Risk management with expected shortfall

Abstract

This article studies optimal, dynamic portfolio and wealth/consumption policies of expected utility-maximizing investors who must also manage market-risk exposure which is measured by expected shortfall (ES). We find that ES managers can incur larger losses when losses occur, compared to benchmark managers. A general-equilibrium analysis reveals that the presence of ES managers increases the market volatility during periods of significant financial market stress. We propose weighted shortfall, a coherent and moreover spectral risk measure, that can rectify the shortcomings of ES.

Appendices

Appendix

General results under WVaR

In this appendix, we will summarize the main result in [33] which solves the utility maximization under a WVaR constraint. The WVaR of a portfolio X is defined as

$$\begin{aligned} \text {WVaR} _{\varPhi } (X)=-\int _{[0,1]} G_X(z) \varPhi (dz), \end{aligned}$$

where $\varPhi$ is a probability measure on [0, 1] (see [20] and [33]).

Consider the following problem

$$\begin{aligned} \begin{aligned} \max _{X(T)} ~&{\mathbb {E}}[u(X(T))]\\ \text {subject to} ~&{\mathbb {E}}[\xi X(T)]\le X(0),\\&\text {WVaR} _{\varPhi }(X(T))\le -\underline{X}. \end{aligned} \end{aligned}$$

(22)

We impose the following assumption to exclude a trivial case.

Assumption 10

$\text {WVaR} _{\varPhi }(X_B(T)) > -\underline{X}$.

This assumption stipulates that the benchmark agent’s optimal terminal wealth does not satisfy the risk constraint. Otherwise, the benchmark agent’s terminal wealth is already optimal to (22) and there is no need to impose the risk constraint.

Let us introduce the following notations. Define $\varphi _{\mu }(z) = - \int _{[z,1)} F_{\xi } ^{-1}(1-s) ds + \mu \varPhi ((z,1]), ~ z \in [0,1),$ with $\varphi _{\mu }(1) = 0$, and its left-continuous version $\varphi _{\mu }(z-) = - \int _{[z,1)} F_{\xi } ^{-1}(1-s) ds + \mu \varPhi ([z,1]), ~ z \in [0,1),$ with $\varphi _{\mu }(1-) = 0$, and it is assumed $\varPhi (\{ 1 \})=0$.

Denote the concave envelope of $\varphi _{\mu }(\cdot -)$ by $\delta _{\mu }(\cdot )$, in other words

$$\begin{aligned} \delta _{\mu }(z) = \sup _{0 \le a \le z \le b \le 1} \frac{(b-z)\varphi _{\mu }(a-) + (z-a)\varphi _{\mu }(b-)}{b-a}, ~z \in [0,1]. \end{aligned}$$

Let $\mathcal {A}_{\varPhi } =\{ \mu :\mu>0 ,\delta _{\mu }^{'} (z+)> 0, ~ \forall z \in [0,1) \}=\{ \mu :\mu >0 , \varphi _{\mu } (z-) < 0, ~ \forall z \in [0,1) \},$ where $\delta _{\mu }^{'} (\cdot +)$ is the right derivative of $\delta _{\mu }$. We impose the following assumption.

Assumption 11

There exist $\lambda$ and $\mu$ such that

$$\begin{aligned} \left\{ \begin{aligned}&\int _{[0,1)}I(\lambda \delta _{\mu }^{'}(z +))F_{\xi } ^{-1}(1-z) dz=X(0),\\&\int _{[0,1)} I(\lambda \delta _{\mu }^{'}(z +)) \varPhi (dz) = \underline{X}. \end{aligned}\right. \end{aligned}$$

We then have the following result which is based on Proposition B.1 in [33].

Proposition 11

Under Assumption 11, the optimal solution to (22) is given by

$$ X^{*}(T)=I(\lambda \delta _{\mu }^{'} ((1-F_{\xi }(\xi ))+)),$$

where $\lambda$ and $\mu \in \mathcal {A}_{\phi }$ solve ${\mathbb {E}}[\xi X^{*}(T)] = X(0)$ and $\rho _{\varPhi }(X^{*}(T))=-\underline{X}$.

Proofs

1.1 Proof of Proposition 3

We will make use of Proposition 11 to solve the ES agent’s optimization problem. Note that $\text {ES}_{\alpha }$ is a special case of WVaR with $\varPhi ([0,z]) = \frac{z}{\alpha } \wedge 1, z \in [0,1]$. Define

$$\begin{aligned} \varphi _{\text {ES}, y }(z)=\left\{ {\begin{array}{ll} - \int _z^1 F_{\xi } ^{-1}(1-s) ds +y \frac{\alpha - z}{\alpha } &{} ~ ~ ~ ~ z \in [0,\alpha ],\\ - \int _z^1 F_{\xi } ^{-1}(1-s) ds &{} ~ ~ ~ ~ z \in (\alpha ,1]. \end{array}} \right. \end{aligned}$$

Let $\delta _{\text {ES}, y }(\cdot )$ be the concave envelope of $\varphi _{\text {ES}, y }(\cdot )$ and $\mathcal {A}_{\text {ES}} = \{ y :y>0 ,\delta _{\text {ES}, y }^{'} (z+) > 0, ~z \in [0,1) \}$ where $\delta _{\text {ES}, y }^{'}(\cdot + )$ is the right derivative of $\delta _{\text {ES}, y }$. According to Proposition 11, the optimal solution to the ES agent’s problem is given by

$$\begin{aligned} X_{\text {ES}}(T) = I(\lambda \delta _{\text {ES}, \mu }^{'}( (1 - F_{\xi }(\xi ))+ )), \end{aligned}$$

(23)

where $\lambda >0$ and $\mu \in \mathcal {A}_{\text {ES}}$ are determined by ${\mathbb {E}}[\xi X_{\text {ES}}(T)] =X(0)$ and $\text {ES}_{\alpha }(X_{\text {ES}}(T)) = -\underline{X}$.

First, we derive $\mathcal {A}_{\text {ES}} = \{ y :y>0 ,\delta _{y }^{'} (z+) > 0, ~z \in [0,1) \}$.

Lemma 1

$$\begin{aligned} \mathcal {A}_{\text {ES}} = \left\{ y :y >0 \text { and } \varphi _{\text {ES}, y } \left( 1 - F_{\xi }\left( \frac{y }{\alpha }\right) \right) <0 \right\} . \end{aligned}$$

Proof

According to Lemma B.3 in [33], $\mathcal {A}_{\text {ES}} =\{ y :y >0 , \varphi _{\text {ES}, y } (z-) < 0, ~ \forall z \in [0,1) \}$. Note that $\varphi _{\text {ES}, y }(z)$ is continuous, it suffices to show $\{ y :y>0 , \varphi _{\text {ES}, y } (z)< 0, ~ \forall z \in [0,1) \} = \{ y :y >0 \text { and } \varphi _{\text {ES}, y } (1 - F_{\xi }(\frac{y }{\alpha })) <0 \}$.

First, if $y \le \alpha F_{\xi }^{-1}(1-\alpha )$, then $\varphi _{\text {ES}, y }^{'}(z+)>0, ~z \in [0,1)$ and consequently $\varphi _{\text {ES}, y }(z)<0, ~z \in [0,1)$.

Next, if $y > \alpha F_{\xi }^{-1}(1-\alpha )$, we have $\varphi _{\text {ES}, y }^{'}(z)>0$ for $z \in (0,1 - F_{\xi }(\frac{y }{\alpha }))$, $\varphi _{\text {ES}, y }^{'}(1 - F_{\xi }(\frac{y }{\alpha }))=0$, $\varphi _{\text {ES}, y }^{'}(z)<0$ for $z \in (1 - F_{\xi }(\frac{y }{\alpha }),\alpha )$, and $\varphi _{\text {ES}, y }^{'}(z)>0$ for $z \in (\alpha ,1)$. Therefore, $\varphi _{\text {ES}, y }(z)<0, ~z \in [0,1)$ if and only if $\varphi _{\text {ES}, y } (1 - F_{\xi }(\frac{y }{\alpha })) <0$. The rest of the proof follows from the fact that $\varphi _{\text {ES}, y }(z)$ is continuous and $\varphi _{\text {ES}, y }(1) = 0$. $\square$

Next, we derive an analytical expression for $\delta _{\text {ES}, y }(\cdot )$, the concave envelope of $\varphi _{\text {ES}, y }(\cdot )$. Suppose $y \in \mathcal {A}_{\text {ES}}$. Let $\overline{t}_{\text {ES} }:= 1 - F_{\xi }((F_{\xi }^{-1}(1-\alpha )- \frac{y }{\alpha }) \vee 0)$ and we have $\alpha < \overline{t}_{\text {ES} } \le 1$. For $\alpha< t < \overline{t}_{\text {ES} }$, define

$$\begin{aligned} s_{\text {ES},y }(t) =1-F_{\xi }(F_{\xi }^{-1}(1-t)+ \frac{y }{\alpha }). \end{aligned}$$

We have $0 \le s_{\text {ES},y }(t) <\alpha$ and $\varphi _{\text {ES}, y }^{'}(s_{\text {ES},y }(t) )=\varphi _{\text {ES}, y }^{'}(t)$ for $\alpha<t < \overline{t}_{\text {ES} }$. Define

$$\begin{aligned} h_{\text {ES},y }(t)= \varphi _{\text {ES}, y }(t)-\varphi _{\text {ES}, y }(s_{\text {ES},y }(t) )- \varphi _{\text {ES}, y }^{'}(t)(t-s_{\text {ES},y }(t) ). \end{aligned}$$

(24)

The following lemma is crucial in deriving $\delta _{\text {ES}, y }(\cdot )$.

