Appendix: Proofs
1.1 A.1 Preliminaries
The next proof relies on the definition of consumption and investment policies \(\mathcal{C}\) and \(\mathcal{P}\), respectively, which combine into admissible strategies.
Definition A.1
In all markets (segmented or integrated), the set \(\mathcal{C}\) of consumption policies consists of all adapted processes \((c_{t})_{t\ge 0}\) with \(c_{t}>0\) a.s. for all \(t\ge 0\) and \(\mathbb{E}[\int ^{T}_{0} c_{t} dt]<\infty \) for all \(T >0\). In each segmented market, the set of investment policies \(\mathcal{P}^{i}\), \(i=1,2\), consists of all predictable processes \((\phi _{t})_{t\ge 0}\) such that \(\mathbb{E}[\int ^{T}_{0} \phi _{t}^{2} d \langle P^{(i)} \rangle _{t}]< \infty \) for all \(T>0\), where \(P^{(i)}\) is the respective asset price. In the integrated market, the set \(\overline{\mathcal{P}}\) of investment policies consists of \(\mathbb{R}^{2}\)-valued, predictable processes \((\phi _{t})_{t\ge 0}\) such that \(\mathbb{E}[\langle \int ^{\cdot }_{0} \phi \cdot dP\rangle _{T}]< \infty \) for all \(T>0\). The set \(\mathcal{A}\) of admissible strategies consists of all pairs \((c,\phi )\) such that \(c\in \mathcal{C}\), \(\phi \in \mathcal{P}^{(i)}\) for each segmented market and \(\phi \in \overline{\mathcal{P}}\) for the integrated market, and the corresponding wealth in (3.6) or (3.7) satisfies \(Y_{t}\ge 0\) a.s. for all \(t\ge 0\).
Definition 3.3 of an equilibrium leads to the familiar representation of prices as discounted cash flows and the safe rate as minus the growth rate of the stochastic discount factor. Although the result is often taken for granted and can be informally derived from a perturbation argument, it is not guaranteed to hold in general and the literature does offer counterexamples; see Basak and Cuoco [4, Remark 4 and Corollary 3] and the discussion in Karatzas et al. [31]. Thus we offer a proof that applies to the model considered here.
Proposition A.2
In the segmented markets, the equilibrium asset prices are
$$\begin{aligned} P^{(i)}_{t} = \mathbb{E}_{t}\bigg[ \int ^{\infty }_{t} \frac{M^{(i)}_{s}}{M^{(i)}_{t}} D^{(i)}_{s} ds \bigg], \qquad \textit{where } M^{(i)}_{t} = e^{-\beta t}(D^{(i)}_{t})^{-\gamma }, \end{aligned}$$
while equilibrium safe rates \(r^{(i)}_{t}\)are identified by the local martingale condition for \((M^{(i)}_{t} e^{\int _{0}^{t} r^{(i)}_{s} ds})_{t\geq 0}\). Likewise, in the integrated market, the equilibrium asset prices are
$$\begin{aligned} \bar{P}^{(i)}_{t} = \mathbb{E}_{t}\bigg[ \int ^{\infty }_{t} \frac{\bar{M}_{s}}{\bar{M}_{t}} D^{(i)}_{s} ds \bigg], \qquad \textit{where } \bar{M}_{t} = e^{-\beta t}(D^{(1)}_{t} + D^{(2)}_{t})^{- \gamma } . \end{aligned}$$
Proof
Note that if \(Y_{0}=P^{(i)}_{0}\), \(c_{t}=D^{(i)}_{t}\) and \(\phi _{t}\equiv 1\), then (3.6) implies that \(Y_{t}=P^{(i)}_{t}\) for all \(t \geq 0\). Fix \(i=1,2\), \(t_{0}>0\), \(\vartheta > 0\) and \(K>1\). Define the stopping time
$$ \tau _{0}:=(t_{0}+\vartheta )\wedge \inf \bigg\{ t>t_{0} \,: \, P^{(i)}_{t} \leq \frac{1}{K} P^{(i)}_{t_{0}} \text{ and } D^{(i)}_{t}\geq K D^{(i)}_{t_{0}} \bigg\} . $$
For any \(\delta \in (-\vartheta , \vartheta )\), consider an event \(A \in \mathcal{F}_{t_{0}}\) on which we adopt an alternative strategy, where from time \(t_{0}\) to \(\tau _{0}\) the consumption changes from \(D^{(i)}_{t}\) to \({c^{\delta }_{t}=(1-\frac{\delta }{\vartheta })D^{(i)}_{t}}\) with the difference in consumption invested in the risky asset, so that the number of shares changes from 1 to \({\phi ^{\delta }_{t}:=1+ \frac{\delta }{\vartheta }\Delta _{t}}\) for some process \((\Delta _{t})_{t\geq 0}\). Thus the corresponding wealth on \(A\) becomes
$$ {Y^{\delta }_{t}=\left (1+ \frac{\delta }{\vartheta }\Delta _{t}\right )P^{(i)}_{t}}. $$
To satisfy the budget constraints (3.6), \((\Delta _{t})\) satisfies on \(A\)
$$\begin{aligned} d\bigg(1+\frac{\delta }{\vartheta }\Delta _{t}\bigg)P^{(i)}_{t} &= r^{(i)}_{t} \bigg(\Big(1+\frac{\delta }{\vartheta }\Delta _{t}\Big)P^{(i)}_{t} - \Big(1+\frac{\delta }{\vartheta }\Delta _{t}\Big) P^{(i)}_{t}\bigg) dt \\ &\phantom{=:}+ \bigg(1+ \frac{\delta }{\vartheta } \Delta _{t}\bigg) ({dP^{(i)}_{t}}+D^{(i)}_{t} dt )- \bigg(1-\frac{\delta }{\vartheta }\bigg)D^{(i)}_{t} dt, \end{aligned}$$
therefore
$$\begin{aligned} \frac{\delta }{\vartheta } P^{(i)}_{t} d\Delta _{t}+ \frac{\delta }{\vartheta } d \langle \Delta , P^{(i)} \rangle _{t} = \bigg(\frac{\delta }{\vartheta }+\frac{\delta }{\vartheta }\Delta _{t} \bigg) D^{(i)}_{t} dt \end{aligned}$$
and so
$$\begin{aligned} d\Delta _{t} = (1+\Delta _{t} )\frac{D^{(i)}_{t}}{P^{(i)}_{t}}dt . \end{aligned}$$
Hence we get on \(A\) that
After \(\tau _{0}\), the investor holds \(\phi ^{\delta }_{t}\equiv 1+\frac{\delta }{\vartheta }\Delta _{\tau _{0}}\) unit of risky asset and consumes \({c^{\delta }_{t}=(1+ \frac{\delta }{\vartheta } \Delta _{\tau _{0}})D^{(i)}_{t}}\). To ensure that consumption remains positive, assume also that
$$ {\delta >-\frac{1}{K^{2}} e^{-\vartheta K^{2} \frac{D^{(i)}_{t_{0}}}{P^{(i)}_{t_{0}}}} \frac{P^{(i)}_{t_{0}}}{D^{(i)}_{t_{0}}}} . $$
The change in expected utility from \((c_{t},\phi _{t})_{t\geq 0}\) to \((c^{\delta }_{t},\phi ^{\delta }_{t})_{t\geq 0}\) is thus
$$\begin{aligned} \Delta ^{\delta }J&= \mathbb{E}\bigg[1_{A}\bigg(\int ^{\tau _{0}}_{t_{0}} \! \frac{ e^{-\beta t} ( (1-\frac{\delta }{\vartheta } )D^{(i)}_{t} )^{1-\gamma }}{1-\gamma } dt +\!\int _{\tau _{0}}^{+\infty } \! \frac{ e^{-\beta t} ( (1+\frac{\delta }{\vartheta }\Delta _{\tau _{0}} )D^{(i)}_{t} )^{1-\gamma }}{1-\gamma } dt \\ &\phantom{=:} \qquad \qquad -\int _{t_{0}}^{+\infty } \frac{ e^{-\beta t} (D^{(i)}_{t} )^{1-\gamma }}{1-\gamma } dt\bigg) \bigg]. \end{aligned}$$
Because \(\frac{x^{1-\gamma }}{1-\gamma }\) is concave, we have \(y^{-\gamma }(y-x) \leq \frac{y^{1-\gamma }}{1-\gamma }- \frac{x^{1-\gamma }}{1-\gamma }\) for any \(x,y>0\), whence
$$\begin{aligned} \Delta ^{\delta }J\geq \mathbb{E}\bigg[1_{A}\bigg( -&\frac{\delta }{\vartheta } \int ^{\tau _{0}}_{t_{0}} e^{-\beta t}D^{(i)}_{t} \bigg(\Big(1- \frac{\delta }{\vartheta }\Big)D^{(i)}_{t}\bigg)^{-\gamma } dt \\ +& \frac{\delta }{\vartheta }\Delta _{\tau _{0}}\int _{\tau _{0}}^{+ \infty } D^{(i)}_{t} e^{-\beta t}\bigg(\Big(1+\frac{\delta }{\vartheta } \Delta _{\tau _{0}}\Big)D^{(i)}_{t}\bigg)^{-\gamma } dt\bigg)\bigg]. \end{aligned}$$
