
\begin{proof}
Using Lemma~\ref{lem:split}, the stage-3 continuation is $(1+\nu)\ln W_3 + K$. For a renter the problem is to maximise $\ln(b+y_1-\rho-n_1) + \beta\ln(y_2-\rho+n_1R-n_2) + \beta^2(1+\nu)\mathbb E\ln(y_3+n_2R)$ over $n_1,n_2\ge 0$ ($n_1$ is chosen at stage~1 and $n_2$ at stage~2, but since $c_2$ is deterministic given $n_1$ the two-stage problem equals this static problem). The objective is strictly concave in $(n_1,n_2)$ and the constraint set is convex, so the Karush--Kuhn--Tucker conditions are necessary and sufficient. At $n_1 = n_2 = 0$ they read $1/c^R_1\ge\beta R/c^R_2$ and $1/c^R_2\ge\beta R(1+\nu)\mathbb E[1/W^R_3]$, which hold strictly by~\eqref{eq:euler}. For an owner, the choice variables are $(B,n_1,n_2)$ with $0\le B\le(1-\theta)p$; lowering $B$ by $\epsilon$ lowers $c_1$ by $\epsilon$ and raises $c_2$ by $\epsilon\,\delta_2/(1-\theta)$ and $W_3$ by $\epsilon\,\delta_3/(1-\theta)$. The KKT condition for $B = (1-\theta)p$ is
\[
\frac{1}{c^O_1}\;\ge\;\beta\frac{\delta_2}{1-\theta}\frac{1}{c^O_2} + \beta^2(1+\nu)\frac{\delta_3}{1-\theta}\,\mathbb E\Big[\frac{1}{W^O_3}\Big].
\]
By~\eqref{eq:euler}, $\beta/c^O_2 < 1/(Rc^O_1)$ and $\beta^2(1+\nu)\mathbb E[1/W^O_3] < \beta/(Rc^O_2) < 1/(R^2c^O_1)$, so the right-hand side is strictly below $\frac{1}{c^O_1}\cdot\frac{\delta_2/R+\delta_3/R^2}{1-\theta} = \frac{1}{c^O_1}$ by~\eqref{eq:fisher}. The conditions for $n_1 = n_2 = 0$ are as for the renter. Substituting the corner allocations gives \eqref{eq:app_VO}--\eqref{eq:app_VR}, and the difference follows from \eqref{eq:Delta}.
\end{proof}

\subsection{Threshold, comparative statics and tilt}

\begin{proof}[Proof of Proposition~\ref{prop:threshold}]
Let $D(b)\equiv V^O(b)-V^R(b) = \ln c^O_1(b) - \ln c^R_1(b) + \Delta$ on $\mathcal B$ (Lemma~\ref{lem:htm}). $D$ is continuous, $D(b)\to-\infty$ as $b\downarrow dp-y_1$ (since $c^O_1\to 0$ while $c^R_1\to dp-\rho>0$), and $D'(b) = 1/c^O_1(b) - 1/c^R_1(b) > 0$ because $c^O_1(b) = c^R_1(b) - (dp-\rho) < c^R_1(b)$ by Assumption~\ref{ass:interior}. As $b\to\infty$, $\ln(c^O_1/c^R_1)\to 0$ so $D(b)\to\Delta>0$. Hence $D$ has exactly one zero $\hat b$, at which $c^O_1/c^R_1 = \Lambda$, i.e.\ $b+y_1-dp = \Lambda(b+y_1-\rho)$, which gives~\eqref{eq:bhat}; the two identities follow by substitution. If $dp\le\rho$, $c^O_1\ge c^R_1$ so $D(b)\ge\Delta>0$ for all feasible $b$.
\end{proof}

\begin{remark}[Proof gap: single crossing outside the hand-to-mouth region]
\label{rem:gap_single}
Proposition~\ref{prop:threshold} characterises the tenure decision on $\mathcal B$, where Assumption~\ref{ass:htm} holds. For $b>\bar b$ the household saves and the values are not the closed forms \eqref{eq:app_VO}--\eqref{eq:app_VR}; the envelope derivative is still $1/c^o_1(b)$ for each tenure, so $V^O-V^R$ is increasing wherever $c^O_1(b)<c^R_1(b)$, but this has not been shown at every crossing point. The numerical benchmark finds a single crossing on $[dp-y_1,3]$ with the general optimiser, and the quantitative model finds exactly one sign change of the buy-minus-rent value along cash at every age and productivity state (Section~\ref{sec:baseline}); a general proof is left open.
\end{remark}