Lemma 2

For $y \in \mathcal {A}_{\text {ES}}$, there exist $\alpha< t^{*}_{\text {ES}} < 1$ such that $h_{\text {ES},y }(t)<0$ for $\alpha<t < t^{*}_{\text {ES}}$, $h_{\text {ES},y }(t^{*}_{\text {ES}})=0$, and $h_{\text {ES},y }(t)>0, ~t^{*}_{\text {ES}}< t < \overline{t}_{\text {ES} }$.

Proof

Note that

$$\begin{aligned} \begin{aligned} h_{\text {ES},y }(t)&= \varphi _{\text {ES}, y }(t)-\varphi _{\text {ES}, y }(s_{\text {ES},y }(t) )- \varphi _{\text {ES}, y }^{'}(t)(t-s_{\text {ES},y }(t) )\\&= \int _{s_{\text {ES},y }(t) }^t F_{\xi }^{-1}(1-z)dz - y \frac{\alpha -s_{\text {ES},y }(t) }{\alpha } - F_{\xi }^{-1}(1-t)(t-s_{\text {ES},y }(t) ), \end{aligned} \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} h_{\text {ES},y }(t)&> F_{\xi }^{-1}(1-t)(t-s_{\text {ES},y }(t) ) - y \frac{\alpha -s_{\text {ES},y }(t) }{\alpha } - F_{\xi }^{-1}(1-t)(t-s_{\text {ES},y }(t) )\\&= - y \frac{\alpha -s_{\text {ES},y }(t) }{\alpha } , \end{aligned} \end{aligned}$$

and

$$\begin{aligned} h_{\text {ES},y }(t)< & {} F_{\xi }^{-1}(1-s_{\text {ES},y }(t) )(t-s_{\text {ES},y }(t) ) - y \frac{\alpha -s_{\text {ES},y }(t) }{\alpha } \\&- (F_{\xi }^{-1}(1-s_{\text {ES},y }(t) ) - \frac{y }{\alpha })(t-s_{\text {ES},y }(t) )\\= & {} \frac{y }{\alpha }(t - \alpha ). \end{aligned}$$

Thus, $h_{\text {ES},y }(\alpha +) \le 0$. If $\alpha F_{\xi }^{-1}(1-\alpha ) \ge y$, we have $\overline{t}_{\text {ES} }=1 - F_{\xi }(F_{\xi }^{-1}(1-\alpha )- \frac{y }{\alpha })$, $s_{\text {ES},y }(\overline{t}_{\text {ES} }-)=\alpha$, and $h_{\text {ES},y }(\overline{t}_{\text {ES} }-) \ge 0$. If $\alpha F_{\xi }^{-1}(1-\alpha ) < y$, we have $\overline{t}_{\text {ES} }=1$ and $s_{\text {ES},y }(\overline{t}_{\text {ES} }-)=1-F_{\xi }(\frac{y }{\alpha })$. By Lemma 1, we have

$$\begin{aligned} \begin{aligned} h_{\text {ES},y }(1-)&= \varphi _{\text {ES}, y }(1)-\varphi _{\text {ES}, y }\left( 1-F_{\xi }\left( \frac{y }{\alpha }\right) \right) - \varphi _{\text {ES}, y }^{'}(1-)\left( 1-1+F_{\xi }\left( \frac{y }{\alpha }\right) \right) \\&= - \varphi _{\text {ES}, y }\left( 1-F_{\xi }\left( \frac{y }{\alpha }\right) \right) >0. \end{aligned} \end{aligned}$$

Thus, $h_{\text {ES},y }(\overline{t}_{\text {ES} }-) \ge 0$. Since $h_{\text {ES},y }(\cdot )$ is continuous, there exists at least one t such that $h_{\text {ES},y }(t)=0$.

Next, for $\alpha<t_1< t_2 < \overline{t}_{\text {ES} }$,

$$\begin{aligned} \begin{aligned}&h_{\text {ES},y }(t_1) - h_{\text {ES},y }(t_2)\\&\quad = [\varphi _{\text {ES}, y }(t_1) - \varphi _{\text {ES}, y }(t_2)] - [\varphi _{\text {ES}, y }(s_{\text {ES},y }(t_1)) - \varphi _{\text {ES}, y }(s_{\text {ES},y }(t_2))] \\&\quad - [\varphi _{\text {ES}, y }^{'}(t_1)t_1 - \varphi _{\text {ES}, y }^{'}(t_2)t_2] + [\varphi _{\text {ES}, y }^{'}(s_{\text {ES},y }(t_1))s_{\text {ES},y }(t_1) - \varphi _{\text {ES}, y }^{'}(s_{\text {ES},y }(t_2))s_{\text {ES},y }(t_2)]\\&\quad = \int _{t_2}^{t_1} \varphi _{\text {ES}, y }^{'}(z)dz - \int _{s_{\text {ES},y }(t_2)}^{s_{\text {ES},y }(t_1)} \varphi _{\text {ES}, y }^{'}(z)dz - \left[ \int _{t_2}^{t_1} \varphi _{\text {ES}, y }^{'}(z)dz + \int _{t_2}^{t_1} zd\varphi _{\text {ES}, y }^{'}(z)\right] \\&\quad + \left[ \int _{s_{\text {ES},y }(t_2)}^{s_{\text {ES},y }(t_1)} \varphi _{\text {ES}, y }^{'}(z)dz + \int _{s_{\text {ES},y }(t_2)}^{s_{\text {ES},y }(t_1)} zd\varphi _{\text {ES}, y }^{'}(z)\right] \\&\quad = \int _{s_{\text {ES},y }(t_2)}^{s_{\text {ES},y }(t_1)} zd\varphi _{\text {ES}, y }^{'}(z) - \int _{t_2}^{t_1} zd\varphi _{\text {ES}, y }^{'}(z) \\&\quad = \int _{t_2}^{t_1} s_{\text {ES},y }(z)d\varphi _{\text {ES}, y }^{'}(s_{\text {ES},y }(z)) - \int _{t_2}^{t_1} zd\varphi _{\text {ES}, y }^{'}(z) \\&\quad = -\int _{t_1}^{t_2} [s_{\text {ES},y }(z) - z]d\varphi _{\text {ES}, y }^{'}(z) < 0. \end{aligned} \end{aligned}$$

Thus, $h_{\text {ES},y }(\cdot )$ is strictly increasing. This completes the proof. $\square$

Proposition 12

For $y \in \mathcal {A}_{\text {ES}}$,

$$\begin{aligned} \delta _{\text {ES}, y }(z)=\left\{ {\begin{array}{ll} \varphi _{\text {ES}, y }(z) &{} ~ ~ ~ ~ z \in [0,s_{\text {ES},y }(t^{*}_{\text {ES},y })),\\ \varphi _{\text {ES}, y }(s_{\text {ES},y }(t^{*}_{\text {ES},y })) +\varphi _{\text {ES}, y }^{'}(s_{\text {ES},y }(t^{*}_{\text {ES},y }))(z -s_{\text {ES},y }(t^{*}_{\text {ES},y })) &{} ~ ~ ~ ~ z \in [s_{\text {ES},y }(t^{*}_{\text {ES},y }),t^{*}_{\text {ES},y }],\\ \varphi _{\text {ES}, y }(z) &{} ~ ~ ~ ~ z \in (t^{*}_{\text {ES},y } ,1], \end{array}} \right. \end{aligned}$$

and

$$\begin{aligned} \delta _{\text {ES}, y }^{'}(z)=\left\{ {\begin{array}{ll} F_{\xi } ^{-1}(1-z) - \frac{y }{\alpha } &{} ~ ~ ~ ~ z \in [0,s_{\text {ES},y }(t^{*}_{\text {ES},y })),\\ F_{\xi } ^{-1}(1-t^{*}_{\text {ES},y }) &{} ~ ~ ~ ~ z \in [s_{\text {ES},y }(t^{*}_{\text {ES},y }),t^{*}_{\text {ES},y }],\\ F_{\xi } ^{-1}(1-z) &{} ~ ~ ~ ~ z \in (t^{*}_{\text {ES},y } ,1], \end{array}} \right. \end{aligned}$$

where $\alpha< t^{*}_{\text {ES},y } <1$ is the unique root of $h_{\text {ES},y }(t)=0$, and $s_{\text {ES},y }(t^{*}_{\text {ES},y })=1-F_{\xi }(F_{\xi }^{-1}(1-t^{*}_{\text {ES},y })+ \frac{y }{\alpha }) \in (0,\alpha )$.