As \((c_{t},\phi _{t})_{t\geq 0}\) is optimal, it follows that \(\lim _{\delta \downarrow 0}\frac{ \Delta ^{\delta }J}{\delta }\leq 0\) and \(\lim _{\delta \uparrow 0}\frac{ \Delta ^{\delta }J}{\delta }\geq 0\), whence
$$\begin{aligned} \lim _{\delta \rightarrow 0} \frac{1}{\vartheta } \mathbb{E}\bigg[1_{A} \bigg(&-\int ^{\tau _{0}}_{t_{0}} e^{-\beta t}D^{(i)}_{t} \bigg(\Big(1- \frac{\delta }{\vartheta }\Big)D^{(i)}_{t}\bigg)^{-\gamma } dt \\ & +\Delta _{\tau _{0}}\int _{\tau _{0}}^{+\infty } D^{(i)}_{t} e^{- \beta t}\bigg(\Big(1+\frac{\delta }{\vartheta }\Delta _{\tau _{0}}\Big)D^{(i)}_{t} \bigg)^{-\gamma } dt\bigg)\bigg] = 0 \end{aligned}$$
and therefore
$$ \frac{\mathbb{E} [1_{A} (-\int ^{\tau _{0}}_{t_{0}} e^{-\beta t}D^{(i)}_{t} (D^{(i)}_{t} )^{-\gamma } dt+\Delta _{\tau _{0}}\int _{\tau _{0}}^{+\infty } D^{(i)}_{t} e^{-\beta t} (D^{(i)}_{t} )^{-\gamma } dt ) ]}{\vartheta }= 0. $$
(A.1)
Because (A.1) holds for all \(\vartheta \neq 0\),
$$\begin{aligned} \lim _{\vartheta \downarrow 0} \frac{\mathbb{E} [1_{A} (-\int ^{\tau _{0}}_{t_{0}} e^{-\beta t}D^{(i)}_{t} (D^{(i)}_{t} )^{-\gamma } dt+\Delta _{\tau _{0}}\int _{\tau _{0}}^{+\infty } D^{(i)}_{t} e^{-\beta t} (D^{(i)}_{t} )^{-\gamma } dt ) ]}{\vartheta } = 0. \end{aligned}$$
(A.2)
Note that
$$\begin{aligned} \lim _{\vartheta \downarrow 0} \frac{\int ^{\tau _{0}}_{t_{0}} e^{-\beta t}D^{(i)}_{t} (D^{(i)}_{t} )^{-\gamma } dt}{\vartheta } =&e^{- \beta t_{0}} (D^{(i)}_{t_{0}} )^{1-\gamma }, \\ \lim _{\vartheta \downarrow 0} \int _{\tau _{0}}^{+\infty } D^{(i)}_{t} e^{-\beta t} (D^{(i)}_{t} )^{-\gamma } dt =& \int _{t_{0}}^{+\infty } D^{(i)}_{t} e^{-\beta t} (D^{(i)}_{t} )^{-\gamma } dt \end{aligned}$$
and
$$ \lim _{\vartheta \downarrow 0}\frac{\Delta _{\tau _{0}}}{\vartheta }= \lim _{\vartheta \downarrow 0}\frac{1}{\vartheta }\int ^{\tau _{0}}_{t_{0}} e^{\int _{s}^{\tau _{0}} \frac{D^{(i)}_{u}}{P^{(i)}_{u}} du} \frac{D^{(i)}_{s}}{P^{(i)}_{s}} ds= \frac{D^{(i)}_{t_{0}}}{P^{(i)}_{t_{0}}}. $$
Thus an application of the dominated convergence theorem to (A.2) yields
$$ \mathbb{E}\bigg[1_{A}\bigg(-e^{-\beta t_{0}} (D^{(i)}_{t_{0}} )^{1- \gamma }+\frac{D^{(i)}_{t_{0}}}{P^{(i)}_{t_{0}}}\mathbb{E}_{t_{0}} \Big[\int _{t_{0}}^{+\infty } D^{(i)}_{t} e^{-\beta t} (D^{(i)}_{t} )^{- \gamma } dt\Big]\bigg)\bigg]= 0. $$
As \(A \in \mathcal{F}_{t_{0}}\) is arbitrary, it follows that
$$ P^{(i)}_{t_{0}}= \mathbb{E}_{t_{0}}\bigg[\int _{t_{0}}^{+\infty } D^{(i)}_{t} e^{-\beta (t-t_{0})}\bigg(\frac{D^{(i)}_{t}}{D^{(i)}_{t_{0}}}\bigg)^{- \gamma } dt\bigg]. $$
(A.3)
To identify the safe rate \(r^{(i)}\), fix \(i \in \{1,2\}\), \(t_{1}>t_{0}>0\), \(\vartheta \in (0, t_{1}-t_{0})\) and \({K>1}\), and define stopping times
$$\begin{aligned} \eta _{K}&:=\inf \bigg\{ t>t_{0} \,:\, e^{\int ^{t}_{t_{0}} \vert r^{(i)}_{u} \vert du} \geq K \text{ and } P^{(i)}_{t} \leq \frac{1}{K} P^{(i)}_{t_{0}} \bigg\} , \\ \tau _{0}&:= (t_{0}+\vartheta )\wedge \eta _{K}\wedge \inf \bigg\{ t>t_{0} : D^{(i)}_{t} \geq K D^{(i)}_{t_{0}} \text{ and} \\ &\qquad \qquad \qquad \qquad \qquad \qquad \quad \frac{1}{P^{(i)}_{t} } \int ^{t}_{t_{0}} e^{\int ^{t}_{s} \vert r^{(i)}_{u} \vert du}D_{s}^{(i)}ds \geq \frac{K}{2} \bigg\} . \end{aligned}$$
For any \(0 < |\delta |<\frac{\vartheta }{K}\), further define another stopping time
$$ \tau _{1}:= t_{1}\wedge \eta _{K}\wedge \inf \bigg\{ t>\tau _{0} \,:\, \bigg\vert \frac{\delta }{\vartheta }\frac{1}{P^{(i)}_{t}}\int ^{\tau _{0}}_{t_{0}} e^{\int ^{t}_{s} r^{(i)}_{u} du}D_{s}^{(i)}ds\bigg\vert \geq \frac{1}{2} \bigg\} . $$
Consider an event \(A \in \mathcal{F}_{t_{0}}\) on which we adopt an alternative strategy in which, as before, from \(t_{0}\) to \(\tau _{0}\) the consumption changes from \(D^{(i)}_{t}\) to \({c^{\delta }_{t}=(1-\frac{\delta }{\vartheta })D^{(i)}_{t}}\), while the difference now is invested in the safe asset. Thus the corresponding wealth on \(A\) is \({Y^{\delta }_{t}=P^{(i)}_{t}+\frac{\delta }{\vartheta }\Delta _{t}}\), where \({\Delta _{t}=\int ^{t}_{t_{0}} e^{\int ^{t}_{s} r^{(i)}_{u} du}D_{s}^{(i)} ds}\) on \(A\). From time \(\tau _{0}\) to \(\tau _{1}\), the strategy remains \((c^{\delta }_{t},\phi ^{\delta }_{t})=(D^{(i)}_{t},1)\), and thus wealth on \(A\) becomes \({Y^{\delta }_{t}=P^{(i)}_{t}+\frac{\delta }{\vartheta }\Delta _{t}}\), where \({\Delta _{t}=\int ^{\tau _{0}}_{t_{0}} e^{\int ^{t}_{s} r^{(i)}_{u} du} D^{(i)}_{s} ds}\) on \(A\). After \(\tau _{1}\), the lump sum \({\frac{\delta }{\vartheta }\Delta _{\tau _{1}}}\) is used to buy \({\frac{\delta }{\vartheta } \frac{\Delta _{\tau _{1}}}{P^{(i)}_{\tau _{1}}}}\) units of stock so that the number of shares on \(A\) becomes \({\phi ^{\delta }_{t}=1+\frac{\delta }{\vartheta } \frac{\Delta _{\tau _{1}}}{P^{(i)}_{\tau _{1}}}}\) and the consumption rate is \({c^{\delta }_{t}=(1+ \frac{\delta }{\vartheta } \frac{\Delta _{\tau _{1}}}{P^{(i)}_{\tau _{1}}})D^{(i)}_{t}}\) for \(t \geq \tau _{1}\) on \(A\). Then the change in expected utility from \((c_{t},\phi _{t})_{t\geq 0}\) to \((c^{\delta }_{t},\phi ^{\delta }_{t})_{t\geq 0}\) is
$$\begin{aligned} \Delta ^{\delta }J = \mathbb{E}\bigg[1_{A}\bigg(&\int ^{\tau _{0}}_{t_{0}} \frac{ e^{-\beta t} ( (1-\frac{\delta }{\vartheta } )D^{(i)}_{t} )^{1-\gamma }}{1-\gamma } dt \\ & +\int _{\tau _{1}}^{+\infty } \frac{ e^{-\beta t} ( (1+\frac{\delta }{\vartheta }\frac{\Delta _{\tau _{1}}}{P^{(i)}_{\tau _{1}}} )D^{(i)}_{t} )^{1-\gamma }}{1-\gamma } dt \!-\!\int _{t_{0}}^{+\infty } \frac{ e^{-\beta t} (D^{(i)}_{t} )^{1-\gamma }}{1-\gamma } dt\!\bigg) \bigg]. \end{aligned}$$
Again, concavity implies that \(y^{-\gamma }(y-x) \leq \frac{y^{1-\gamma }}{1-\gamma }- \frac{x^{1-\gamma }}{1-\gamma } \text{ for any } x\), \(y>0\); thus
$$\begin{aligned} \Delta ^{\delta }J \geq \mathbb{E}\bigg[1_{A}\bigg(&- \frac{\delta }{\vartheta }\int ^{\tau _{0}}_{t_{0}} e^{-\beta t} D^{(i)}_{t} \bigg(\Big(1-\frac{\delta }{\vartheta }\Big)D^{(i)}_{t}\bigg)^{-\gamma } dt \\ & +\frac{\delta }{\vartheta } \frac{\Delta _{\tau _{1}}}{P^{(i)}_{\tau _{1}}}\int _{\tau _{1}}^{+ \infty } e^{-\beta t}D^{(i)}_{t}\bigg(\Big(1+\frac{\delta }{\vartheta } \frac{\Delta _{\tau _{1}}}{P^{(i)}_{\tau _{1}}}\Big)D^{(i)}_{t} \bigg)^{-\gamma } dt\bigg)\bigg]. \end{aligned}$$