\begin{proof}[Proof of Proposition~\ref{prop:cs}]
$\hat b$ is defined by $D(\hat b;\xi) = 0$ for a parameter $\xi$, with $\partial D/\partial b = 1/c^O_1 - 1/c^R_1 > 0$ at $\hat b$ (proof of Proposition~\ref{prop:threshold}). Because the Euler inequalities are strict, a small change in $\xi$ leaves the corner allocation optimal, so the derivative of each value function with respect to $\xi$ is the partial derivative of \eqref{eq:app_VO}--\eqref{eq:app_VR} at the fixed allocation. By the implicit function theorem $\operatorname{sign}(\partial\hat b/\partial\xi) = -\operatorname{sign}(\partial D/\partial\xi)$. The benchmark has no later purchase, so $V^R$ depends on none of $\theta,\kappa_b,\kappa_s,m,\bar\pi$; this is the restriction named in the statement. We compute $\partial D/\partial\xi$ for each parameter, using $E_3\equiv\mathbb E[1/W^O_3(y_3)]$ and the bounds implied by~\eqref{eq:euler}:
\begin{equation}
\frac{\beta}{c^O_2} < \frac{1}{Rc^O_1}, \qquad
\beta^2(1+\nu)E_3 < \frac{\beta}{Rc^O_2} < \frac{1}{R^2c^O_1}.
\label{eq:app_bounds}
\end{equation}

\emph{Rent.} $\partial V^R/\partial\rho = -1/c^R_1 - \beta/c^R_2 < 0$ and $V^O$ does not depend on $\rho$, so $\partial D/\partial\rho>0$.

\emph{Premium and stage-1 income.} $\partial D/\partial\chi = 1$. $\partial D/\partial y_1 = 1/c^O_1 - 1/c^R_1 = \partial D/\partial b$, hence $\partial\hat b/\partial y_1 = -1$.

\emph{Costs.} $\partial V^O/\partial\kappa_b = -p/c^O_1 < 0$; $\partial V^O/\partial m = -p/c^O_1 - \beta p/c^O_2 < 0$; $\partial V^O/\partial\kappa_s = -\beta^2(1+\nu)\,p\,E_3 < 0$; $V^R$ is unaffected.

\emph{Price.} $\partial V^O/\partial p = -d/c^O_1 - \beta(\delta_2+m)/c^O_2 + \beta^2(1+\nu)e_3E_3$. If $e_3\le 0$ all three terms are non-positive and the first is strictly negative. If $e_3>0$, use the second Euler inequality in the form $\beta/c^O_2 > \beta^2R(1+\nu)E_3$ on the middle term:
\[
\frac{\partial V^O}{\partial p} < -\frac{d}{c^O_1} + \beta^2(1+\nu)E_3\big[e_3 - R(\delta_2+m)\big], \qquad e_3 - R(\delta_2+m) = (1-\kappa_s) - (1-\theta)R^2 - Rm,
\]
where the last equality uses $\delta_3 + R\delta_2 = (1-\theta)R^2$ from~\eqref{eq:fisher}. If the bracket is non-positive we are done. Otherwise apply~\eqref{eq:app_bounds}:
\[
\frac{\partial V^O}{\partial p} < -\frac{1}{c^O_1}\Big[d - \frac{(1-\kappa_s)-(1-\theta)R^2-Rm}{R^2}\Big]
= -\frac{1}{c^O_1}\Big[1+\kappa_b+m+\frac{m}{R}-\frac{1-\kappa_s}{R^2}\Big] < 0,
\]
because $R\ge 1$ implies $(1-\kappa_s)/R^2<1$. Hence $\partial D/\partial p<0$.

\emph{Down payment.} With the loan-to-value constraint binding, $B = (1-\theta)p$ and the payments are $\delta_2p$, $\delta_3p$. Let $\tilde\delta_2\equiv-\partial\delta_2/\partial\theta = R-(1-a)/(1+\bar\pi)>0$ and $\tilde\delta_3\equiv-\partial\delta_3/\partial\theta = (1-a)R/(1+\bar\pi)$, so that $\tilde\delta_2+\tilde\delta_3/R = R$. Then
\[
\frac{\partial V^O}{\partial\theta} = p\Big[-\frac{1}{c^O_1} + \frac{\beta\tilde\delta_2}{c^O_2} + \beta^2(1+\nu)\tilde\delta_3E_3\Big]
< \frac{p}{c^O_1}\Big[-1+\frac{\tilde\delta_2}{R}+\frac{\tilde\delta_3}{R^2}\Big] = 0,
\]
by~\eqref{eq:app_bounds}. A higher down payment is present-value neutral and hurts only through liquidity.