Proof

$\delta _{\text {ES}, y } (\cdot )$ is obviously concave. Note that $\delta _{\text {ES}, y }(z)=\varphi _{\text {ES}, y }(z), ~z \in [0,s_{\text {ES},y }(t^{*}_{\text {ES},y })] \cup [t^{*}_{\text {ES},y } ,1]$, $\delta ^{'} _{\text {ES},y }(z) > \varphi ^{'} _{\text {ES},y }(z), ~z \in (s_{\text {ES},y }(t^{*}_{\text {ES},y }),\alpha )$, and $\delta ^{'} _{\text {ES},y }(z) < \varphi ^{'} _{\text {ES},y }(z), ~z \in (\alpha ,t^{*}_{\text {ES},y })$, we have $\delta _{\text {ES}, y }(z) > \varphi _{\text {ES}, y }(z), ~ z \in (s_{\text {ES},y }(t^{*}_{\text {ES},y }), t^{*}_{\text {ES},y })$. Moreover, $\delta ^{'}_{\text {ES},y }(\cdot )$ is constant on $(s_{\text {ES},y }(t^{*}),t^{*})$. We claim that $\delta _{\text {ES}, y }$ is the concave envelope of $\varphi _{\text {ES}, y }$, i.e., the smallest concave function that dominates $\varphi _{\text {ES}, y }$. For any concave function $\eta$ such that $\eta (z) \ge \varphi _{\text {ES}, y } (z), ~z \in [0,1]$, we show that $\eta (z) \ge \delta _{\text {ES}, y }(z), ~z \in [0,1]$. Obviously, $\eta (z) \ge \delta _{\text {ES}, y }(z), ~z \in [0,s_{\text {ES},y }(t^{*}_{\text {ES},y })] \cup [t^{*}_{\text {ES},y } ,1]$. For $z \in (s_{\text {ES},y }(t^{*}_{\text {ES},y }), t^{*}_{\text {ES},y })$, we have

$$\begin{aligned} \begin{aligned} \eta (z)&\ge \frac{t^{*}_{\text {ES},y } - z}{ t^{*}_{\text {ES},y } - s_{\text {ES},y }(t^{*}_{\text {ES},y })} \eta (s_{\text {ES},y }(t^{*}_{\text {ES},y })) + \frac{z - s_{\text {ES},y }(t^{*}_{\text {ES},y })}{ t^{*}_{\text {ES},y } - s_{\text {ES},y }(t^{*}_{\text {ES},y })} \eta (t^{*}_{\text {ES},y }) \\&\ge \frac{t^{*}_{\text {ES},y } - z}{ t^{*}_{\text {ES},y } - s_{\text {ES},y }(t^{*}_{\text {ES},y })} \varphi _{\text {ES}, y } (s_{\text {ES},y }(t^{*}_{\text {ES},y })) + \frac{z - s_{\text {ES},y }(t^{*}_{\text {ES},y })}{ t^{*}_{\text {ES},y } - s_{\text {ES},y }(t^{*}_{\text {ES},y })} \varphi _{\text {ES}, y } (t^{*}_{\text {ES},y }) \\&= \frac{t^{*}_{\text {ES},y } - z}{ t^{*}_{\text {ES},y } - s_{\text {ES},y }(t^{*}_{\text {ES},y })} \delta _{\text {ES}, y } (s_{\text {ES},y }(t^{*}_{\text {ES},y })) + \frac{z - s_{\text {ES},y }(t^{*}_{\text {ES},y })}{ t^{*}_{\text {ES},y } - s_{\text {ES},y }(t^{*}_{\text {ES},y })} \delta _{\text {ES}, y } (t^{*}_{\text {ES},y }) \\&= \delta _{\text {ES}, y } (z). \end{aligned} \end{aligned}$$

This completes the proof. $\square$

Plugging the expression of $\delta _{\text {ES}, \mu }$ into (23), we have

$$\begin{aligned} X_{\text {ES}}(T)=\left\{ \begin{array}{ll} I(\lambda \xi - \frac{1}{\alpha }\lambda \mu ) &{} ~ ~ ~ ~ F_{\xi }^{-1}(1-t^{*}_{\text {ES},\mu })+ \frac{\mu }{\alpha } \le \xi ,\\ I(\lambda F_{\xi }^{-1}(1-t^{*}_{\text {ES},\mu })) &{} ~ ~ ~ ~ F_{\xi }^{-1}(1-t^{*}_{\text {ES},\mu }) \le \xi< F_{\xi }^{-1}(1-t^{*}_{\text {ES},\mu })+ \frac{\mu }{\alpha } ,\\ I(\lambda \xi ) &{} ~ ~ ~ ~ \xi < F_{\xi }^{-1}(1-t^{*}_{\text {ES},\mu }). \end{array} \right. \end{aligned}$$

Substitute $\lambda _{\text {ES}}$ for $\lambda$, $\mu _{\text {ES}}$ for $\mu$, $\underline{\xi }_{\text {ES}}$ for $F_{\xi }^{-1}(1-t^{*}_{\text {ES},\mu })$, and $\overline{\xi }_{\text {ES}}$ for $F_{\xi }^{-1}(1-t^{*}_{\text {ES},\mu })+ \frac{\mu }{\alpha }$, we arrive at (5). Moreover, $h_{\mu _{\text {ES}} }(t^{*}_{\text {ES},\mu _{\text {ES}}})=0$ is equivalent to $h_{\mu _{\text {ES}}}(1- F_{\xi }(\underline{\xi }_{\text {ES}}))=0$.

We now show the last six assertions. First, because $s_{\text {ES},\mu _{\text {ES}}}(t^{*}_{\text {ES},\mu _{\text {ES}}})< \alpha < t^{*}_{\text {ES},\mu _{\text {ES}}}$ and ${\mathbb {P}}(\xi > \overline{\xi }_{\text {VaR}} ) = \alpha$, we have $\underline{\xi }_{\text {ES}}< \overline{\xi }_{\text {VaR}} < \overline{\xi }_{\text {ES}}$. Next, if $I(\lambda _{\text {ES}} \underline{\xi }_{\text {ES}}) \le \underline{X}$, then the ES constraint cannot be satisfied as $X_{\text {ES}}(T) < I(\lambda _{\text {ES}} \underline{\xi }_{\text {ES}})\le \underline{X}$ when $\xi > \overline{\xi }_{\text {ES}}$. Thus, $I(\lambda _{\text {ES}} \underline{\xi }_{\text {ES}}) > \underline{X}$.

$\lambda _{\text {VaR}} > \lambda _{\text {B}}$ is due to [7] or [33]. If $\lambda _{\text {ES}} \le \lambda _{\text {VaR}}$, then $\underline{\xi }_{\text {ES}} > \underline{\xi }_{\text {VaR}}$, due to the fact that $I(\lambda _{\text {ES}} \underline{\xi }_{\text {ES}})> \underline{X}=I(\lambda _{\text {VaR}} \underline{\xi }_{\text {VaR}})$. Moreover, since $I(\lambda _{\text {ES}} \xi ) \ge I(\lambda _{\text {VaR}} \xi ), I(\lambda _{\text {ES}} \xi - \frac{1}{\alpha } \lambda _{\text {ES}} \mu _{\text {ES}}) > I(\lambda _ {\text {VaR}} \xi )$, we have $X_{\text {ES}}(T) \ge X_{\text {VaR}}(T), ~ a.s.$, and the inequality is strict whenever $\xi > \underline{\xi }_{\text {VaR}}$, violating the budget constraint. Therefore, $\lambda _{\text {ES}} > \lambda _{\text {VaR}}$.

Similarly, as $I(\lambda _{\text {VaR}} \underline{\xi }_{\text {ES}})> I(\lambda _{\text {ES}} \underline{\xi }_{\text {ES}})> \underline{X}=I(\lambda _{\text {VaR}} \underline{\xi }_{\text {VaR}})$, we must have $\underline{\xi }_{\text {ES}} < \underline{\xi }_{\text {VaR}}$. The rest of the proof follows from direct comparison.

1.2 Proof of Proposition 4

Proof

The claims for the benchmark agent is standard. We show the ES agent part.

• 1. It is well-known that $\xi (t) X _{\text {ES}}(t)$ is a martingale and thus

  $$\begin{aligned} \begin{aligned} \xi (t) X _{\text {ES}}(t) = {\mathbb {E}}[\xi X _{\text {ES}}(T) |\mathcal {F}_t]=&{\mathbb {E}}[\xi I(\lambda _{\text {ES}} \xi ) \mathbf {1} _ {\xi< \underline{\xi }_{\text {ES}}} |\mathcal {F}_t] + {\mathbb {E}}[\xi I(\lambda _{\text {ES}} \underline{\xi }_{\text {ES}}) \mathbf {1} _ {\underline{\xi }_{\text {ES}} \le \xi < \overline{\xi }_{\text {ES}}} |\mathcal {F}_t] \\&+ {\mathbb {E}}[\xi I(\lambda _{\text {ES}} \xi - \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}}) \mathbf {1} _ {\xi \ge \overline{\xi }_{\text {ES}}} |\mathcal {F}_t]. \end{aligned} \end{aligned}$$

  When r and $\theta$ are constant, conditional on $\mathcal {F}_t$, $\ln \xi$ is normally distributed with mean $\ln \xi (t) - (r + \frac{\Vert \theta \Vert ^2 }{2} )(T-t)$ and variance $\Vert \theta \Vert ^2 (T-t)$. We have

  $$\begin{aligned} \begin{aligned} {\mathbb {E}}[\xi I(\lambda _{\text {ES}} \xi ) \mathbf {1} _ {\xi< \underline{\xi }_{\text {ES}}} |\mathcal {F}_t] =&\xi (t) \frac{e^{\varGamma (t)}}{(\lambda _{\text {ES}} \xi (t))^{\frac{1}{\gamma }}} N (d_1( \underline{\xi }_{\text {ES}})),\\ {\mathbb {E}}[\xi I(\lambda _{\text {ES}} \underline{\xi }_{\text {ES}}) \mathbf {1} _ {\underline{\xi }_{\text {ES}} \le \xi < \overline{\xi }_{\text {ES}}} |\mathcal {F}_t] =&\xi (t) \underline{X}_{\text {ES}} e^{-r(T-t)}[N(-d_2 (\underline{\xi }_{\text {ES}})) - N(-d_2 (\overline{\xi }_{\text {ES}}))], \end{aligned} \end{aligned}$$
  $$\begin{aligned} \begin{aligned}&{\mathbb {E}}[\xi I(\lambda _{\text {ES}} \xi - \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}}) \mathbf {1} _ {\xi \ge \overline{\xi }_{\text {ES}}} |\mathcal {F}_t] \\ =&\xi (t) e^{-r(T-t)} \int _{d_2(\overline{\xi }_{\text {ES}})}^{+\infty } \phi (z) \frac{1}{(\lambda _{\text {ES}} \xi (t) e^{\Vert \theta \Vert \sqrt{T-t}z - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)} - \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}} )^{\frac{1}{\gamma }}}dz. \end{aligned} \end{aligned}$$

  Rearranging terms gives the expression for $X _{\text {ES}}(t)$.