As \((c_{t},\phi _{t})_{t\geq 0}\) is optimal, we have \(\lim _{\delta \downarrow 0}\frac{ \Delta ^{\delta }J}{\delta }\leq 0\) and \(\lim _{\delta \uparrow 0}\frac{ \Delta ^{\delta }J}{\delta }\geq 0\) and thus
$$\begin{aligned} \lim _{\delta \rightarrow 0} \frac{1}{\vartheta }\mathbb{E}\bigg[1_{A} \bigg(&-\int ^{\tau _{0}}_{t_{0}} e^{-\beta t} D^{(i)}_{t} \bigg( \Big(1-\frac{\delta }{\vartheta }\Big)D^{(i)}_{t}\bigg)^{-\gamma } dt \\ &+\frac{\Delta _{\tau _{1}}}{ P^{(i)}_{\tau _{1}}}\int _{\tau _{1}}^{+ \infty } D^{(i)}_{t} e^{-\beta t}\bigg(\Big(1+\frac{\delta }{\vartheta } \frac{\Delta _{\tau _{1}}}{P^{(i)}_{\tau _{1}}}\Big)D^{(i)}_{t} \bigg)^{-\gamma } dt\bigg)\bigg] = 0. \end{aligned}$$
Because \(\lim _{\delta \rightarrow 0}\tau _{1}=\tau _{2}:=\min \{t_{1},\eta _{K} \}\) almost surely and \(\frac{\Delta _{\tau _{1}}}{P^{(i)}_{\tau _{1}}} \leq (\tau _{1}-t_{0})K^{3} \frac{D^{(i)}_{t_{0}}}{P^{(i)}_{t_{0}}}\), dominated convergence yields
$$ \lim _{\vartheta \downarrow 0} \frac{\mathbb{E} [1_{A} (-\int ^{\tau _{0}}_{t_{0}} e^{-\beta t} (D^{(i)}_{t} )^{1-\gamma } dt+\frac{\Delta _{\tau _{2}}}{ P^{(i)}_{\tau _{2}}}\int _{\tau _{2}}^{+\infty } e^{-\beta t} (D^{(i)}_{t} )^{1-\gamma } dt ) ]}{\vartheta }= 0. $$
(A.4)
Recalling that \({\Delta _{\tau _{2}}=\int ^{\tau _{0}}_{t_{0}} e^{\int ^{\tau _{2}}_{s} r^{(i)}_{u} du} D^{(i)}_{s} ds}\), it follows that
$$\begin{aligned} \lim _{\vartheta \downarrow 0} \frac{\Delta _{\tau _{2}}}{\vartheta } =&e^{ \int ^{\tau _{2}}_{t_{0}} r^{(i)}_{u} du} D^{(i)}_{t_{0}} . \end{aligned}$$
Applying the dominated convergence theorem to (A.4), (A.3) implies that
$$\begin{aligned} &\mathbb{E}\bigg[1_{A}\bigg(-e^{-\beta t_{0}} (D^{(i)}_{t_{0}} )^{1- \gamma }+ \frac{e^{\int ^{\tau _{2}}_{t_{0}} r^{(i)}_{u} du} D^{(i)}_{t_{0}} }{P^{(i)}_{\tau _{2}}} \mathbb{E}_{\tau _{2}}\Big[\int _{\tau _{2}}^{+\infty } D^{(i)}_{t} e^{- \beta t} (D^{(i)}_{t} )^{-\gamma } dt\Big]\bigg)\bigg] \\ &= e^{-\beta t_{0}}\mathbb{E}\big[1_{A} D^{(i)}_{t_{0}}\big(- (D^{(i)}_{t_{0}} )^{-\gamma }+e^{\int ^{\tau _{2}}_{t_{0}} r^{(i)}_{u} du} e^{-\beta ( \tau _{2}-t_{0})} (D^{(i)}_{\tau _{2}} )^{-\gamma }\big)\big] =0 . \end{aligned}$$
As \(A \in \mathcal{F}_{t_{0}}\) is arbitrary, we get
$$\begin{aligned} e^{\int ^{t_{0}}_{0} (r^{(i)}_{u}-\beta ) du} (D^{(i)}_{t_{0}} )^{- \gamma }= \mathbb{E}_{t_{0}} \big[e^{\int ^{\min \{t_{1},\eta _{K} \}}_{0} (r^{(i)}_{u}-\beta ) du} (D^{(i)}_{\min \{t_{1},\eta _{K} \}} )^{- \gamma } \big], \end{aligned}$$
which implies that \(( e^{\int ^{t}_{0} (r^{(i)}_{u}-\beta ) du} (D^{(i)}_{t})^{-\gamma })_{t \geq 0}\) is a local martingale. □
1.2 A.2 Proof of Theorem 4.1
The proof focuses on region 1, as the argument for region 2 is analogous.
(i) Recalling that \(e^{\int ^{t}_{0} r^{(1)}_{s} ds}M^{(1)}_{t}=e^{\int ^{t}_{0} (r^{(1)}_{s}- \beta ) ds} c_{t}^{\gamma }=e^{\int ^{t}_{0} (r^{(1)}_{s}-\beta ) ds}(D_{t}^{(1)})^{ \gamma }\) and applying Itô’s lemma, it follows that
$$\begin{aligned} &d \big(e^{\int ^{t}_{0} (r^{(1)}_{s}-\beta ) ds} (D_{t}^{(1)} )^{- \gamma }\big) \\ &= (r^{(1)}_{t}-\beta ) e^{\int ^{t}_{0} (r^{(1)}_{s}-\beta ) ds} (D_{t}^{(1)} )^{-\gamma } dt \\ &\phantom{=:}- \gamma e^{\int ^{t}_{0} (r^{(1)}_{s}-\beta ) ds} (D^{(1)}_{t} )^{- \gamma -1} \big((\mu -\kappa ) D^{(1)}_{t}+\kappa w_{1} D_{t} \big)dt \\ &\phantom{=:}+\frac{\gamma (1+\gamma )\sigma ^{2}}{2}e^{\int ^{t}_{0} (r^{(1)}_{s}- \beta ) ds} (D^{(1)}_{t} )^{-\gamma -1}D_{t} dt+ dL_{t} \\ &= e^{\int ^{t}_{0} (r^{(1)}_{s}-\beta ) ds} (D^{(1)}_{t} )^{-\gamma } \\ &\phantom{=:}\times \bigg(r^{(1)}_{t}-\beta -\gamma (\mu -\kappa )-\gamma \kappa w_{1} \frac{D_{t}}{D^{(1)}_{t}}+\frac{\gamma (\gamma +1) \sigma ^{2}}{2} \frac{D_{t}}{D^{(1)}_{t}} \bigg) dt+dL_{t} \end{aligned}$$
for some local martingale \(L\) (a stochastic integral with respect to Brownian motion, whose expression is inconsequential). Thus the equilibrium safe rate is
$$ {r^{(1)}_{t}=\beta +\gamma (\mu -\kappa )+\gamma \mu w_{1} \frac{D_{t}}{D^{(1)}_{t}}-\frac{\gamma (\gamma +1) \sigma ^{2}}{2} \frac{D_{t}}{D^{(1)}_{t}}} . $$
Write the price of asset 1 in terms of \(X_{t}\) and \(D_{t}\) as
$$\begin{aligned} P^{(1)}_{t}&= \mathbb{E}_{t}\bigg[\int ^{\infty }_{t} e^{-\beta (s-t)} \frac{ (D_{s}^{(1)} )^{1-\gamma }}{ (D_{t}^{(1)} )^{-\gamma }} ds\bigg] \\ &=\mathbb{E}_{t}\bigg[\int ^{\infty }_{t} e^{-\beta (s-t)} \frac{ D_{s}^{1-\gamma }}{ D_{t}^{-\gamma }} \frac{X_{s}^{1-\gamma }}{X_{t}^{-\gamma }} ds\bigg] \\ &= \int ^{\infty }_{t} e^{-\beta (s-t)} \mathbb{E}_{t}\bigg[ \frac{ D_{s}^{1-\gamma }}{ D_{t}^{-\gamma }} \frac{X_{s}^{1-\gamma }}{X_{t}^{-\gamma }}\bigg] ds \\ &= \int ^{\infty }_{t} e^{-\beta (s-t)} \mathbb{E}_{t}\bigg[ \frac{ D_{s}^{1-\gamma }}{D_{t}^{-\gamma }}\bigg] \mathbb{E}_{t}\bigg[ \frac{X_{s}^{1-\gamma }}{X_{t}^{-\gamma }}\bigg] ds \\ &= D_{t} X_{t}^{\gamma } \int ^{\infty }_{t} e^{-\beta (s-t)} \mathbb{E}_{t} \bigg[\bigg(\frac{D_{s}}{D_{t}}\bigg)^{1-\gamma }\bigg]\mathbb{E}_{t} [X_{s}^{1- \gamma } ] ds , \end{aligned}$$
(A.5)
where the fourth equality follows by the independence of \((B^{D}_{t})_{t\geq 0}\) and \((B^{X}_{t})_{t\geq 0}\). As \((D_{t})_{t\geq 0}\) is a geometric Brownian motion, it satisfies
$$ \mathbb{E}\bigg[\bigg(\frac{D_{s}}{D_{t}}\bigg)^{1-\gamma } \bigg]= \mathbb{E} \big[e^{(1-\gamma ) (\mu -\frac{\sigma ^{2}}{2} )(s-t)+(1- \gamma )\sigma (B^{D}_{s}-B^{D}_{t} )} \big] =e^{(1-\gamma ) (\mu - \frac{\gamma \sigma ^{2}}{2} )(s-t)}. $$
Then (A.5) becomes
$$\begin{aligned} P^{(1)}_{t} =&D_{t} X_{t}^{\gamma }\int ^{\infty }_{t} e^{-\theta (s-t)} \mathbb{E}_{t} [X_{s}^{1-\gamma } ] ds =D_{t} X_{t}^{\gamma }\mathbb{E}_{t} \bigg[\int ^{\infty }_{t} e^{-\theta (s-t)} X_{s}^{1-\gamma } ds \bigg] \\ =& D_{t} X_{t}^{\gamma }\mathbb{E}\bigg[\int ^{\infty }_{t} e^{-\theta (s-t)} X_{s}^{1-\gamma } ds \bigg\vert X_{t} \bigg]= D_{t} X_{t}^{\gamma } f^{(1)}(X_{t}), \end{aligned}$$
where the third equality follows by the Markov property of \((X_{t})_{t\geq 0}\) and
$$\begin{aligned} f^{(1)}(x) :=\mathbb{E}\bigg[\int ^{\infty }_{t} e^{-\theta (s-t)} X_{s}^{1- \gamma } ds \bigg\vert X_{t}=x \bigg] =\mathbb{E}\bigg[\int ^{\infty }_{0} e^{-\theta s} X_{s}^{1-\gamma } ds \bigg\vert X_{0}=x\bigg]. \end{aligned}$$
Note that the second equality holds true because of the Markov property.