\emph{Anticipated inflation.} $\partial\delta_2/\partial\bar\pi = (1-\theta)(1-a)/(1+\bar\pi)^2\equiv x>0$ and $\partial\delta_3/\partial\bar\pi = -Rx$. Hence
\[
\frac{\partial V^O}{\partial\bar\pi} = xp\Big[-\frac{\beta}{c^O_2} + \beta^2R(1+\nu)E_3\Big] < 0
\]
by the second inequality in~\eqref{eq:euler}; $V^R$ does not depend on $\bar\pi$.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:tilt}]
(i) The signs of $\partial\delta_2/\partial\bar\pi$, $\partial\delta_3/\partial\bar\pi$ and the present-value identity are in the proof of Proposition~\ref{prop:cs}; the threshold result is the last item of that proposition. (ii) Let the stage-2 real payment rise by $x$ and the stage-3 payment fall by $Rx$ (a change in $\bar\pi$). Replace $n_2$ by $n_2-x$, feasible for $x<n_2$: $c_2$ is unchanged, and $W_3 = y_3+(1-\kappa_s)p-D_3+n_2R$ changes by $+Rx-Rx = 0$. The original allocation of $(c_1,c_2,c_3,b')$ is therefore still feasible; conversely any allocation feasible after the change is feasible before it with $n_2+x$, so the feasible sets of consumption plans coincide and the optimum is unchanged. For $\theta$: an increase by $x$ raises the down payment by $px$ and lowers the stage-2 and stage-3 payments by $\tilde\delta_2px$ and $\tilde\delta_3px$. Replace $n_1$ by $n_1-px$ and $n_2$ by $n_2-\tilde\delta_3px/R$ (feasible for small $x$): $c_1$ is unchanged; $c_2$ changes by $+\tilde\delta_2px-Rpx+\tilde\delta_3px/R = 0$ by $\tilde\delta_2+\tilde\delta_3/R = R$; $W_3$ changes by $+\tilde\delta_3px-\tilde\delta_3px = 0$. The same argument in reverse shows the feasible sets coincide.
\end{proof}

\subsection{Leverage}

\begin{proof}[Proof of Proposition~\ref{prop:leverage}]
(1) $W^O_3 = y_3+(1-\kappa_s)(1+g)p-\delta_3p$, so $\partial W^O_3/\partial[(1+g)p] = 1-\kappa_s$; dividing by the stage-1 cash outlay $dp$ gives the multiplier, whose monotonicity is immediate. (2) The owner's housing-related stage-3 wealth is $(1-\kappa_s)(1+g)p-\delta_3p$. The renter-investor pays $\rho$ at stages 1 and 2 instead of $dp$ and $(\delta_2+m)p$, so saves $dp-\rho$ at stage~1 and $(\delta_2+m)p-\rho$ at stage~2, worth $(dp-\rho)R^2+((\delta_2+m)p-\rho)R$ at stage~3. The owner does at least as well iff
\[
(1-\kappa_s)(1+g)\ge dR^2+(\delta_2+m)R+\delta_3-\frac{\rho}{p}(R^2+R) = R^2(1+\kappa_b+m)+mR-\frac{\rho}{p}(R^2+R),
\]
using $dR^2+\delta_2R+\delta_3 = (\theta+\kappa_b+m)R^2+(1-\theta)R^2$ from~\eqref{eq:fisher}; $\theta$ cancels. (3) Housing equity at resale is $(1-\kappa_s)(1-x)p-\delta_3p\le 0$ iff $x\ge x^*$.
\end{proof}