• 2. Applying Ito’s lemma to $X _{\text {ES}}(t)$, we have

  $$\begin{aligned} \begin{aligned} d X _{\text {ES}}(t) =- \frac{\partial X _{\text {ES}}(t)}{\partial \xi (t)} \xi (t) \theta ^{\top } dW(t) + \ldots dt = X _{\text {ES}}(t)\pi ^{\top }_{_{\text {ES}}}(t)\sigma dW(t) + \ldots dt. \end{aligned} \end{aligned}$$

  Comparing the coefficient of the dW(t) term, we have

  $$\begin{aligned} \begin{aligned} \pi _{_{\text {ES}}}(t) =&\frac{1}{\gamma }(\sigma ^{'})^{-1} \theta \frac{1}{X _{\text {ES}}(t)} [ \frac{e^{\varGamma (t)}}{(\lambda _{\text {ES}} \xi (t))^{\frac{1}{\gamma }}} - \frac{e^{\varGamma (t)}}{(\lambda _{\text {ES}} \xi (t))^{\frac{1}{\gamma }}} N (-d_1( \underline{\xi }_{\text {ES}})) \\&+ e^{-r(T-t)}\int _{d_2(\overline{\xi }_{\text {ES}})}^{+\infty } \phi (z) \frac{\lambda _{\text {ES}} e^{\Vert \theta \Vert \sqrt{T-t}z - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)}}{(\lambda _{\text {ES}} \xi (t) e^{\Vert \theta \Vert \sqrt{T-t}z - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)} - \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}} )^{\frac{1 + \gamma }{\gamma }}}dz]. \end{aligned} \end{aligned}$$

  Dividing it by $\pi _{B}(t)$ yields

  $$\begin{aligned} \begin{aligned} q_{\text {ES}}(t) =&\frac{1}{X _{\text {ES}}(t)} [ \frac{e^{\varGamma (t)}}{(\lambda _{\text {ES}} \xi (t))^{\frac{1}{\gamma }}} - \frac{e^{\varGamma (t)}}{(\lambda _{\text {ES}} \xi (t))^{\frac{1}{\gamma }}} N (-d_1( \underline{\xi }_{\text {ES}})) \\&+ e^{-r(T-t)}\int _{d_2(\overline{\xi }_{\text {ES}})}^{+\infty } \phi (z) \frac{\lambda _{\text {ES}} e^{\Vert \theta \Vert \sqrt{T-t}z - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)}}{(\lambda _{\text {ES}} \xi (t) e^{\Vert \theta \Vert \sqrt{T-t}z - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)} - \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}} )^{\frac{1 + \gamma }{\gamma }}}dz]\\ =&\frac{1}{X _{\text {ES}}(t)} [X _{\text {ES}}(t) - \underline{X}_{\text {ES}}e^{-r(T-t)}(N(-d_2( \underline{\xi }_{\text {ES}})) - N(-d_2( \overline{\xi }_{\text {ES}})))\\&+ \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}}e^{-r(T-t)}G(\overline{\xi }_{\text {ES}}, \frac{\gamma }{1 + \gamma })]\\ =&1 -\frac{\underline{X}_{\text {ES}}e^{-r(T-t)}(N(-d_2( \underline{\xi }_{\text {ES}})) - N(-d_2( \overline{\xi }_{\text {ES}})))}{X _{\text {ES}}(t)} + \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}} \frac{e^{-r(T-t)}}{X _{\text {ES}}(t)}G(\overline{\xi }_{\text {ES}}, \frac{\gamma }{1 + \gamma }). \end{aligned} \end{aligned}$$

  (25)

• 3. The first line in (25) reveals that it is non-negative. We now verify the limits. Because $X _{\text {ES}}(t) \rightarrow \infty$ as $\xi (t) \rightarrow 0$ and

  $$- \underline{X}_{\text {ES}}(N(-d_2( \underline{\xi }_{\text {ES}})) - N(-d_2( \overline{\xi }_{\text {ES}}))) + \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}}G(\overline{\xi }_{\text {ES}}, \frac{\gamma }{1 + \gamma })$$

  is bounded, $\lim _{\xi (t) \rightarrow 0} q_{\text {ES}}(t) =1$. For $\epsilon >0$, we have

  $$\begin{aligned} \begin{aligned}&\frac{e^{-r(T-t)}(N(-d_2( \underline{\xi }_{\text {ES}})) - N(-d_2( \overline{\xi }_{\text {ES}})))}{X _{\text {ES}}(t)}\\&\quad \le \frac{N(d_2( \overline{\xi }_{\text {ES}})) - N(d_2( \underline{\xi }_{\text {ES}}))}{\int _{d_2(\overline{\xi }_{\text {ES}})}^{+\infty } \phi (z) \frac{1}{(\lambda _{\text {ES}} \xi (t) e^{\Vert \theta \Vert \sqrt{T-t}z - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)} - \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}} )^{\frac{1}{\gamma }}}dz}\\&\quad \le \frac{N(d_2( \overline{\xi }_{\text {ES}})) - N(d_2( \underline{\xi }_{\text {ES}}))}{\int _{d_2(\overline{\xi }_{\text {ES}})}^{d_2(\overline{\xi }_{\text {ES}}) + \epsilon } \phi (z) \frac{1}{(\lambda _{\text {ES}} \xi (t) e^{\Vert \theta \Vert \sqrt{T-t}(d_2(\overline{\xi }_{\text {ES}}) + \epsilon )- (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)} - \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}} )^{\frac{1}{\gamma }}}dz}\\&\quad = C_1 \frac{N(d_2( \overline{\xi }_{\text {ES}})) - N(d_2( \underline{\xi }_{\text {ES}}))}{N(d_2( \overline{\xi }_{\text {ES}}) + \epsilon ) - N(d_2( \overline{\xi }_{\text {ES}}))} \rightarrow 0 \text { as } \xi (t) \rightarrow \infty , \end{aligned} \end{aligned}$$

  where $C_1$ is a positive constant. For $\lambda >0$ such that $\lambda _{\text {ES}} \overline{\xi }_{\text {ES}} - \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}} > \lambda \overline{\xi }_{\text {ES}}$, we have

  $$ \lambda _{\text {ES}} \xi (t) e^{\Vert \theta \Vert \sqrt{T-t}d_2( \overline{\xi }_{\text {ES}}) - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)} - \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}} > \lambda \xi (t) e^{\Vert \theta \Vert \sqrt{T-t}d_2( \overline{\xi }_{\text {ES}}) - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)}.$$

  Consequently,

  $$\begin{aligned} \begin{aligned} \frac{e^{-r(T-t)}G(\overline{\xi }_{\text {ES}}, \frac{\gamma }{1 + \gamma })}{X _{\text {ES}}(t)}< \frac{G(\overline{\xi }_{\text {ES}}, \frac{\gamma }{1 + \gamma })}{G(\overline{\xi }_{\text {ES}}, \gamma )}<&\frac{\int _{d_2(\overline{\xi }_{\text {ES}})}^{+\infty } \phi (z) \frac{1}{(\lambda \xi (t) e^{\Vert \theta \Vert \sqrt{T-t}z - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)})^{\frac{1}{\gamma }+1}}dz}{\int _{d_2(\overline{\xi }_{\text {ES}})}^{+\infty } \phi (z) \frac{1}{(\lambda _{\text {ES}} \xi (t) e^{\Vert \theta \Vert \sqrt{T-t}z - (r - \frac{\Vert \theta \Vert ^2}{2})(T-t)})^{\frac{1}{\gamma }}}dz}\\ =&\frac{C_2}{\xi (t)} \rightarrow 0, \text { as } \xi (t) \rightarrow \infty , \end{aligned} \end{aligned}$$

  where $C_2$ is a positive constant. Therefore, $\lim _{\xi (t) \rightarrow \infty } q_{\text {ES}}(t) =1$.