(ii) For the welfare, recall that
$$ \mathcal{U}^{(1)}_{t} =\mathbb{E}_{t}\bigg[\int ^{\infty }_{t} e^{- \beta (s-t)}\frac{c_{s}^{1-\gamma }}{1-\gamma } ds\bigg]=\mathbb{E}_{t} \bigg[\int ^{\infty }_{t} e^{-\beta (s-t)} \frac{ (D_{s}^{(1)} )^{1-\gamma }}{1-\gamma } ds\bigg]. $$
Comparing this expression to the formula for \(P_{t}^{(1)}\) in (A.5), note that
$$\begin{aligned} \mathcal{U}^{(1)}_{t} &= \frac{1}{1-\gamma } (D^{(1)}_{t} )^{-\gamma } P^{(1)}_{t}= \frac{1}{1-\gamma } (D^{(1)}_{t} )^{1-\gamma }X_{t}^{\gamma -1}f^{(1)}(X_{t}) \\ &= \frac{1}{1-\gamma } D_{t}^{1-\gamma }f^{(1)}(X_{t}). \end{aligned}$$
(iii) As \(P^{(1)}_{t}=D_{t} X_{t}^{\gamma }f^{(1)}(X_{t})\), where \(f^{(1)}\) is defined in (4.1), Itô’s formula yields
$$\begin{aligned} &dP^{(1)}_{t} \\ &= X_{t}^{\gamma }f^{(1)}(X_{t}) dD_{t}+ D_{t} X_{t}^{\gamma -1}\big( \gamma f^{(1)}(X_{t})+X_{t} (f^{(1)} )'(X_{t}) \big)dX_{t} \\ &\quad {} +X_{t}^{\gamma -1}\big(\gamma f^{(1)}(X_{t})+X_{t} (f^{(1)} )'(X_{t}) \big) d \langle D,X \rangle _{t} \\ &\quad {} +\frac{1}{2} D_{t} X_{t}^{\gamma -2}\big(\gamma (\gamma -1) f^{(1)}(X_{t})+2 \gamma X_{t} (f^{(1)} )'(X_{t}) +X_{t}^{2} (f^{(1)} )''(X_{t}) \big) d \langle X \rangle _{t} \\ &=D_{t} X_{t}^{\gamma -1} \bigg(\Big(\mu X_{t} f^{(1)}(X_{t})+\kappa (w_{1}-X_{t}) \big(\gamma f^{(1)}(X_{t})+X_{t} (f^{(1)} )'(X_{t}) \big) \\ & \qquad \qquad \ \ \,\quad {}+\frac{\sigma ^{2}}{2}(1-X_{t})\big( \gamma (\gamma -1) f^{(1)}(X_{t})+2\gamma X_{t} (f^{(1)} )'(X_{t}) \\ &\phantom{=:} \qquad \qquad \qquad \qquad \qquad \,\ \ {} +X_{t}^{2} (f^{(1)} )''(X_{t}) \big)\Big) dt +\sigma X_{t} f^{(1)}(X_{t}) dB^{D}_{t} \\ & \phantom{=:}\qquad \quad \quad {} +\sigma \sqrt{X_{t}(1-X_{t})}\big(\gamma f^{(1)}(X_{t})+X_{t} (f^{(1)} )'(X_{t}) \big) dB^{X}_{t}\bigg) \\ &= \sigma D_{t} X_{t}^{\gamma }f^{(1)} ( X_{t} )dB^{D}_{t} + A^{1}_{t} dt \\ & \phantom{=:}+ \sigma D_{t} X_{t}^{\gamma -1}\big(\gamma f^{(1)}(X_{t})+X_{t}({f^{(1)}})'(X_{t}) \big)\sqrt{X_{t} (1-X_{t})}dB^{X}_{t} \\ &=\sigma D_{t} X_{t}^{\gamma }f^{(1)} ( X_{t} )dB^{D}_{t} + \sigma D_{t} X_{t}^{\gamma -1}h^{(1)} ( X_{t} )\sqrt{X_{t} (1-X_{t})}dB^{X}_{t} + A^{1}_{t} dt \end{aligned}$$
(A.6)
for an adapted process \(A^{1}\). An analogous computation for region 2 yields
$$\begin{aligned} &dP^{(2)}_{t} \\ &= \sigma D_{t} (1-X_{t})^{\gamma }f^{(2)} ( X_{t} ) dB^{D}_{t} + A^{2}_{t} dt \\ &\phantom{=:}+ \sigma D_{t} (1-X_{t})^{\gamma -1} \big(-\gamma f^{(2)}(X_{t})+(1-X_{t})({f^{(2)}})'(X_{t}) \big)\sqrt{X_{t} (1-X_{t})}dB^{X}_{t} \\ &= \sigma D_{t} (1-X_{t})^{\gamma }f^{(2)} ( X_{t} ) dB^{D}_{t} \\ & \phantom{=:}+\sigma D_{t} (1-X_{t})^{\gamma -1}h^{(2)} (X_{t} )\sqrt{X_{t} (1-X_{t})} dB^{X}_{t}+ A^{2}_{t} dt \end{aligned}$$
for another adapted process \(A^{2}\). Hence
$$\begin{aligned} \frac{d \langle P^{(1)} \rangle _{t}}{dt} &= \sigma ^{2} D^{2}_{t} X_{t}^{2( \gamma -1)} \Big( \big(X_{t} f^{(1)}(X_{t})\big)^{2} +\big(h^{(1)}(X_{t}) \big)^{2} X_{t}(1-X_{t})\Big), \\ \frac{d \langle P^{(2)} \rangle _{t}}{dt}& = \sigma ^{2} D^{2}_{t} (1-X_{t})^{2( \gamma -1)} \\ &\phantom{=:}\times \Big( \big((1-X_{t}) f^{(2)}(X_{t})\big)^{2} +\big(h^{(2)}(X_{t}) \big)^{2} X_{t}(1-X_{t})\Big), \\ \frac{d \langle P^{(1)}, P^{(2)} \rangle _{t}}{dt} &= \sigma ^{2} D^{2}_{t} X_{t}^{\gamma }(1-X_{t})^{\gamma } \big( f^{(1)}(X_{t}) f^{(2)}(X_{t}) +h^{(1)}(X_{t})h^{(2)}(X_{t}) \big), \end{aligned}$$
and the claim follows by (4.5).
(iv) Again, the proof focuses on region 1, as the argument for region 2 is analogous. Following the derivation of (A.6), we get
$$\begin{aligned} &\frac{dP^{(1)}_{t}}{P^{(1)}_{t}} - \sigma dB^{D}_{t} - \sigma \sqrt{ \frac{1-X_{t}}{X_{t}}}\bigg(\gamma +X_{t} \frac{ (f^{(1)} )'(X_{t})}{f^{(1)}(X_{t})} \bigg) dB^{X}_{t} \\ &= \bigg(\mu +\kappa \frac{w_{1}-X_{t}}{X_{t}}\Big(\gamma +X_{t} \frac{ (f^{(1)} )'(X_{t})}{f^{(1)}(X_{t})} \Big) \\ &\qquad {}+ \frac{\sigma ^{2}}{2}\frac{1-X_{t}}{X_{t}}\Big( \gamma (\gamma -1) +2\gamma X_{t} \frac{ (f^{(1)} )'(X_{t})}{f^{(1)}(X_{t})}+X_{t}^{2} \frac{ (f^{(1)} )''(X_{t})}{f^{(1)}(X_{t})} \Big)\bigg) \\ &=: \mu ^{(1)}_{t}-\frac{D^{(1)}_{t}}{P^{(1)}_{t}} dt \end{aligned}$$
which proves the claim. □
Next, under Assumption 4.2, closed-form expressions for the \(f^{(i)}\) in terms of hypergeometric functions follow (cf. the formulas derived by Hurd and Kuznetsov [30] for related functionals of the Jacobi process).
Proposition A.3
Let Assumptions 3.4and 4.2hold. Then
$$\begin{aligned} f^{(1)}(x) &= \frac{2}{\sigma ^{2} \omega ^{(1)}}\bigg( \frac{\Gamma ^{(1)}_{1}}{1+c_{1}-\gamma } F^{(1)}_{1}(x) G^{(1)}_{1}(1)+ \frac{\Gamma ^{(1)}_{2}}{1+c_{1}-\gamma } x^{2-\gamma } F^{(1)}_{2}(x) G^{(1)}_{1}(x) \\ &\qquad \qquad \quad {}+ \frac{\Gamma ^{(1)}_{2}}{2-\gamma } F^{(1)}_{1}(x) \big(G^{(1)}_{2}(1)-x^{2-\gamma } G^{(1)}_{2}(x)\big) \bigg), \\ f^{(2)}(x) &= \frac{2}{\sigma ^{2} \omega ^{(2)}}\bigg( \frac{\Gamma ^{(2)}_{1}}{1+c_{2}-\gamma } F^{(2)}_{1}(1-x) G^{(2)}_{1}(1) \\ &\qquad \qquad \quad {}+ \frac{\Gamma ^{(2)}_{2} (1-x)^{2-\gamma }}{1+c_{2}-\gamma } F^{(2)}_{2}(1-x) G^{(2)}_{1}(1-x) \\ &\qquad \qquad \quad {}+ \frac{\Gamma ^{(2)}_{2}}{2-\gamma } F^{(2)}_{1}(1-x) \big(G^{(2)}_{2}(1)-(1-x)^{2-\gamma } G^{(2)}_{2}(1-x)\big) \bigg), \end{aligned}$$
where
$$\begin{aligned} F^{(i)}_{1}(x)&:={}_{2}F_{1}(a,b;c_{i};x), \\ G^{(i)}_{1}(x)&:={}_{3}F_{2}(c_{i}-a,c_{i}-b,1+c_{i}-\gamma ;c_{i},2+c_{i}- \gamma ;x), \\ F^{(i)}_{2}(x)&:={}_{2}F_{1}(b+1-c_{i},a+1-c_{i};2-c_{i};x), \\ G^{(i)}_{2}(x)&:={}_{3}F_{2}(1-b,1-a,2-\gamma ;2-c_{i},3-\gamma ;x), \\ a&:= \frac{- (1-\frac{2\kappa }{\sigma ^{2}} )+\sqrt{ (1-\frac{2\kappa }{\sigma ^{2}} )^{2}-\frac{8\theta }{\sigma ^{2}}}}{2}, \\ b&:= \frac{- (1-\frac{2\kappa }{\sigma ^{2}} )-\sqrt{ (1-\frac{2\kappa }{\sigma ^{2}} )^{2}-\frac{8\theta }{\sigma ^{2}}}}{2}, \\ c_{i}& :=\frac{2\kappa w_{i} }{\sigma ^{2}}, \end{aligned}$$
\(\Gamma ^{(i)}_{1}:= \frac{\Gamma (a+b+1-c_{i})\Gamma (1-c_{i})}{\Gamma (a+1-c_{i})\Gamma (b+1-c_{i})}\), \(\Gamma ^{(i)}_{2}:= \frac{\Gamma (a+b+1-c_{i})\Gamma (c_{i}-1)}{\Gamma (a)\Gamma (b)}\), and the Wronskian constant \(\omega ^{(i)}\)is
$$ \omega ^{(i)}:=\big(\varphi ^{(i)} (y) (F_{1}^{(i)} )' (y)- (\varphi ^{(i)} )'(y)F_{1}^{(i)}(y)\big)(1-y)^{\frac{2\kappa (1-w_{i})}{\sigma ^{2}}} y^{ \frac{2\kappa w_{i} }{\sigma ^{2}}} $$
with \(\varphi ^{(i)}(x)={}_{2}F_{1}(a,b;a+b+1-c_{i};1-x)\)and the hypergeometric functions \({}_{2}F_{1}\)and \({}_{3}F_{2}\)given by
$$\begin{aligned} {}_{2}F_{1}(a,b;c;x)&:=\sum ^{\infty }_{n=0} \frac{(a)_{n} (b)_{n}}{(c)_{n} }\frac{x^{n}}{n!}, \\ {}_{3}F_{2}(a,b,p;c,q;x)&:=\sum ^{\infty }_{n=0} \frac{(a)_{n} (b)_{n} (p)_{n}}{(c)_{n} (q)_{n}}\frac{x^{n}}{n!}, \end{aligned}$$
where \((p)_{n}:=p (p+1) \cdots (p+n-2) (p+n-1)\).