\subsection{Dynasties and persistence}

\begin{proof}[Proof of Proposition~\ref{prop:persist}]
(1) By Lemmas~\ref{lem:split} and~\ref{lem:htm}, $b' = \nu W_3/(1+\nu)$ with $W^R_3 = y_3$ and $W^O_3 = y_3+e_3p$; the child owns iff $b'\ge\hat b$ (Proposition~\ref{prop:threshold}), which gives the two cutoffs. Independence of $y_3$ across generations makes the tenure sequence Markov. (2) The transition matrix $\big(\begin{smallmatrix}1-q_{RO} & q_{RO}\\ 1-q_{OO} & q_{OO}\end{smallmatrix}\big)$ has trace $1-q_{RO}+q_{OO}$ and determinant $q_{OO}-q_{RO}$, hence eigenvalues $1$ and $\lambda$. For a two-state chain in its stationary distribution, $\mathbb E[O'\mid O = 1]-\mathbb E[O'\mid O = 0] = q_{OO}-q_{RO} = \lambda$, and $\operatorname{Corr}(O,O') = \lambda$ (direct computation with $\Pr(O = 1) = \mu^*$). If $\mu_t$ is the owner share of a group, $\mu_{t+1}-\mu^* = \lambda(\mu_t-\mu^*)$. The sign and monotonicity statements follow from \eqref{eq:lambda} with $G$ non-decreasing and strictly increasing where $g>0$. (3) Standard for an irreducible aperiodic two-state chain: $\mu^*$ solves $\mu^* = \mu^*q_{OO}+(1-\mu^*)q_{RO}$. (4) Immediate from $W^O_3-W^R_3 = e_3p$ and Lemma~\ref{lem:split}.
\end{proof}

\begin{remark}[No income risk: absorbing classes and history dependence]
\label{rem:norisk}
If $G$ is degenerate at $\bar y_3$, then $\lambda\in\{0,1\}$: when $\nu\bar y_3/(1+\nu)<\hat b\le\nu(\bar y_3+e_3p)/(1+\nu)$ the two tenure classes are absorbing, every initial owner share $\mu_0$ compatible with that inequality at $p = S^{-1}(\mu_0)$ is a stationary equilibrium, and the long-run ownership rate is determined by the initial distribution of transfers. This continuum of history-dependent steady states is a knife-edge of the no-risk case: any $G$ with positive density on $(\hat y_O,\hat y_R)$ gives $\lambda<1$ and a unique stationary distribution given prices (part~3). Persistence is therefore stated as a half-life, $\ln 2/(-\ln\lambda)$, not as multiplicity.
\end{remark}

\subsection{Stationary equilibrium}

\begin{proof}[Proof of Proposition~\ref{prop:exist}]
\emph{Existence.} $S^{-1}$ and $S_r^{-1}$ are continuous on $[0,1]$; $\Delta$, $\Lambda$ and $\hat b$ are continuous in $(p,\rho)$ where Assumption~\ref{ass:feas} holds (compositions of continuous functions and of the expectation of a continuous function over a compact support), with $\hat b\to+\infty$ as $\Delta\downarrow 0$ when $dp>\rho$, so the convention $\mu^* = 0$ at $\Delta\le 0$ makes $\mu^*$ continuous; $G$ is continuous. Hence $f:[0,1]\to[0,1]$ is continuous. At $\mu = 0$, $p = 0$, so $dp = 0\le\rho$ and every feasible household owns ($\hat b = -y_1<0$): $f(0) = 1\ge 0$. At $\mu = 1$, $f(1)\le 1$. By the intermediate value theorem $f(\mu)-\mu$ has a zero on $[0,1]$, which is a stationary equilibrium with $p^* = S^{-1}(\mu)$, $\rho^* = S_r^{-1}(1-\mu)$.

\emph{Uniqueness.} $\mu^* = q_{RO}/(1-q_{OO}+q_{RO})$ is increasing in $q_{RO}$ and in $q_{OO}$; $q_{RO} = 1-G(\hat y_R)$ is non-increasing in $\hat y_R = (1+\nu)\hat b/\nu$ and $q_{OO} = 1-G(\hat y_O)$ is non-increasing in $\hat y_O = \hat y_R-e_3p$. Along the clearing curve, $dp/d\mu>0$ and $d\rho/d\mu<0$, so by Proposition~\ref{prop:cs}
\[
\frac{d\hat b}{d\mu} = \frac{\partial\hat b}{\partial p}\frac{dp}{d\mu}+\frac{\partial\hat b}{\partial\rho}\frac{d\rho}{d\mu}\;\ge\;\frac{\partial\hat b}{\partial p}\frac{dp}{d\mu}>0,
\qquad
\frac{d\hat y_O}{d\mu} = \frac{1+\nu}{\nu}\frac{d\hat b}{d\mu}-e_3\frac{dp}{d\mu}\;\ge\;\Big[\frac{1+\nu}{\nu}\frac{\partial\hat b}{\partial p}-e_3\Big]\frac{dp}{d\mu}\ge 0
\]
under (C). Thus both cutoffs are non-decreasing in $\mu$, $f$ is non-increasing, and strictly decreasing where $g(\hat y_R)>0$. Since $f$ is non-increasing and $\mu\mapsto\mu$ is strictly increasing, $f(\mu)-\mu$ is strictly decreasing on $[0,1]$ and has at most one zero; existence gives exactly one.
\end{proof}