• 4. Define

  $$\begin{aligned} F(\xi (t)) = - \underline{X}_{\text {ES}}(N(-d_2( \underline{\xi }_{\text {ES}})) - N(-d_2( \overline{\xi }_{\text {ES}}))) + \frac{1 }{\alpha }\lambda _{\text {ES}} \mu _{\text {ES}}G(\overline{\xi }_{\text {ES}}, \frac{\gamma }{1 + \gamma }), \end{aligned}$$

  then $q_{\text {ES}}(t) = 1 + e^{-r(T-t)} F(\xi (t)) / X _{\text {ES}}(t)$. Following similar steps as in (3), we can show $\lim _{\xi (t) \rightarrow 0} F(\xi (t) ) =0$ and $\lim _{\xi (t) \rightarrow +\infty } F(\xi (t) ) =0$. Moreover, we have

  $$\begin{aligned} F^{'}(\xi (t)) = \frac{\phi (d_2( \overline{\xi }_{\text {ES}}))}{(\lambda _{\text {ES}} \underline{\xi }_{\text {ES}})^{\frac{1}{\gamma }} \Vert \theta \Vert \sqrt{T-t} \xi (t) }g(d_2( \overline{\xi }_{\text {ES}})), \end{aligned}$$

  where

  $$\begin{aligned} \begin{aligned} g(x) =&a - e^{-\frac{(\ln a )^2}{2 \Vert \theta \Vert ^2 (T-t)}} a^{\frac{x}{\Vert \theta \Vert \sqrt{T-t}}} - \frac{(1 + \frac{1}{\gamma })\Vert \theta \Vert \sqrt{T-t}a(a-1) }{\phi (x)} \\&\cdot \int _{x}^{+\infty } \phi (z) \frac{e^{\Vert \theta \Vert \sqrt{T-t}(z-x)}}{(a e^{\Vert \theta \Vert \sqrt{T-t}(z-x)} + 1 -a )^{2+\frac{1}{\gamma }}}dz ,\\ a =&\frac{\overline{\xi }_{\text {ES}}}{\underline{\xi }_{\text {ES}}}. \end{aligned} \end{aligned}$$

  It is a simple exercise to show $\lim _{x \rightarrow +\infty } g(x) = - \infty$. Because

  $$\begin{aligned} \begin{aligned} \int _{x}^{+\infty } \frac{\phi (z)}{\phi (x)} \cdot \frac{e^{\Vert \theta \Vert \sqrt{T-t}(z-x)}}{(a e^{\Vert \theta \Vert \sqrt{T-t}(z-x)} + 1 -a )^{2+\frac{1}{\gamma }}}dz>&\int _{x}^{+\infty } \frac{\phi (z)}{\phi (x)} \cdot \frac{e^{\Vert \theta \Vert \sqrt{T-t}(z-x)}}{(a e^{\Vert \theta \Vert \sqrt{T-t}(z-x)})^{2+\frac{1}{\gamma }}}dz\\ =&\frac{ \int _{x}^{+\infty } \frac{\phi (z)}{a(a e^{\Vert \theta \Vert \sqrt{T-t}z})^{1+\frac{1}{\gamma }}}dz}{\phi (x) e^{- (1+\frac{1}{\gamma }) \Vert \theta \Vert \sqrt{T-t}x}} \rightarrow +\infty \end{aligned} \end{aligned}$$

  as $x \rightarrow +\infty$, we have $\lim _{x \rightarrow -\infty } g(x) = - \infty$. It is then easy to see that there exist deterministic processes $\underline{\xi } _{\text {ES}} (t)$ and $\overline{\xi } _{\text {ES}} (t)$ such that $F^{'}(\xi (t))<0$ if $\xi (t) < \underline{\xi } _{\text {ES}} (t)$ or $\xi (t) > \overline{\xi } _{\text {ES}} (t)$, thereby completing the proof.

$\square$

1.3 Proof of Proposition 6

We will make use of Proposition 11 to solve the WS agent’s optimization problem. Note that $\text {WS}_{\alpha }$ is a special case of WVaR with $\varPhi ([0,z]) = \frac{\int _0^{z} F_{\xi } ^{-1}(1-s)ds}{\int _0^{\alpha } F_{\xi } ^{-1}(1-s)ds} \wedge 1, z \in [0,1]$. Define

$$\begin{aligned} \varphi _{\text {WS}, y }(z)=\left\{ {\begin{array}{ll} - \int _z^1 F_{\xi } ^{-1}(1-s) ds +y \int _z^{\alpha } F_{\xi } ^{-1}(1-s)ds &{} ~ ~ ~ ~ z \in [0,\alpha ],\\ - \int _z^1 F_{\xi } ^{-1}(1-s) ds &{} ~ ~ ~ ~ z \in (\alpha ,1], \end{array}} \right. \end{aligned}$$

where we have substituted y for $\frac{y}{\int _0^{\alpha } F_{\xi } ^{-1}(1-s)ds}$ to simplify the notation.

Let $\delta _{\text {WS}, y }(\cdot )$ be the concave envelope of $\varphi _{\text {WS}, y }(\cdot )$ and $\mathcal {A}_{\text {WS}} = \{ y :y>0 ,\delta _{\text {WS}, y }^{'} (z+) > 0, ~z \in [0,1) \}$ where $\delta _{\text {WS}, y }^{'}(\cdot + )$ is the right derivative of $\delta _{\text {WS}, y }$. According to Proposition 11, the optimal solution to the WS agent’s problem is given by

$$\begin{aligned} X_{\text {WS}}(T) = I(\lambda \delta _{\text {WS}, \mu }^{'}( (1 - F_{\xi }(\xi ))+ )), \end{aligned}$$

(26)

where $\lambda >0$ and $\mu \in \mathcal {A}_{\text {WS}}$ are determined by

$$\begin{aligned} \left\{ {\begin{array}{ll} {\mathbb {E}}[\xi X_{\text {WS}}(T)]&{}=X(0) \\ \text {WS}_{\alpha }(X_{\text {WS}}(T)) &{}= -\underline{X}. \end{array}} \right. \end{aligned}$$

First, we derive $\mathcal {A}_{\text {WS}} = \{ y :y>0 ,\delta _{y }^{'} (z+) > 0, ~z \in [0,1) \}$.

Lemma 3

$$\begin{aligned} \mathcal {A}_{\text {WS}} = \{ y : 0< y < \frac{\int _{0}^1 F_{\xi } ^{-1}(1-s)ds}{\int _0^{\alpha } F_{\xi } ^{-1}(1-s)ds}\}. \end{aligned}$$

Proof

According to Lemma B.3 in [33], $\mathcal {A}_{\text {WS}} =\{ y :y >0 , \varphi _{\text {WS}, y } (z-) < 0, ~ \forall z \in [0,1) \}$. Note that $\varphi _{\text {WS}, y }(z)$ is continuous, it suffices to show $\{ y :y >0 , \varphi _{\text {WS}, y } (z)< 0, ~ \forall z \in [0,1) \} = \{ y : 0< y < \frac{\int _{0}^1 F_{\xi } ^{-1}(1-s)ds}{\int _0^{\alpha } F_{\xi } ^{-1}(1-s)ds}\}$.

First, if $y < 1$, then $\varphi _{\text {WS}, y }^{'}(z+)>0, ~z \in [0,1)$ and consequently $\varphi _{\text {WS}, y }(z)<0, ~z \in [0,1)$.

Next, if $y \ge 1$, we have $\varphi _{\text {WS}, y }^{'}(z) \le 0$ for $z \in (0,\alpha )$ and $\varphi _{\text {WS}, y }^{'}(z) > 0$ for $z \in (\alpha , 1 )$. Therefore, $\varphi _{\text {WS}, y }(z)<0, ~z \in [0,1)$ if and only if $\varphi _{\text {WS}, y } (0) <0$ which is equivalent to $y < \frac{\int _{0}^1 F_{\xi } ^{-1}(1-s)ds}{\int _0^{\alpha } F_{\xi } ^{-1}(1-s)ds}$. $\square$

We now derive an analytical expression for $\delta _{\text {WS}, y }(\cdot )$, the concave envelope of $\varphi _{\text {WS}, y }(\cdot )$. Suppose $y \in \mathcal {A}_{\text {WS}}$. Let $\overline{t}_{\text {WS} }:= 1 - F_{\xi }( 0 \vee (1-y)F_{\xi }^{-1}(1-\alpha ))$ and we have $\alpha < \overline{t}_{\text {WS} } \le 1$. For $\alpha< t < \overline{t}_{\text {WS} }$, define

$$\begin{aligned} s_{\text {WS},y }(t) =\left\{ {\begin{array}{ll} 1-F_{\xi }(\frac{F_{\xi }^{-1}(1-t)}{1-y}) &{} ~ ~ ~ ~ 0< y< 1,\\ 0 &{} ~ ~ ~ ~ 1 \le y < \frac{\int _{0}^1 F_{\xi } ^{-1}(1-s)ds}{\int _0^{\alpha } F_{\xi } ^{-1}(1-s)ds}.\\ \end{array}} \right. \end{aligned}$$

We have $0 \le s_{\text {WS},y }(t) <\alpha$. Moreover, if $0<y<1$, then $\varphi _{\text {WS}, y }^{'}(s_{\text {WS},y }(t) )=\varphi _{\text {WS}, y }^{'}(t)$. Define

$$\begin{aligned} h_{\text {WS},y }(t)= \varphi _{\text {WS}, y }(t)-\varphi _{\text {WS}, y }(s_{\text {WS},y }(t) )- \varphi _{\text {WS}, y }^{'}(t)(t-s_{\text {WS},y }(t) ), ~ \alpha< t < \overline{t}_{\text {WS} }. \end{aligned}$$

(27)

We have the following lemma which is analogous to Lemma 2.

Lemma 4

For $y \in \mathcal {A}_{\text {WS}}$, there exist $\alpha< t^{*}_{\text {WS}} < \overline{t}_{\text {WS} }$ such that $h_{\text {WS},y }(t)<0$ for $\alpha<t < t^{*}_{\text {WS}}$, $h_{\text {WS},y }(t^{*}_{\text {WS}})=0$, and $h_{\text {WS},y }(t)>0, ~t^{*}_{\text {WS}}< t < \overline{t}_{\text {WS} }$.