Proof
The proof focuses on region 1, as the formulae for region 2 follow by replacing \(x\) with \(1-x\) and \(w_{1}\) with \(w_{2}\). First, we can characterise the function \(f^{(1)}\) in terms of the density function and speed measure of the process \(X\): we have
$$\begin{aligned} \mathbb{E}\bigg[\int ^{\infty }_{0} e^{-\theta s} X_{s}^{1-\gamma } ds \bigg\vert X_{0}=x\bigg] =&\int ^{\infty }_{0} e^{-\theta s} \bigg( \int ^{1}_{0} y^{1-\gamma } p(s;x,y) m(y) dy \bigg) ds \\ =& \int ^{1}_{0} y^{1-\gamma } \bigg(\int ^{\infty }_{0} e^{-\theta s} p(s;x,y) ds \bigg) m(y) dy, \end{aligned}$$
(A.7)
where \(p(s;x,y)\) is the transition density of the Markov process \(X_{t}\) at \(y\), conditionally on \(X_{0}=x\), with respect to the speed measure \(m(dy)=m(y)dy\). The representation of the Green function for scalar diffusions (e.g. Borodin and Salminen [9, II.10–11]) yields \({m(x)=\frac{2}{\sigma ^{2}}x^{c_{1}-1}(1-x)^{a+b-c_{1}}}\) and
$$\begin{aligned} \int ^{\infty }_{0} e^{-\theta s} p(s;x,y) ds= \textstyle\begin{cases} \frac{1}{\omega ^{(1)}} F_{1}^{(1)}(x)\varphi ^{(1)}(y),&\quad x\leq y, \\ \frac{1}{\omega ^{(1)}} F_{1}^{(1)}(y)\varphi ^{(1)}(x), &\quad x \geq y, \end{cases}\displaystyle \end{aligned}$$
where \(F_{1}^{(1)}\) and \(\varphi ^{(1)}\) are the fundamental solutions of the ordinary differential equation
$$ (x-x^{2} )g''(x)+\frac{2\mu }{\sigma ^{2}} (w_{1}-x) g'(x)= \frac{2\theta }{\sigma ^{2}} g(x) $$
with the respective boundary conditions
$$ \textstyle\begin{array}{rlrl} F_{1}^{(1)}(0+)&>0, \qquad &(F_{1}^{(1)} )'(0+) s(0+)&=0, \\ \varphi ^{(1)}(0+)&=+\infty , \qquad &(\varphi ^{(1)} )'(0+) s(0+)&>- \infty , \\ \varphi ^{(1)}(1-)&>0, \qquad &(\varphi ^{(1)} )'(1-) s(1-)&=0, \\ F_{1}^{(1)}(1-)&=+\infty , \qquad & (F_{1}^{(1)} )'(1-) s(1-)&>- \infty , \end{array} $$
where \(s(x):=\frac{2}{\sigma ^{2}}x^{c_{1}-1}(1-x)^{a+b-c_{1}}=x(1-x)m(x)\). Thus (A.7) further simplifies to
$$\begin{aligned} &\mathbb{E}\bigg[\int ^{\infty }_{0} e^{-\theta s} X_{s}^{1-\gamma } ds \bigg\vert X_{0}=x\bigg] \\ &= \frac{2}{\sigma ^{2} \omega ^{(1)}} \bigg({}_{2}F_{1}(a,b;a+b+1-c_{1};1-x) \\ &\phantom{=}\qquad \qquad {} \times \int ^{x}_{0} {}_{2}F_{1}(a,b;c_{1};y) (1-y)^{a+b-c_{1}} y^{c_{1}-\gamma } dy \\ &\phantom{=}\qquad \qquad {} + {}_{2}F_{1}(a,b;c_{1};x) \\ &\phantom{=}\qquad \qquad \quad {}\times \int ^{1}_{x} {}_{2}F_{1}(a,b;a+b+1-c_{1};1-y) (1-y)^{a+b-c_{1}} y^{c_{1}-\gamma } dy\bigg). \end{aligned}$$
Note that Assumption 4.2 ensures that the first integral on the right-hand side converges. Applying the identities for hypergeometric functions in Abramowitz and Stegun [1, Chap. 15.3], the expression for \(f^{(1)}\) simplifies to
$$\begin{aligned} &\mathbb{E}\bigg[\int ^{\infty }_{0} e^{-\theta s} X_{s}^{1-\gamma } ds \bigg\vert X_{0}=x\bigg] \\ &= \frac{2}{\sigma ^{2} \omega ^{(1)}}\bigg(\Gamma _{1}^{(1)} {}_{2}F_{1}(a,b;c_{1};x) \int ^{x}_{0} {}_{2}F_{1}(a,b;c_{1};y) (1-y)^{a+b-c_{1}} y^{c_{1}- \gamma } dy \\ &\qquad \qquad \quad {} + \Gamma _{2}^{(1)} x^{1-c_{1}} {}_{2}F_{1}(b+1-c_{1},a+1-c_{1};2-c_{1};x) \\ &\qquad \qquad \qquad {} \times \int ^{x}_{0} {}_{2}F_{1}(a,b;c_{1};y) (1-y)^{a+b-c_{1}} y^{c_{1}-\gamma } dy \\ &\qquad \qquad \quad {}+ \Gamma _{1}^{(1)} {}_{2}F_{1}(a,b;c_{1};x) \int ^{1}_{x} {}_{2}F_{1}(a,b;c_{1};y) (1-y)^{a+b-c_{1}} y^{c_{1}- \gamma } dy \\ &\qquad \qquad \quad {} + \Gamma _{2}^{(1)} {}_{2}F_{1}(a,b;c_{1};x) \\ &\qquad \qquad \qquad {} \times \int ^{1}_{x} {}_{2}F_{1}(b+1-c_{1},a+1-c_{1};2-c_{1};y) (1-y)^{a+b-c_{1}} y^{1-\gamma } dy\bigg) \\ & = \frac{2}{\sigma ^{2} \omega ^{(1)}}\bigg(\Gamma _{1}^{(1)} {}_{2}F_{1}(a,b;c_{1};x) \int ^{1}_{0} {}_{2}F_{1}(c_{1}-a,c_{1}-b;c_{1};y) y^{c_{1}-\gamma } dy \\ & \qquad \qquad \quad {} + \Gamma _{2}^{(1)} x^{1-c_{1}} {}_{2}F_{1}(b+1-c_{1},a+1-c_{1};2-c_{1};x) \\ &\qquad \qquad \qquad {} \times \int ^{x}_{0} {}_{2}F_{1}(c_{1}-a,c_{1}-b;c_{1};y) y^{c-\gamma } dy \\ & \qquad \qquad \quad {} + \Gamma _{2}^{(1)} {}_{2}F_{1}(a,b;c_{1};x) \int ^{1}_{x} {}_{2}F_{1}(1-b,1-a;2-c_{1};y) y^{1-\gamma } dy\bigg) \\ & = \frac{2}{\sigma ^{2} \omega ^{(1)}}\bigg( \frac{\Gamma _{1}^{(1)} {}_{2}F_{1}(a,b;c_{1};x)}{1+c_{1}-\gamma } {}_{3}F_{2}(c_{1}-a,c_{1}-b,1+c_{1}- \gamma ;c_{1},2+c_{1}-\gamma ;1) \\ &\qquad \qquad \quad {} + \Gamma _{2}^{(1)} \frac{x^{2-\gamma }}{1+c_{1}-\gamma } {}_{2}F_{1}(b+1-c_{1},a+1-c_{1};2-c_{1};x) \\ & \qquad \qquad \qquad {} \times {}_{3}F_{2}(c_{1}-a,c_{1}-b,1+c_{1}- \gamma ;c_{1},2+c_{1}-\gamma ;x) \\ &\qquad \qquad \quad {} + \Gamma _{2}^{(1)}\frac{1}{2-\gamma } {}_{2}F_{1}(a,b;c_{1};x) \\ &\qquad \qquad \qquad {} \times \Big({}_{3}F_{2}(1-b,1-a,2- \gamma ;2-c_{1},3-\gamma ;1) \\ & \qquad \qquad \qquad \qquad {} - x^{2-\gamma } {}_{3}F_{2}(1-b,1-a,2- \gamma ;2-c_{1},3-\gamma ;x) \Big)\Bigg), \end{aligned}$$
which yields the claim for \(f^{(1)}\). □
1.3 A.3 Proof of Theorem 4.3
(i) Note that
$$\begin{aligned} \bar{P}^{(1)}_{t} =& \mathbb{E}_{t}\bigg[\int ^{\infty }_{t} e^{-\beta (s-t)} \bigg(\frac{D_{s}}{D_{t}}\bigg)^{1-\gamma } \frac{D_{t}}{D_{s}} D^{(1)}_{s} ds\bigg] \\ =&D_{t} \, \mathbb{E}_{t}\bigg[\int ^{\infty }_{t} e^{-\beta (s-t)} \bigg(\frac{D_{s}}{D_{t}}\bigg)^{1-\gamma } X_{s} ds\bigg] \\ =& D_{t}\int ^{\infty }_{t} e^{-\beta (s-t)} \mathbb{E}_{t}\bigg[ \bigg(\frac{D_{s}}{D_{t}}\bigg)^{1-\gamma }\bigg] \mathbb{E}_{t} [X_{s} ] ds \\ =&D_{t}\int ^{\infty }_{t} e^{-\theta (s-t)} \mathbb{E}_{t} [X_{s} ] ds \\ =&D_{t}\int ^{\infty }_{t} e^{-\theta (s-t)} \mathbb{E} [X_{s}\vert X_{t} ] ds =D_{t}f(X_{t}), \end{aligned}$$
where the fifth equality follows by the Markov property of \((X_{t})_{t \geq 0}\) and \(f\) is defined as
$$ f(x) :=\int ^{\infty }_{t} e^{-\theta (s-t)} \mathbb{E}\left [X_{s} \vert X_{t}=x\right ] ds= \int ^{\infty }_{0} e^{-\theta s} \mathbb{E} \left [X_{s} \vert X_{0}=x\right ] ds, $$
(A.8)
where the second equality again uses the Markov property. Integrating (3.2) and taking the expectation yields \(\mathbb{E}[X_{t}\vert X_{0}=x]=e^{-\kappa t} x+w_{1}(1-e^{-\kappa t})\). Therefore, (A.8) further simplifies to
$$ f(x) =\int ^{\infty }_{0} e^{-\theta s} \big(e^{-\kappa s} x+w_{1} (1-e^{- \kappa s} )\big) ds=\frac{1}{\theta +\kappa }x+ \frac{\kappa }{(\theta +\kappa )\theta }w_{1} , $$
whence the price of asset 1 is
$$\begin{aligned} \bar{P}^{(1)}_{t}&= D_{t} f(X_{t}) \\ &= \frac{1}{\theta +\kappa }X_{t} D_{t} + \frac{\kappa }{(\theta +\kappa )\theta }w_{1} D_{t} \\ &= \frac{1}{\theta +\kappa } D^{(1)}_{t}+ \frac{\kappa w_{1}}{(\theta +\kappa )\theta } (D^{(1)}_{t}+D^{(2)}_{t} ) \\ &=\frac{1}{\theta }\bigg(\frac{\theta + \kappa w_{1}}{\theta +\kappa } D_{t}^{(1)} + \frac{\kappa w_{1}}{\theta +\kappa } D_{t}^{(2)}\bigg), \end{aligned}$$
and the price of asset 2 follows analogously.