\begin{proof}[Proof of Corollary~\ref{cor:fixed}]
Market clearing forces $\mu^* = \bar H$. Under (C), $\mu^*(p,\rho)$ is strictly decreasing in $p$ at fixed $\rho$ (proof of Proposition~\ref{prop:exist} with $d\rho = 0$), so the clearing price is unique and, by the implicit function theorem, moves in the direction of the demand shift.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:noseg}]
With unit demand, every stage-1 and stage-2 household occupies exactly one unit whatever its tenure, so the total demand for units per cohort is $1$ and $S(p^*) = 1$ pins $p^*$. Landlord arbitrage (buy at $p$, pay $(m+\phi)p$, collect $\rho$, sell at $p$ next period, discount at $R$) gives $\rho^* = p^*(1+m+\phi)-p^*/R$. Nothing else enters either equation. The argument uses unit demand at the first step: if the number of units demanded by a household depended on its wealth or on prices, $S(p^*)$ would equal an aggregate demand that depends on the wealth distribution, and the neutrality would fail (Remark~\ref{rem:noseg_scope}).
\end{proof}

\subsection{Surprise inflation}

\begin{proof}[Proof of Proposition~\ref{prop:accounting}]
(1)--(2) Substitute the inflation paths into $D_2 = (i_1+a)(1-\theta)p/(1+\pi_{t+1})$ and $D_3 = (1+i_2)(1-a)(1-\theta)p/((1+\pi_{t+1})(1+\pi_{t+2}))$ with $1+i_1 = R(1+\bar\pi)$ and, for the ARM under (E2), $1+i_2 = R'(1+\pi_{t+2})$: the $(1+\pi_{t+2})$ factors cancel and leave $\delta_3p\,(R'/R)/(1+\pi^u)$; the FRM keeps $1+i_2 = R(1+\bar\pi)$ and the extra $(1+\pi^u)$ in the denominator remains. The real stage-3 payment of the ARM exceeds $\delta_3p$ iff $R'>R(1+\pi^u)$, i.e.\ $1+r+\phi_r\pi^u>(1+r)(1+\pi^u)$, i.e.\ $\phi_r>1+r = R$. The cohort aged~3 has a nominal stage-3 payment fixed before the shock, deflated by the realised price level. (3) Real deposit return $R/(1+\pi^u)$ against $R$; the wage and rent factors are definitions. (4) The lender's nominal claims are the mortgages net of deposits, so its real revaluation is minus the households' aggregate real revaluation. Condition \eqref{eq:gaincond} is the household's net real gain written out, the stage-3 term being the contract-specific erosion of items 1--2; $\Gamma_3\ge 0$ when $\phi_r\le R$, and the right-hand side is $O(\pi^u)$ with coefficient $y_2(1-\eta_w)+nR$ against $\delta_2p+\delta_3p$ on the left.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:impact} and Corollary~\ref{cor:creditor}]
We prove the corollary; the proposition is the case $n^O = n^R = 0$. Stage-3 owner-parents at $t+1$ bought at $p^*$ and sell at $p_{t+1}$, with nominal stage-3 payment fixed before the shock and predetermined deposits $n^O$, so their end-of-life wealth is $W_3 = y_3+E+Rn^O/(1+\pi^u)$ with $E(p_{t+1},\pi^u)\equiv(1-\kappa_s)p_{t+1}-\delta_3p^*/(1+\pi^u)$; renter-parents have $W_3 = y_3+Rn^R/(1+\pi^u)$. Under Assumption~\ref{ass:static} the children's threshold is $\hat b(p_{t+1},\rho_{t+1})$ computed with $p'' = p^*$, $\rho' = \rho^*$; its partial derivative in $p$ is $-\big[d/c^O_1+\beta(\delta_2+m)/c^O_2\big]\big/\big[1/c^O_1-1/c^R_1\big]>0$ (no resale term) and in $\rho$ it is negative as in Proposition~\ref{prop:cs}. Let $Y\equiv(1+\nu)\hat b/\nu$. By Lemma~\ref{lem:split} the child of a renter-parent owns iff $y_3\ge\hat y_R\equiv Y-Rn^R/(1+\pi^u)$ and the child of an owner-parent iff $y_3\ge\hat y_O\equiv Y-E-Rn^O/(1+\pi^u)$, so the stage-1 owner share is