Proof

We first consider the case $1 \le y < \frac{\int _{0}^1 F_{\xi } ^{-1}(1-s)ds}{\int _0^{\alpha } F_{\xi } ^{-1}(1-s)ds}$. We have

$$\begin{aligned} \begin{aligned} h_{\text {WS},y }(t)&= \varphi _{\text {WS}, y }(t)-\varphi _{\text {WS}, y }(0)- \varphi _{\text {WS}, y }^{'}(t)t\\&= \int _0^t F_{\xi }^{-1}(1-z)dz - y \int _0^{\alpha } F_{\xi }^{-1}(1-z)dz - F_{\xi }^{-1}(1-t)t. \end{aligned} \end{aligned}$$

It is easy to see $h_{\text {WS},y }(\alpha +) < 0$ and $h_{\text {WS},y }(\overline{t}_{\text {WS} }-) = h_{\text {WS},y }(1-) >0$. Because $h_{\text {WS},y }(\cdot )$ is continuous, there exists at least one t such that $h_{\text {WS},y }(t)=0$. Moreover, for $\alpha<t_1< t_2 < \overline{t}_{\text {ES} }$,

$$\begin{aligned} \begin{aligned} h_{\text {WS},y }(t_1) - h_{\text {WS},y }(t_2) =&-\int _{t_1}^{t_2} F_{\xi }^{-1}(1-z)dz + F_{\xi }^{-1}(1-t_2)t_2 - F_{\xi }^{-1}(1-t_1)t_1\\ =&-\int _{t_1}^{t_2} F_{\xi }^{-1}(1-z)dz + \int _{t_1}^{t_2} F_{\xi }^{-1}(1-z)dz + \int _{t_1}^{t_2} zdF_{\xi }^{-1}(1-z)\\ =&\int _{t_1}^{t_2} zdF_{\xi }^{-1}(1-z)<0. \end{aligned} \end{aligned}$$

Thus, $h_{\text {WS},y }(\cdot )$ is strictly increasing, thereby proving the claim.

Next, suppose $0<y<1$. Note that

$$\begin{aligned} \begin{aligned} h_{\text {WS},y }(t)&= \varphi _{\text {WS}, y }(t)-\varphi _{\text {WS}, y }(s_{\text {WS},y }(t) )- \varphi _{\text {WS}, y }^{'}(t)(t-s_{\text {WS},y }(t) )\\&= \int _{s_{\text {WS},y }(t) }^t F_{\xi }^{-1}(1-z)dz - y \int _{s_{\text {WS},y }(t) }^{\alpha } F_{\xi }^{-1}(1-z)dz - F_{\xi }^{-1}(1-t)(t-s_{\text {WS},y }(t) ). \end{aligned} \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned} h_{\text {WS},y }(t)&> F_{\xi }^{-1}(1-t)(t-s_{\text {WS},y }(t) ) - y \int _{s_{\text {WS},y }(t) }^{\alpha } F_{\xi }^{-1}(1-z)dz - F_{\xi }^{-1}(1-t)(t-s_{\text {WS},y }(t) )\\&= - y \int _{s_{\text {WS},y }(t) }^{\alpha } F_{\xi }^{-1}(1-z)dz, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} h_{\text {WS},y }(t)<&F_{\xi }^{-1}(1-s_{\text {WS},y }(t) )(t-s_{\text {WS},y }(t) ) - y \int _{s_{\text {WS},y }(t) }^{\alpha } F_{\xi }^{-1}(1-z)dz \\&- (1-y)F_{\xi }^{-1}(1-s_{\text {WS},y }(t) ) (t-s_{\text {WS},y }(t) )\\ <&y [ F_{\xi }^{-1}(1-s_{\text {WS},y }(t) )(t-s_{\text {WS},y }(t) ) - F_{\xi }^{-1}(1-\alpha )(\alpha -s_{\text {WS},y }(t) ) ]. \end{aligned} \end{aligned}$$

Therefore, $h_{\text {WS},y }(\overline{t}_{\text {WS} }-) \ge 0$ and $h_{\text {WS},y }(\alpha +) \le 0$. Because $h_{\text {WS},y }(\cdot )$ is continuous, there exists at least one t such that $h_{\text {WS},y }(t)=0$. Similar to the proof of Lemma 2, we can show $h_{\text {WS},y }(\cdot )$ is strictly increasing. This completes the proof. $\square$

Proposition 13

For $y \in \mathcal {A}_{\text {WS}}$,

$$\begin{aligned} \delta _{\text {WS}, y }(z)=\left\{ {\begin{array}{ll} \varphi _{\text {WS}, y }(z) &{} ~ ~ z \in [0,s_{\text {WS},y }(t^{*}_{\text {WS},y })),\\ \varphi _{\text {WS}, y }(s_{\text {WS},y }(t^{*}_{\text {WS},y }))&{}\\ \quad +\varphi _{\text {WS}, y }^{'}(s_{\text {WS},y }(t^{*}_{\text {WS},y }))(z -s_{\text {WS},y }(t^{*}_{\text {WS},y })) &{} ~ ~ z \in [s_{\text {WS},y }(t^{*}_{\text {WS},y }),t^{*}_{\text {WS},y }],\\ \varphi _{\text {WS}, y }(z) &{} ~ ~ z \in (t^{*}_{\text {WS},y } ,1], \end{array}} \right. \end{aligned}$$

and

$$\begin{aligned} \delta _{\text {WS}, y }^{'}(z)=\left\{ {\begin{array}{ll} (1-y)F_{\xi } ^{-1}(1-z) &{} ~ ~ ~ ~ z \in [0,s_{\text {WS},y }(t^{*}_{\text {WS},y })),\\ F_{\xi } ^{-1}(1-t^{*}_{\text {WS},y }) &{} ~ ~ ~ ~ z \in [s_{\text {WS},y }(t^{*}_{\text {WS},y }),t^{*}_{\text {WS},y }],\\ F_{\xi } ^{-1}(1-z) &{} ~ ~ ~ ~ z \in (t^{*}_{\text {WS},y } ,1], \end{array}} \right. \end{aligned}$$

where $\alpha< t^{*}_{\text {WS},y } <1$ is the unique root of $h_{\text {WS},y }(t)=0$, and $s_{\text {WS},y }(t^{*}_{\text {WS},y })=0$ if $y \ge 1$.

Proof

The proof is analogous to that of Proposition 12 and thus we omit it. $\square$

Plugging the expression of $\delta _{\text {WS}, \mu }$ into (26), we have, for $0<\mu <1$,

$$\begin{aligned} X_{\text {WS}}(T)=\left\{ \begin{array}{ll} I((1- \mu )\lambda \xi ) &{} ~ ~ ~ ~ \frac{F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu })}{1 - \mu } \le \xi ,\\ I(\lambda F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu })) &{} ~ ~ ~ ~ F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu }) \le \xi< \frac{F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu })}{1 - \mu },\\ I(\lambda \xi ) &{} ~ ~ ~ ~ \xi < F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu }), \end{array} \right. \end{aligned}$$

(28)

and, for $\mu \ge 1$,

$$\begin{aligned} X_{\text {WS}}(T)=\left\{ \begin{array}{ll} I(\lambda F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu })) &{} ~ ~ ~ ~ F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu }) \le \xi ,\\ I(\lambda \xi ) &{} ~ ~ ~ ~ \xi < F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu }). \end{array} \right. \end{aligned}$$

(29)

Note that $t^{*}_{\text {WS},\mu } > \alpha$. If (29) is optimal, then the WS constraint $\text {WS}_{\alpha }(X_{\text {WS}}(T)) = -\underline{X}$ implies that $I(\lambda F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu })) = \underline{X}$. In this case, $X_{\text {WS}}(T) = X_{\text {PI}}(T)$. Moreover, $\alpha \le \overline{\alpha }$.

If $\alpha > \overline{\alpha }$, then the optimal terminal wealth is given in the form of (28). Substitute $\lambda _{\text {WS}}$ for $\lambda$, $\mu _{\text {WS}}$ for $\mu$, $\underline{\xi }_{\text {WS}}$ for $F_{\xi }^{-1}(1-t^{*}_{\mu })$, and $\overline{\xi }_{\text {WS}}$ for ${F_{\xi }^{-1}(1-t^{*}_{\text {WS},\mu })}/{(1 - \mu )}$, we arrive at (13). Moreover, $h_{\text {WS},\mu _{\text {WS}} }(t^{*}_{\text {WS},\mu _{\text {WS}}})=0$ is equivalent to $h_{\text {WS},\mu _{\text {WS}}}(1- F_{\xi }(\underline{\xi }_{\text {WS}}))=0$.

We now show the last four assertions. First, for $\alpha > \overline{\alpha }$, the fact that $s_{\text {WS},\mu _{\text {WS}}}(t^{*}_{\text {WS},\mu _{\text {WS}}})< \alpha < t^{*}_{\text {WS},\mu _{\text {WS}}}$ implies that ${\mathbb {P}}(\xi> \overline{\xi }_{\text {WS}})< \alpha < {\mathbb {P}}(\xi > \underline{\xi }_{\text {WS}})$.

Second, for $\alpha > \overline{\alpha }$, if $I(\lambda _{\text {WS}} \underline{\xi }_{\text {WS}}) \le \underline{X}$, then $X_{\text {WS}}(T) < I(\lambda _{\text {WS}} \underline{\xi }_{\text {WS}})\le \underline{X}$ for $\xi > \overline{\xi }_{\text {WS}}$ and $\text {WS}_{\alpha }(X_{\text {WS}}(T)) > -\underline{X}$, which is a contradiction. Therefore, $I(\lambda _{\text {WS}} \underline{\xi }_{\text {WS}}) > \underline{X}$.

Third, if $\lambda _{\text {WS}} \ge \lambda _{\text {B}}$, then $X_{\text {WS}}(T) \ge X_{\text {B}}(T), ~ a.s.$, and the inequality is strict when $\xi > \underline{\xi }_{\text {WS}}$, violating the budget constraint. Consequently, $\lambda _{\text {WS}} > \lambda _{\text {B}}$.