As \(e^{\int ^{t}_{0} {\bar{r}_{s}} ds}\bar{M}_{t}=e^{\int ^{t}_{0} ({ \bar{r_{s}}}-\beta ) ds}D_{t}^{-\gamma }\), \(t \geq 0\), is a local martingale and
$$\begin{aligned} &d \big(e^{\int ^{t}_{0} (\bar{r_{s}}-\beta ) ds}D_{t}^{-\gamma } \big) \\ &=\left (\bar{r_{t}}-\beta \right ) e^{\int ^{t}_{0} (\bar{r_{s}}- \beta ) ds} D_{t}^{-\gamma } dt \\ &\phantom{=:}- \gamma e^{\int ^{t}_{0} (\bar{r_{s}}-\beta ) ds} D_{t}^{-\gamma } \bigg(\mu +\frac{(- \gamma -1) \sigma ^{2}}{2}\bigg)dt+ dL_{t} \\ &= e^{\int ^{t}_{0} (\bar{r_{s}}-\beta ) ds} D_{t}^{-\gamma } \bigg( \bar{r_{t}}-\beta -\gamma \Big(\mu -\frac{(\gamma +1)\sigma ^{2}}{2} \Big)\bigg) dt+ dL_{t}, \end{aligned}$$
for some local martingale \(L\), it follows that \(\bar{r_{t}}=\beta +\gamma \mu - \frac{\gamma (\gamma +1)\sigma ^{2}}{2}\).
(ii) For the welfare, note that
$$\begin{aligned} \overline{\mathcal{U}}_{t} =& \mathbb{E}_{t}\bigg[\int ^{\infty }_{t} e^{- \beta (s-t)}\frac{c_{s}^{1-\gamma }}{1-\gamma } ds\bigg] =\mathbb{E}_{t} \bigg[\int ^{\infty }_{t} e^{-\beta (s-t)} \frac{ D_{s}^{1-\gamma }}{1-\gamma } ds\bigg] \\ =&\frac{D_{t}^{1-\gamma }}{1-\gamma }\mathbb{E}_{t}\bigg[\int ^{\infty }_{t} e^{-\beta (s-t)} \left (\frac{D_{s}}{D_{t}}\right )^{1-\gamma } ds \bigg] \\ =&\frac{D_{t}^{1-\gamma }}{1-\gamma }\int ^{\infty }_{t} e^{-\beta (s-t)} \mathbb{E}_{t}\bigg[\bigg(\frac{D_{s}}{D_{t}}\bigg)^{1-\gamma } \bigg] ds \\ =&\frac{D_{t}^{1-\gamma }}{1-\gamma }\int ^{\infty }_{t} e^{-\theta (s-t)} ds =\frac{D_{t}^{1-\gamma }}{1-\gamma }\frac{1}{\theta }. \end{aligned}$$
(iii) In the integrated market, recalling that \(a_{0}\) and \(b_{i}\) are defined in Theorem 4.3, we have \(\bar{P}^{(1)}_{t}=a_{0} D_{t} X_{t}+b_{1} D_{t}\) and \(\bar{P}^{(2)}_{t}=a_{0} D_{t} (1-X_{t})+b_{2} D_{t}\). By Itô’s formula,
$$\begin{aligned} d\bar{P}^{(1)}_{t}&= \big(\mu (a_{0} X_{t}+b_{1}) D_{t} +a_{0} \kappa (w_{1}-X_{t}) D_{t}\big)dt \\ &\phantom{=:}+\sigma (a_{0} X_{t}+b_{1}) D_{t} dB^{D}_{t}+\sigma a_{0} D_{t} \sqrt{X_{t}(1-X_{t})} dB^{X}_{t}, \end{aligned}$$
(A.9)
$$\begin{aligned} d\bar{P}^{(2)}_{t}&= \Big(\mu \big(a_{0} (1-X_{t})+b_{2}\big) D_{t} -a_{0} \kappa (w_{1}-X_{t}) D_{t}\Big)dt \\ &\phantom{=:}+\sigma \big(a_{0} (1-X_{t})+b_{2}\big) D_{t} dB^{D}_{t}-\sigma a_{0} D_{t} \sqrt{X_{t}(1-X_{t})} dB^{X}_{t}, \end{aligned}$$
(A.10)
whence
$$\begin{aligned} \frac{d \langle \bar{P}^{(1)} \rangle _{t}}{dt}&=\sigma ^{2} D^{2}_{t} \big( (a_{0} X_{t}+b_{1} )^{2}+a_{0}^{2} X_{t}(1-X_{t})\big), \\ \frac{d \langle \bar{P}^{(2)} \rangle _{t}}{dt}&=\sigma ^{2} D^{2}_{t} \Big( \big(a_{0} (1-X_{t})+b_{2} \big)^{2}+a_{0}^{2} X_{t}(1-X_{t}) \Big), \\ \frac{d \langle \bar{P}^{(1)}, \bar{P}^{(2)} \rangle _{t}}{dt}&=\sigma ^{2} D^{2}_{t} \Big( (a_{0} X_{t}+b_{1} )\big(a_{0} (1-X_{t})+b_{2}\big)-a_{0}^{2} X_{t}(1-X_{t})\Big), \end{aligned}$$
and the claim follows by (4.5).
(iv) Equations (A.9) and (A.10) yield
$$\begin{aligned} \frac{d\bar{P}^{(1)}_{t}}{\bar{P}^{(1)}_{t}} =& \bigg(\mu + \frac{a_{0} \kappa (w_{1}-X_{t})}{a_{0} X_{t}+b_{1}} \bigg)dt+\sigma dB^{D}_{t}+ \sigma \frac{a_{0} \sqrt{X_{t}(1-X_{t})}}{a_{0} X_{t}+b_{1}} dB^{X}_{t}, \\ \frac{d\bar{P}^{(2)}_{t}}{\bar{P}^{(2)}_{t}} =& \bigg(\mu - \frac{a_{0} \kappa (w_{1}-X_{t})}{a_{0} (1-X_{t})+b_{2}} \bigg)dt+ \sigma dB^{D}_{t}-\sigma \frac{a_{0} \sqrt{X_{t}(1-X_{t})}}{a_{0} (1-X_{t})+b_{2}} dB^{X}_{t}. \end{aligned}$$
From Theorem 4.3 (iii), recall that \(a_{0}=\frac{1}{\theta +\kappa }\), \(b_{1}=\frac{\kappa w_{1}}{\theta (\theta +\kappa )}\) and \(b_{2}=\frac{\kappa w_{2}}{\theta (\theta +\kappa )}\). Thus
$$\begin{aligned} a_{0} \kappa w_{1}&= \frac{\kappa w_{1}}{\theta +\kappa }=\theta b_{1}, \qquad a_{0} \kappa (1-w_{1})= \frac{\kappa w_{2}}{\theta +\kappa }= \theta b_{2}, \\ 1-a_{0} \kappa &= \frac{\theta }{\theta +\kappa } = \theta a_{0}. \end{aligned}$$
Therefore we get
$$\begin{aligned} a_{0} \kappa (w_{1}-X_{t})+X_{t}&= a_{0} \kappa w_{1} +X_{t}(1-a_{0} \kappa )= \theta (b_{1}+a_{0} X_{t}), \\ -a_{0} \kappa (w_{1}-X_{t})+1-X_{t}&= \theta a_{0}+ \theta b_{2} - X_{t} \theta a_{0} = \theta \big(a_{0}(1-X_{t})+b_{2}\big), \end{aligned}$$
so that the expected returns are
$$\begin{aligned} \bar{\mu }^{(1)}_{t} =&\mu + \frac{a_{0} \kappa (w_{1}-X_{t})}{a_{0} X_{t}+b_{1}}+ \frac{X_{t}}{a_{0} X_{t}+b_{1}}=\mu +\theta , \\ \bar{\mu }^{(2)}_{t} =& \mu - \frac{a_{0} \kappa (w_{1}-X_{t})}{a_{0} (1-X_{t})+b_{2}}+ \frac{1-X_{t}}{a_{0} (1-X_{t})+b_{2}}=\mu +\theta . \end{aligned}$$
□
1.4 A.4 Proof of Proposition 5.1
The proof focuses on region 1, as region 2 is analogous.