\[
\mu^d(p,\rho,\pi^u) = \mu^*\big[1-G(\hat y_O)\big]+(1-\mu^*)\big[1-G(\hat y_R)\big].
\]
Let $Q\equiv(\mu^*+\mu_{t+1})/2$, $p = S^{-1}(Q)$, $\rho = S_r^{-1}(1-Q)$ and $\Psi(Q,\pi^u)\equiv\mu^d(S^{-1}(Q),S_r^{-1}(1-Q),\pi^u)-(2Q-\mu^*)$; an impact equilibrium is a zero of $\Psi$. Differentiating the cutoffs at fixed prices, $\partial\hat y_O/\partial\pi^u = [-\delta_3p^*+Rn^O]/(1+\pi^u)^2$ and $\partial\hat y_R/\partial\pi^u = Rn^R/(1+\pi^u)^2$, so
\[
\frac{\partial\Psi}{\partial\pi^u} = \frac{\mu^*g(\hat y_O)\big(\delta_3p^*-Rn^O\big)-(1-\mu^*)g(\hat y_R)Rn^R}{(1+\pi^u)^2} = \frac{\Xi}{(1+\pi^u)^2},
\]
which is strictly positive under Assumption~\ref{ass:nominal} because $g(\hat y_O)>0$. The deposits do not enter the price derivatives:
\[
\frac{\partial\mu^d}{\partial p} = -\Big[\mu^*g(\hat y_O)+(1-\mu^*)g(\hat y_R)\Big]\frac{1+\nu}{\nu}\frac{\partial\hat b}{\partial p}+\mu^*g(\hat y_O)(1-\kappa_s)\le 0
\]
under (C$'$), while $\partial\mu^d/\partial\rho = -[\cdots]\frac{1+\nu}{\nu}\partial\hat b/\partial\rho\ge 0$ multiplies $d\rho/dQ<0$. Hence $\partial\Psi/\partial Q\le-2<0$ and, by the implicit function theorem, $dQ/d\pi^u = -(\partial\Psi/\partial\pi^u)/(\partial\Psi/\partial Q)$ has the sign of $\Xi$. Items 1--2 of the proposition and the price statements of the corollary follow from $dp/dQ = p/(\varepsilon Q)>0$ for $\varepsilon\in(0,\infty)$, $d\rho/dQ<0$ for $\varepsilon_r<\infty$, and $\mu_{t+1} = 2Q-\mu^*$. If $\varepsilon = 0$ the owner stock is fixed, $\mu_{t+1} = \mu^*$, and $p_{t+1}$ solves $\mu^d(p,\rho^*,\pi^u) = \mu^*$ with $\partial\mu^d/\partial p<0$ (strict because $g(\hat y_O)>0$ and (C$'$) with the $(1-\mu^*)g(\hat y_R)$ term), so $dp/d\pi^u = -(\partial\mu^d/\partial\pi^u)/(\partial\mu^d/\partial p)$ again has the sign of $\Xi$. Item~3 of the proposition: $d\hat b = (\partial\hat b/\partial p)dp+(\partial\hat b/\partial\rho)d\rho\ge 0$ since both products are non-negative when $\Xi>0$; hence $\hat y_R = Y$ rises and $q_{RO}$ falls. Since $\mu_{t+1}-\mu^* = \mu^*(q_{OO}-q^*_{OO})+(1-\mu^*)(q_{RO}-q^*_{RO})\ge 0$ with the second term non-positive, $q_{OO}\ge q^*_{OO}$, and $\lambda = q_{OO}-q_{RO}$ rises. When $\Xi<0$ both products are non-positive, so $\hat b$ falls; the total derivative of $\hat y_R$ is then $\frac{1+\nu}{\nu}d\hat b/d\pi^u+Rn^R/(1+\pi^u)^2$, the sum of a negative price term and a non-negative deposit term, which gives the unsigned statement of the corollary and its sign when $n^R = 0$. Item~4 is the change in $W_3$ of stage-3 owners (both terms non-negative), the lender's side of Proposition~\ref{prop:accounting}, Lemma~\ref{lem:split} applied to the parents' gain minus the higher outlay $d\,dp_{t+1}$ of the child, and the stage-1 budget of a renter. Item~5: with one stock and unit demand, $p_{t+1}$ solves $S(p) = 1$ and does not depend on $\pi^u$.
\end{proof}