The last claim follows from direct comparison.

1.4 Proof of Proposition 7

Proof

The proof is analogous to that of Proposition 4. We first consider the case $\alpha > \overline{\alpha }$.

• 1. Because $\xi (t) X _{\text {WS}}(t)$ is a martingale, we have

  $$\begin{aligned} \begin{aligned} \xi (t) X _{\text {WS}}(t) = {\mathbb {E}}[\xi X _{\text {WS}}(T) |\mathcal {F}_t] =&{\mathbb {E}}[\xi I(\lambda _{\text {WS}} \xi ) \mathbf {1} _ {\xi< \underline{\xi }_{\text {WS}}} |\mathcal {F}_t] + {\mathbb {E}}[\xi I(\lambda _{\text {WS}} \underline{\xi }_{\text {WS}}) \mathbf {1} _ {\underline{\xi }_{\text {WS}} \le \xi < \overline{\xi }_{\text {WS}}} |\mathcal {F}_t] \\&+ {\mathbb {E}}[\xi I(\lambda _{\text {WS}} (1 - \mu _{\text {WS}}) \xi ) \mathbf {1} _ {\xi \ge \overline{\xi }_{\text {WS}}} |\mathcal {F}_t]. \end{aligned} \end{aligned}$$

  When r and $\theta$ are constant, conditional on $\mathcal {F}_t$, $\ln \xi$ is normally distributed with mean $\ln \xi (t) - (r + \frac{\Vert \theta \Vert ^2 }{2} )(T-t)$ and variance $\Vert \theta \Vert ^2 (T-t)$. We have

  $$\begin{aligned} \begin{aligned} {\mathbb {E}}[\xi I(\lambda _{\text {WS}} \xi ) \mathbf {1} _ {\xi< \underline{\xi }_{\text {WS}}} |\mathcal {F}_t] =&\xi (t) \frac{e^{\varGamma (t)}}{(\lambda _{\text {WS}} \xi (t))^{\frac{1}{\gamma }}} N (d_1( \underline{\xi }_{\text {WS}})),\\ {\mathbb {E}}[\xi I(\lambda _{\text {WS}} \underline{\xi }_{\text {WS}}) \mathbf {1} _ {\underline{\xi }_{\text {WS}} \le \xi < \overline{\xi }_{\text {WS}}} |\mathcal {F}_t] =&\xi (t) \underline{X}_{\text {WS}} e^{-r(T-t)}[N(-d_2 (\underline{\xi }_{\text {WS}})) - N(-d_2 (\overline{\xi }_{\text {WS}}))], \end{aligned} \end{aligned}$$

  and

  $$\begin{aligned} \begin{aligned} {\mathbb {E}}[\xi I(\lambda _{\text {WS}} (1 - \mu _{\text {WS}}) \xi ) \mathbf {1} _ {\xi \ge \overline{\xi }_{\text {WS}}} |\mathcal {F}_t] = \xi (t) \frac{e^{\varGamma (t)}}{(\lambda _{\text {WS}} (1 - \mu _{\text {WS}}) \xi (t))^{\frac{1}{\gamma }}} N (-d_1( \overline{\xi }_{\text {WS}})). \end{aligned} \end{aligned}$$

  Rearranging terms gives the expression for $X _{\text {WS}}(t)$.

• 2. Applying Ito’s lemma to $X _{\text {WS}}(t)$, we have

  $$\begin{aligned} \begin{aligned} d X _{\text {WS}}(t) = - \frac{\partial X _{\text {WS}}(t)}{\partial \xi (t)} \xi (t) \theta ^{\top } dW(t) + \ldots dt = X _{\text {WS}}(t)\pi ^{\top }_{_{\text {WS}}}(t)\sigma dW(t) + \ldots dt. \end{aligned} \end{aligned}$$

  Comparing the coefficient of the dW(t) term, we have

  $$\begin{aligned} \begin{aligned} \pi _{_{\text {WS}}}(t) =&- (\sigma ^{'})^{-1} \theta \frac{\partial X _{\text {WS}}(t)}{\partial \xi (t)} \frac{\xi (t)}{X _{\text {WS}}(t)}\\ =&\frac{1}{\gamma }(\sigma ^{'})^{-1} \theta \frac{1}{X _{\text {WS}}(t)} [ \frac{e^{\varGamma (t)}}{(\lambda _{\text {WS}} \xi (t))^{\frac{1}{\gamma }}} - \frac{e^{\varGamma (t)}}{(\lambda _{\text {WS}} \xi (t))^{\frac{1}{\gamma }}} N (-d_1( \underline{\xi }_{\text {WS}})) \\&+ \frac{e^{\varGamma (t)}}{(\lambda _{\text {WS}} (1 - \mu _{\text {WS}}) \xi (t))^{\frac{1}{\gamma }}} N (-d_1( \overline{\xi }_{\text {WS}}))]. \end{aligned} \end{aligned}$$

  Dividing it by $\pi _{B}(t)$ yields

  $$\begin{aligned} \begin{aligned} q_{\text {WS}}(t) =&\frac{1}{X _{\text {WS}}(t)} [ \frac{e^{\varGamma (t)}}{(\lambda _{\text {WS}} \xi (t))^{\frac{1}{\gamma }}} - \frac{e^{\varGamma (t)}}{(\lambda _{\text {WS}} \xi (t))^{\frac{1}{\gamma }}} N (-d_1( \underline{\xi }_{\text {WS}})) \\&+ \frac{e^{\varGamma (t)}}{(\lambda _{\text {WS}} (1 - \mu _{\text {WS}}) \xi (t))^{\frac{1}{\gamma }}} N (-d_1( \overline{\xi }_{\text {WS}}))] \\ =&\frac{1}{X _{\text {WS}}(t)} [X _{\text {WS}}(t) - \underline{X}_{\text {WS}}e^{-r(T-t)}(N(-d_2( \underline{\xi }_{\text {WS}})) - N(-d_2( \overline{\xi }_{\text {WS}})))]\\ =&1 -\frac{\underline{X}_{\text {WS}}e^{-r(T-t)}(N(-d_2( \underline{\xi }_{\text {WS}})) - N(-d_2( \overline{\xi }_{\text {WS}})))}{X _{\text {WS}}(t)}. \end{aligned} \end{aligned}$$

• 3. The claim is straightforward to verify.

The case $\alpha \le \overline{\alpha }$ can be seen as the limiting case of $\alpha > \overline{\alpha }$ by letting $\overline{\xi }_{\text {WS}} \rightarrow +\infty$ and $\underline{X}_{\text {WS}} \rightarrow \underline{X}$. The limit is also easy to verify. $\square$

1.5 Proof of Proposition 9

Proof

The benchmark agent’s optimal consumption and time-T wealth are standard in the literature. For the ES agent, we first consider the following problem:

$$\begin{aligned} \begin{aligned} \max _{c_{\text {ES}}(t),t \in [T, +\infty ) } ~&{\mathbb {E}}[\int _T^{+\infty } e^{-\rho (t - T)} \ln (c_{\text {ES}}(t))dt |\mathcal {F}_{T-}]\\ \text {subject to} ~&{\mathbb {E}}[ \int _T^{+\infty } \xi (t) c_{\text {ES}}(t)dt | \mathcal {F}_{T-}] \le \xi (T-) X_{\text {ES}}(T-), \text { almost surely.} \end{aligned} \end{aligned}$$

The optimal consumption is

$$ c_{\text {ES}}(t) = \frac{e^{-\rho (t - T)}}{\lambda _{\text {ES},2} \xi (t)}, t \in [T, +\infty ), $$

where $\lambda _{\text {ES},2} = { 1 }/{ (\rho \xi (T-)X_{\text {ES}}(T-) )},$and the optimal value is

$$\begin{aligned} \frac{1}{\rho } \ln (X_{\text {ES}}(T-)) + {\mathbb {E}}[\int _T^{+\infty } e^{- \rho (t-T)}\ln \big ( \rho e^{- \rho (t-T)} \frac{\xi (T-)}{\xi (t)} \big )dt |\mathcal {F}_T]. \end{aligned}$$

Consequently, we can consider the following problem:

$$\begin{aligned} \begin{aligned} \max _{c_{\text {ES}}(t),t \in [0,T],X_{\text {ES}}(T-)} ~&{\mathbb {E}}[\int _0^{T} e^{-\rho t} \ln (c_{\text {ES}}(t))dt + \frac{e^{- \rho T}}{\rho } \ln (X_{\text {ES}}(T-))]\\ \text {subject to} ~&{\mathbb {E}}[ \int _0^T \xi (t) c_{\text {ES}}(t)dt + \xi (T-) X_{\text {ES}}(T-)] \le X_{\text {ES}}(0),\\&\text {ES}_{\alpha }(X_{\text {ES}}(T-))\le -\underline{X}. \end{aligned} \end{aligned}$$

The rest of the proof is a straightforward extension of Proposition 3. The case of WS agent can be proved in a similar fashion. $\square$

1.6 Proof of Proposition 9

Proof

Clearing the consumption good market gives (18), (19), and (20). The proof that clearing the good market implies all other markets are cleared appears in [5]. Morevoer, it is straightforward to verify (18), (19), and (20) satisfy Assumption 5 and 6. r and $\theta$ are determined by applying Ito’s lemma to the state price density. $\square$

1.7 Proof of Proposition 10

Proof

• 1. We omit the proof as the benchmark economy is a special case of the ES economy when the ES constraint is not binding, i.e., $\mu _{\text {ES}}=0$.