In the segmented market, the pricing formula in Proposition A.3 yields
$$\begin{aligned} &\lim _{T\rightarrow \infty }\mathbb{E}\bigg[ \frac{P_{T}^{(1)}}{D_{T}^{(1)}}\bigg] \\ & = \lim _{T \rightarrow \infty }\mathbb{E} [X_{T}^{\gamma -1}f^{(1)}(X_{T}) ]= \int ^{1}_{0} x^{\gamma -1}f^{(1)}(x) m(x) dx \\ &= \int ^{1}_{0} \frac{2 x^{c_{1}+\gamma -2} (1-x)^{a+b-c_{1}}}{\sigma ^{2} \omega ^{(1)} B(a+b-c_{1}+1,c_{1})} \\ &\phantom{=:} \qquad \times \bigg(\Gamma _{1}^{(1)} {}_{2}F_{1}(a,b;c_{1};x)\int ^{1}_{0} {}_{2}F_{1}(c_{1}-a,c_{1}-b;c_{1};y) y^{c_{1}-\gamma } dy \\ &\phantom{=:}\qquad \qquad + \Gamma _{2}^{(1)} x^{1-c_{1}} {}_{2}F_{1}(b+1-c_{1},a+1-c_{1};2-c_{1};x) \\ &\phantom{=:}\qquad \qquad \quad \times \int ^{x}_{0} {}_{2}F_{1}(c_{1}-a,c_{1}-b;c_{1};y) y^{c_{1}-\gamma } dy \\ & \phantom{=:}\qquad \qquad + \Gamma _{2}^{(1)} {}_{2}F_{1}(a,b;c_{1};x) \int ^{1}_{x} {}_{2}F_{1}(1-b,1-a;2-c_{1};y) y^{1-\gamma } dy\bigg) dx \\ & = \frac{2}{\sigma ^{2} \omega ^{(1)} B(a+b-c_{1}+1,c_{1})} \\ &\phantom{=:} \times \bigg(\Gamma _{1}^{(1)} \int ^{1}_{0} {}_{2}F_{1}(c_{1}-a,c_{1}-b;c_{1};x) x^{c_{1}+\gamma -2} dx \\ & \phantom{=:}\qquad \times \int ^{1}_{0} {}_{2}F_{1}(c_{1}-a,c_{1}-b;c_{1};y) y^{c_{1}- \gamma } dy \\ & \phantom{=:}\qquad + \Gamma _{2}^{(1)} \int ^{1}_{0} x^{\gamma -1} {}_{2}F_{1}(1-b,1-a;2-c_{1};x) \\ & \phantom{=:}\qquad \qquad \qquad {} \times \Big(\int ^{x}_{0} {}_{2}F_{1}(c_{1}-a,c_{1}-b;c_{1};y) y^{c_{1}-\gamma } dy \Big) dx \\ & \phantom{=:}\qquad + \Gamma _{2}^{(1)} \int ^{1}_{0} y^{1-\gamma } {}_{2}F_{1}(1-b,1-a;2-c_{1};y) \\ & \phantom{=:} \qquad \qquad \qquad {} \times \Big( \int ^{y}_{0} x^{c_{1}+ \gamma -2} {}_{2}F_{1}(a,b;c_{1};x) dx\Big)dy\bigg) \\ & = \frac{2}{\sigma ^{2} \omega ^{(1)} B(a+b-c_{1}+1,c_{1})} \\ &\phantom{=:}\times \bigg(\Gamma _{1}^{(1)} \frac{G_{1}^{(1)}(1)}{c_{1}+1-\gamma } \frac{G_{3}^{(1)}(1)}{c_{1}+\gamma -1} \\ &\quad \quad \quad \,\,+ \Gamma _{2}^{(1)} \int ^{1}_{0} x^{c_{1}} F_{3}^{(1)}(x) \Big(\frac{G_{1}^{(1)}(x)}{c_{1}+1-\gamma }+ \frac{G_{3}^{(1)}(x)}{c_{1}+\gamma -1}\Big) dx\bigg). \end{aligned}$$
On the other hand, in the integrated market, Theorem 4.3 implies that
$$\begin{aligned} &\lim _{T\rightarrow \infty }\mathbb{E}\bigg[ \frac{\bar{P}_{T}^{(1)}}{D_{T}^{(1)}}\bigg] \\ &=\lim _{T\rightarrow \infty }\mathbb{E}\left [ \frac{1}{\theta +\kappa }\left (1+\frac{\kappa w_{1}}{\theta } \frac{1}{X_{T}}\right )\right ]= \frac{1}{\theta +\kappa }\bigg(1+ \frac{\kappa w_{1}}{\theta }\int ^{1}_{0} \frac{1}{x} m(x) dx\bigg) \\ &= \frac{1}{\theta +\kappa }\bigg(1+\frac{\kappa w_{1}}{\theta }\int ^{1}_{0} \frac{ x^{c_{1}-2} (1-x)^{a+b-c_{1}}}{ B(a+b-c_{1}+1,c_{1})} dx \bigg)=\frac{1}{\theta +\kappa }\left (1+\frac{\kappa w_{1}}{\theta } \frac{a+b}{c_{1}-1}\right ) \\ &=\frac{1}{\theta +\kappa }\bigg(1+\frac{\kappa w_{1}}{\theta } \frac{2\frac{\kappa }{\sigma ^{2}}-1}{2\frac{\kappa w_{1}}{\sigma ^{2}}-1} \bigg), \end{aligned}$$
where \(m(x):=\frac{ x^{c_{1}-1} (1-x)^{a+b-c_{1}}}{ B(a+b-c_{1}+1,c_{1})}\). □
1.5 A.5 Proof of Proposition 5.2
The real exchange rate is the ratio between the stochastic discount factors in the two countries, i.e., \(p_{t}=\frac{M_{t}^{(1)}}{M_{t}^{(2)}}\). Under segmentation, Proposition A.2 yields that \(p_{t}= \frac{e^{-\beta t} (D^{(1)}_{t})^{-\gamma }}{e^{-\beta t} (D^{(2)}_{t})^{-\gamma }}\), and Proposition 5.2 follows by recalling that \(D^{(1)}_{t}=D_{t} X_{t}\) and \(D^{(2)}_{t}=D_{t} (1-X_{t})\). □
1.6 A.6 Proof of Proposition 6.1
Because of the market clearing conditions (6.1)–(6.3) and the obtained optimal consumption (6.4), the stochastic discount factor \(\bar{M}_{t}\) solves
$$ D_{t}=c^{(1)}_{t}+c^{(2)}_{t}= (y^{(1)}e^{\beta _{1} t} \bar{M}_{t} )^{- \frac{1}{\gamma }} + (y^{(2)}e^{\beta _{2} t} \bar{M}_{t} )^{- \frac{1}{2\gamma }}. $$
Solving this quadratic equation gives
$$\begin{aligned} \bar{M}_{t}^{-\frac{1}{2\gamma }} =& \frac{- (y^{(2)}e^{\beta _{2} t} )^{-\frac{1}{2\gamma }}+\sqrt{ (y^{(2)}e^{\beta _{2} t} )^{-\frac{1}{\gamma }}+4D_{t} (y^{(1)}e^{\beta _{1} t} )^{-\frac{1}{\gamma }} }}{2 (y^{(1)}e^{\beta _{1} t} )^{-\frac{1}{\gamma }}} \\ =& \frac{ (y^{(2)} )^{-\frac{1}{2\gamma }}}{2 (y^{(1)} )^{-\frac{1}{\gamma }}}e^{ (\frac{\beta _{1}}{\gamma }-\frac{\beta _{2}}{2\gamma } )t}\bigg(-1+ \sqrt{1+ \frac{4 (y^{(1)} )^{-\frac{1}{\gamma }}}{ (y^{(2)} )^{-\frac{1}{\gamma }}} D_{t} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }t}}\bigg). \end{aligned}$$
With the stochastic discount factor \(\bar{M}_{t}\), the wealth process \(Y^{(i)}\) of agent \(i\) is
$$\begin{aligned} \bar{M}_{t} Y_{t}^{(i)}= \mathbb{E}_{t} \left [\int ^{\infty }_{t} \bar{M}_{u} c_{u}^{(i)} du \right ]. \end{aligned}$$
Thus the ratio of wealths becomes
$$\begin{aligned} \frac{Y_{t}^{(2)}}{Y_{t}^{(1)}} =& \frac{\mathbb{E}_{t} [\int ^{\infty }_{t} \bar{M}_{u} c_{u}^{(2)} du ]}{\mathbb{E}_{t} [\int ^{\infty }_{t} \bar{M}_{u} c_{u}^{(1)} du ]}= \frac{ (y^{(2)} )^{-\frac{1}{2\gamma }}}{ (y^{(1)} )^{-\frac{1}{\gamma }}} \frac{\mathbb{E}_{t} [\int ^{\infty }_{t} \bar{M}_{u}^{1-\frac{1}{2\gamma }} e^{-\frac{\beta _{2}}{2\gamma }u} du ]}{\mathbb{E}_{t} [\int ^{\infty }_{t} \bar{M}_{u}^{1-\frac{1}{\gamma }} e^{-\frac{\beta _{1}}{\gamma }u} du ]} \\ =& 2 \frac{\mathbb{E}_{t} [\int ^{\infty }_{t} (-1+\sqrt{1+K D_{u} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }u}} )^{1-2\gamma } e^{ (\frac{\beta _{1}}{\gamma }(1-2\gamma )-\frac{\beta _{2}}{\gamma }(1-\gamma ) )u} du ]}{\mathbb{E}_{t} [\int ^{\infty }_{t} (-1+\sqrt{1+K D_{u} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }u}} )^{2-2\gamma } e^{ (\frac{\beta _{1}}{\gamma }(1-2\gamma )-\frac{\beta _{2}}{\gamma }(1-\gamma ) )u} du ]}, \qquad \quad \end{aligned}$$
(A.11)
where \(K\) is a positive constant depending on \(y^{(i)}\) and \(\gamma \) only.