• 2. We have

  $$\begin{aligned} \begin{aligned} X_{\text {M}}^{\text {ES}} (t) = X_{\text {B}1}(t) + X_{\text {ES}}(t) =&\frac{1}{\xi (t) } {\mathbb {E}}[ \int _t^T \xi (s) (c_{\text {B}1}(s) + c_{\text {ES}}(s))ds|\mathcal {F}_t] \\&+\frac{1}{\xi (t) } {\mathbb {E}}[ \xi (T-) (X_{\text {B}1}(T-) + X_{\text {ES}}(T-))|\mathcal {F}_t], t \in [0,T). \end{aligned} \end{aligned}$$

  By Propositions 4, 8 and 9, we have

  $$\begin{aligned} \begin{aligned} X_{\text {M}}^{\text {ES}} (t) =&\frac{\delta (t)}{\rho } - \frac{\lambda _{\text {B}1}}{\lambda _{\text {B}1} + \lambda _{\text {ES},1}} \frac{e^{- \rho (T-t)}}{\rho } \delta (t) N (-\hat{d}_1( \overline{\delta }_{\text {ES}} ))\\&+ \frac{\lambda _{\text {B}1}}{\lambda _{\text {B}1} + \lambda _{\text {ES},1}} \frac{e^{- \rho T}}{\rho } \overline{\delta }_{\text {ES}} e^{-(\mu _{\delta } + \rho - |\sigma _{\delta }|^2)(T-t)}[N(-\hat{d}_2 (\overline{\delta }_{\text {ES}})) - N(-\hat{d}_2 (\underline{\delta }_{\text {ES}}))] \\&+ \frac{e^{- \rho T}}{\rho } e^{-(\mu _{\delta } + \rho - |\sigma _{\delta }|^2)(T-t)}\hat{G}(\underline{\delta }_{\text {ES}},1), \end{aligned} \end{aligned}$$

  (30)

  where

  $$\begin{aligned} \begin{aligned} \overline{\delta }_{\text {ES}}&= \frac{\frac{1}{\lambda _{\text {B}1}} + \frac{1}{\lambda _{\text {ES},1}}}{\underline{\xi }_{\text {ES}}}, \\ \underline{\delta }_{\text {ES}}&= \frac{\frac{1}{\lambda _{\text {B}1}} + \frac{1}{\lambda _{\text {ES},1}}}{\overline{\xi }_{\text {ES}}}, \\ \hat{d}_1 (x)&= \frac{\ln \frac{\delta (t) e^{ \rho t} }{x} + (\mu _{\delta } + \rho - \frac{1}{2}|\sigma _{\delta } |^2 )(T-t) }{|\sigma _{\delta } | \sqrt{T-t}}, \\ \hat{d}_2 (x)&= \hat{d}_1 (x) - |\sigma _{\delta } | \sqrt{T-t} , \\ \hat{G}(x,y)&= \int _{\hat{d}_2(x)}^{+\infty }\phi (z) \frac{1}{(\frac{ (\lambda _{\text {B}1} + \lambda _{\text {ES},1}) e^{- \rho t}}{\lambda _{\text {B}1} \delta (t)} e^{|\sigma _{\delta } |\sqrt{T-t}z - (\mu _{\delta } + \rho - \frac{3}{2}|\sigma _{\delta } |^2)(T-t)} - \frac{1 }{\alpha }\lambda _{\text {ES},1} \mu _{\text {ES}} )^{y}}dz. \end{aligned} \end{aligned}$$

  From (14) and Propositions 4, we have

  $$\begin{aligned} \begin{aligned} dX_{\text {M}}^{\text {ES}} (t) =&dX_{\text {B}}(t) + dX_{\text {ES}}(t) \\ =&rX_{\text {M}}^{\text {ES}} (t) dt - \delta (t)dt +X_{\text {B}1}(t) \theta ^2 dt + X_{\text {B}1}(t) \theta dW(t) \\&+X_{\text {ES}}(t) \theta ^2 q_{\text {ES}}(t)dt + X_{\text {ES}}(t) q_{\text {ES}}(t) \theta dW(t), t \in [0,T). \end{aligned} \end{aligned}$$

  Plugging in relevant quantities from Proposition 4, 8, and 9, we have the expression for $|\sigma _{\text {M}}^{\text {ES}} (t)|$ and $\mu _{\text {M}}^{\text {ES}}(t)$. Moreover,

  $$\begin{aligned} \begin{aligned} p_{\text {ES}}(t)&= 1 - \frac{\lambda _{\text {B}1} }{(\lambda _{\text {B}1} + \lambda _{\text {ES},1}) X_{\text {M}}^{\text {ES}} (t)} \frac{e^{- \rho T}}{\rho } \overline{\delta } e^{-(\mu _{\delta } + \rho - |\sigma _{\delta }|^2)(T-t)}[N(-\hat{d}_2 (\overline{\delta }_{\text {ES}})) - N(-\hat{d}_2 (\underline{\delta }_{\text {ES}}))] \\&\quad + \frac{ \lambda _{\text {ES},1} \mu _{\text {ES} } }{\alpha } \frac{e^{- \rho T}}{\rho } \frac{ e^{-(\mu _{\delta } + \rho - |\sigma _{\delta }|^2)(T-t)}}{X_{\text {M}}^{\text {ES}} (t)}\hat{G}(\underline{\delta }_{\text {ES}},2). \end{aligned} \end{aligned}$$

  (31)

  Similarly, we can show the case of $t \ge T$.

• 3. The WS economy can be proved similarly. In particular,

  $$\begin{aligned} \begin{aligned} X_{\text {M}}^{\text {WS}} (t) =&\frac{\delta (t)}{\rho } - \frac{\lambda _{\text {B}1}}{\lambda _{\text {B}1} + \lambda _{\text {WS},1}} \frac{e^{- \rho (T-t)}}{\rho } \delta (t) N (-\hat{d}_1( \overline{\delta }_{\text {WS}}))\\&+ \frac{\lambda _{\text {B}1} }{\lambda _{\text {B}1} + \lambda _{\text {WS},1}} \frac{e^{- \rho T}}{\rho } \overline{\delta }_{\text {WS}} e^{-(\mu _{\delta } + \rho - |\sigma _{\delta }|^2)(T-t)}[N(-\hat{d}_2 (\overline{\delta }_{\text {WS}})) - N(-\hat{d}_2 (\underline{\delta }_{\text {WS}}))] \\&+ \frac{\lambda _{\text {B}1} }{(\lambda _{\text {B}1} + \lambda _{\text {WS},1})(1 - \mu _{\text {WS}})} \frac{e^{- \rho (T-t)}}{\rho } \delta (t) e^{-(\mu _{\delta } + \rho - |\sigma _{\delta }|^2)(T-t)}N (-\hat{d}_1( \underline{\delta }_{\text {WS}})), \end{aligned} \end{aligned}$$

  (32)

  where

  $$\begin{aligned} \overline{\delta }_{\text {WS}} = \frac{\frac{1}{\lambda _{\text {B}1}} + \frac{1}{\lambda _{\text {WS},1}}}{\underline{\xi }_{\text {WS}}} , \quad \underline{\delta }_{\text {WS}} = \frac{\frac{1}{\lambda _{\text {B}1}} + \frac{1}{\lambda _{\text {WS},1}}}{\overline{\xi }_{\text {WS}}} , \end{aligned}$$

  and

  $$\begin{aligned} |\sigma _{\text {M}}^{\text {WS}} (t)| =p_{\text {WS}}(t)|\sigma _{\delta }|, \quad \mu _{\text {M}}^{\text {WS}}(t) - r =p_{\text {WS}}(t) |\sigma _{\delta }|^2, \end{aligned}$$

  where

  $$\begin{aligned} p_{\text {WS}}(t) = 1 - \frac{\lambda _{\text {B}1} }{(\lambda _{\text {B}1} + \lambda _{\text {WS},1}) X_{\text {M}}^{\text {WS}} (t)} \frac{e^{- \rho T}}{\rho } \overline{\delta } e^{-(\mu _{\delta } + \rho - |\sigma _{\delta }|^2)(T-t)}[N(-\hat{d}_2 (\overline{\delta }_{\text {WS}})) - N(-\hat{d}_2 (\underline{\delta }_{\text {WS}}))]. \end{aligned}$$

  (33)

  Here, $\underline{\delta }_{\text {WS}} = 0$ and $\frac{\lambda _{\text {B}1} }{\lambda _{\text {B}1} + \lambda _{\text {WS},1}} \frac{e^{- \rho T}}{\rho } \overline{\delta }_{\text {WS}} = \underline{X}$ if $\alpha \le \overline{\alpha }$.

• 4. Note that if $X_{\text {ES}}(T-) > \frac{e^{-\rho T}}{ \rho \lambda _{\text {ES},1} \underline{\xi }_{ES}}$ then $X_{\text {B}1}(T-) + X_{\text {ES}}(T-)= \frac{\delta (T-)}{\rho }$, and if $X_{\text {ES}}(T-) \le \frac{e^{-\rho T}}{ \rho \lambda _{\text {ES},1} \underline{\xi }_{ES}}$ then $X_{\text {B}1}(T-) + X_{\text {ES}}(T-) > \frac{\delta (T-)}{\rho }$. Hence, $X_{\text {M}}^{\text {ES}} (T-) \ge X_{\text {M}}^{\text {B}} (T-)$ and the inequality is strict with a non-zero probability, thereby proving the result. The WS case can be proved similarly.

• 5. The proof is as of Proposition 4.

• 6. The proof is as of Proposition 7.

$\square$