To establish an upper bound (respectively, lower bound) for the expectation in the numerator (respectively, denominator), consider first the case of \(\mu \!-\!\frac{1}{2}\sigma ^{2}\!+\!\frac{\beta _{2}\!-\!\beta _{1}}{\gamma }>\!0\). On the one hand,
$$\begin{aligned} &\mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \bigg(-1+\sqrt{1+K D_{u} e^{ \frac{\beta _{2}-\beta _{1}}{\gamma }u}}\bigg)^{1-2\gamma } e^{ ( \frac{\beta _{1}}{\gamma }(1-2\gamma )-\frac{\beta _{2}}{\gamma }(1- \gamma ) )u} du \bigg] \\ &\leq \mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \bigg( \frac{2+\sqrt{K D_{u} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }u}}}{K D_{u} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }u}} \bigg)^{2\gamma -1} e^{ (\frac{\beta _{1}}{\gamma }(1-2\gamma )- \frac{\beta _{2}}{\gamma }(1-\gamma ) )u} du \bigg] \\ &\leq 2^{2\gamma -2}\mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \Big( 2^{2 \gamma -1} \big(K D_{u} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }u} \big)^{1-2\gamma } + \big(K D_{u} e^{ \frac{\beta _{2}-\beta _{1}}{\gamma }u} \big)^{\frac{1-2\gamma }{2}} \Big) \\ & \qquad \qquad \qquad \qquad {}\times e^{ ( \frac{\beta _{1}}{\gamma }(1-2\gamma )-\frac{\beta _{2}}{\gamma }(1- \gamma ) )u} du \bigg] \\ &= \mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \big( K_{1} e^{-\beta _{2} u} D_{u}^{1-2\gamma } +K_{2} e^{ (\frac{\beta _{1}}{2\gamma }(1-2\gamma )- \frac{\beta _{2}}{2\gamma } )u} D_{u}^{\frac{1-2\gamma }{2}} \big) du \bigg] \\ &= K_{1} D_{t}^{1-2\gamma } \int ^{\infty }_{t} e^{-\beta _{2} u} e^{ ((1-2 \gamma )\mu - \gamma (1-2\gamma ) \sigma ^{2} )(u-t)} du \\ &\phantom{=:}+ K_{2} D_{t}^{\frac{1-2\gamma }{2}} \int ^{\infty }_{t} e^{ ( \frac{\beta _{1}}{2\gamma }(1-2\gamma )-\frac{\beta _{2}}{2\gamma } )u} e^{ (\frac{1-2\gamma }{2}\mu - \frac{1}{2} \frac{1-2\gamma }{2} \frac{1+2\gamma }{2} \sigma ^{2} )(u-t)} du \\ &= K_{1} e^{-\beta _{2} t} D_{t}^{1-2\gamma }+ K_{2} e^{ ( \frac{\beta _{1}}{2\gamma }(1-2\gamma )-\frac{\beta _{2}}{2\gamma } )t} D_{t}^{ \frac{1-2\gamma }{2}}, \end{aligned}$$
(A.12)
where \(K_{1}\) and \(K_{2}\) are some positive constants independent of \(t\). Above, the first inequality follows from \((\sqrt{1+x}-1)^{-\delta }=(\frac{\sqrt{1+x}+1}{x})^{\delta }\leq ( \frac{2+\sqrt{x}}{x})^{\delta }\) for all \(x>0\), \(\delta \geq 0\), and the second inequality from Jensen’s inequality \((x+y)^{\delta } \leq 2^{\delta -1 }(x^{\delta }+y^{\delta })\) for all \(x,y \geq 0\), \(\delta \geq 1\). On the other hand,
$$\begin{aligned} &\mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \bigg(-1+\sqrt{1+K D_{u} e^{ \frac{\beta _{2}-\beta _{1}}{\gamma }u}}\bigg)^{2-2\gamma } e^{ ( \frac{\beta _{1}}{\gamma }(1-2\gamma )-\frac{\beta _{2}}{\gamma }(1- \gamma ) )u} du \bigg] \\ &\geq \mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \big(K D_{u} e^{ \frac{\beta _{2}-\beta _{1}}{\gamma }u}\big)^{1-\gamma } e^{ ( \frac{\beta _{1}}{\gamma }(1-2\gamma )-\frac{\beta _{2}}{\gamma }(1- \gamma ) )u} du \bigg] \\ &= K^{1-\gamma }\mathbb{E}_{t} \bigg[\int ^{\infty }_{t} e^{-\beta _{1} u} D_{u}^{1-\gamma } du \bigg] = K_{3} e^{-\beta _{1} t} D_{t}^{1-\gamma }, \end{aligned}$$
(A.13)
where \(K_{3}\) is a positive constant independent of \(t\) and the inequality follows from \(-1+\sqrt{1+x} \leq \sqrt{x}\). With (A.12) and (A.13), (A.11) becomes
$$\begin{aligned} \frac{Y_{t}^{(2)}}{Y_{t}^{(1)}}&\leq 2 \frac{K_{1} e^{-\beta _{2} t} D_{t}^{1-2\gamma }+ K_{2} e^{ (\frac{\beta _{1}}{2\gamma }(1-2\gamma )-\frac{\beta _{2}}{2\gamma } )t} D_{t}^{\frac{1-2\gamma }{2}}}{K_{3} e^{-\beta _{1} t} D_{t}^{1-\gamma }} \\ &= K_{1} e^{(\beta _{1}-\beta _{2})t}D_{t}^{-\gamma }+K_{2} e^{ \frac{\beta _{1}-\beta _{2}}{2\gamma }t} D_{t}^{-\frac{1}{2}} \\ &= K_{1} D_{0}^{-\gamma } e^{-\gamma (\mu -\frac{1}{2}\sigma ^{2}+ \frac{\beta _{2}-\beta _{1}}{\gamma } )t-\gamma \sigma B^{D}_{t}}+K_{2} D_{0}^{-\frac{1}{2}} e^{-\frac{1}{2} (\mu -\frac{1}{2}\sigma ^{2}+ \frac{\beta _{2}-\beta _{1}}{\gamma } )t-\frac{1}{2} \sigma B^{D}_{t}}, \end{aligned}$$
which tends to zero as \(t \rightarrow \infty \) because \(\mu -\frac{1}{2}\sigma ^{2}+\frac{\beta _{2}-\beta _{1}}{\gamma }>0\).
Now consider the case of \(\mu -\frac{1}{2}\sigma ^{2}+\frac{\beta _{2}-\beta _{1}}{\gamma }<0\). On the one hand,
$$\begin{aligned} &\mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \bigg(-1+\sqrt{1+K D_{u} e^{ \frac{\beta _{2}-\beta _{1}}{\gamma }u}}\bigg)^{2-2\gamma } e^{ ( \frac{\beta _{1}}{\gamma }(1-2\gamma )-\frac{\beta _{2}}{\gamma }(1- \gamma ) )u} du \bigg] \\ &\leq \mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \bigg( \frac{2+\sqrt{K D_{u} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }u}}}{K D_{u} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }u}} \bigg)^{2\gamma -2} e^{ (\frac{\beta _{1}}{\gamma }(1-2\gamma )- \frac{\beta _{2}}{\gamma }(1-\gamma ) )u} du \bigg] \\ &\leq \mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \Big( K_{1} \big( D_{u} e^{ \frac{\beta _{2}-\beta _{1}}{\gamma }u}\big)^{2-2\gamma } + K_{2} \big( D_{u} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }u}\big)^{1-\gamma } \Big) \\ & \phantom{=:}\qquad \qquad {} \times e^{ (\frac{\beta _{1}}{\gamma }(1-2 \gamma )-\frac{\beta _{2}}{\gamma }(1-\gamma ) )u} du \bigg] \\ &= \mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \big( K_{1} e^{ ( \frac{\beta _{2}}{\gamma }(1-\gamma )-\frac{\beta _{1}}{\gamma } )u} D_{u}^{2-2 \gamma } + K_{2} e^{-\beta _{1} u} D_{u}^{1-\gamma }\big) du \bigg] \\ &= K_{1} e^{ (\frac{\beta _{2}}{\gamma }(1-\gamma )- \frac{\beta _{1}}{\gamma } )t} D_{t}^{2-2\gamma } + K_{2} e^{-\beta _{1} t} D_{t}^{1-\gamma }, \end{aligned}$$
where \(K_{1}\) and \(K_{2}\) are positive constants independent of \(t\). On the other hand,
$$\begin{aligned} &\mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \bigg(-1+\sqrt{1+K D_{u} e^{ \frac{\beta _{2}-\beta _{1}}{\gamma }u}}\bigg)^{1-2\gamma } e^{ ( \frac{\beta _{1}}{\gamma }(1-2\gamma )-\frac{\beta _{2}}{\gamma }(1- \gamma ) )u} du \bigg] \\ &\geq \mathbb{E}_{t} \bigg[\int ^{\infty }_{t} \big(K D_{u} e^{ \frac{\beta _{2}-\beta _{1}}{\gamma }u}\big)^{1-2\gamma } e^{ ( \frac{\beta _{1}}{\gamma }(1-2\gamma )-\frac{\beta _{2}}{\gamma }(1- \gamma ) )u} du \bigg] \\ &= K^{1-2\gamma } \mathbb{E}_{t} \bigg[\int ^{\infty }_{t} e^{-\beta _{2} u} D_{u}^{1-2\gamma } du \bigg] = K_{3} e^{-\beta _{2} t} D_{t}^{1-2 \gamma }, \end{aligned}$$
where \(K_{3}\) is a positive constant independent of \(t\) and the inequality holds due to \(-1+\sqrt{1+x}= \frac{x}{1+\sqrt{1+x}} \leq x\) and the fact that \(1-2\gamma <0\). Then (A.11) becomes
$$\begin{aligned} \frac{Y_{t}^{(2)}}{Y_{t}^{(1)}} \geq & 2 \frac{K_{3} e^{-\beta _{2} t} D_{t}^{1-2\gamma }}{K_{1} e^{ (\frac{\beta _{2}}{\gamma }(1-\gamma )-\frac{\beta _{1}}{\gamma } )t} D_{t}^{2-2\gamma } + K_{2} e^{-\beta _{1} t} D_{t}^{1-\gamma }} \\ =& \frac{1}{K_{1} e^{\frac{\beta _{2}-\beta _{1}}{\gamma }t}D_{t}+K_{2} e^{(\beta _{2}-\beta _{1})t} D_{t}^{\gamma }} \\ =& \frac{1}{K_{1} D_{0} e^{ (\mu -\frac{1}{2}\sigma ^{2}+\frac{\beta _{2}-\beta _{1}}{\gamma } )t+ \sigma B^{D}_{t}}+K_{2} D_{0}^{\gamma } e^{\gamma (\mu -\frac{1}{2}\sigma ^{2}+\frac{\beta _{2}-\beta _{1}}{\gamma } )t+\gamma \sigma B^{D}_{t}}}, \end{aligned}$$
which tends to \(+\infty \) as \(t \rightarrow \infty \) because \(\mu -\frac{1}{2}\sigma ^{2}+\frac{\beta _{2}-\beta _{1}}{\gamma }<0\).
