Appendix: Total payoff
Note that the formulas in “Appendix Total payoff” hold for all \(\eta \in [0,1]\). However, the case \(t^*=\tau\) is not relevant for the special case \(\eta =0\).
1.1 Payoff with support by sponsor
If there is only sponsor support, the total payoff to the beneficiary in \(t^*=T\) is given by
$$\begin{aligned} {}^{\{S\}}\Phi (T)&=\Phi _{PF}(T) + \Phi _{S}(T)\\&= {\left\{ \begin{array}{ll} L_{T} + \beta \alpha \left( X_{T}- \frac{L_{}}{\alpha } \right) ^+,&{}\quad X_{T} \ge L_{T} \\ L_{T} ,&{}\quad X_{T} < L_{T} \quad {\text {and}}\quad C_{{T}}> \psi C_0 e^{g {T}} +(L_{T} -X_{{T}}) \\ X_{T} + (C_{T}-\psi C_0 e^{g {T}}), &{}\quad X_{T} < L_{T} \quad {\text {and}} \quad \psi C_0 e^{g {T}}<C_{{T}}< \psi C_0 e^{g {T}} + (L_{T} -X_{{T}})\\ X_{T}, &{}\quad X_{T}<L_{T} \quad {\text {and}} \quad C_{{T}}<\psi C_0 e^{g {T}}\end{array}\right. }\\&= X_{T}+ \beta \alpha \left( X_{T}- \frac{L_{T}}{\alpha } \right) ^+ + \min \{L_{T}-X_{T},(C_{T} - \psi C_0 e^{g {T}})^+\}. \end{aligned}$$
For \(t^*=\tau\) this reduces to
$$\begin{aligned} {}^{\{S\}}\Phi (\tau )&= {\left\{ \begin{array}{ll} L_{{\tau }}, &{}\quad C_{{\tau }} > \psi C_0 e^{g {\tau }} + (1-\eta )L_{\tau } \\ \eta L_{{\tau }} + (C_{{\tau }}-\psi C_0 e^{g {\tau }}), &{} \quad \psi C_0 e^{g {\tau }} < C_{{\tau }}< \psi C_0 e^{g {\tau }} + (1-\eta )L_{\tau } \\ \eta L_{{\tau }}, &{} \quad C_{{\tau }}< \psi C_0 e^{g {\tau }}\end{array}\right. }\\&= \eta L_{\tau } + \min \{(1- \eta ) L_{\tau }, (C_{\tau } - \psi C_0 e^{g {\tau }})^+\}. \end{aligned}$$
1.2 Payoff with support by PGF
If there is only a PGF support, for \(t^*=T\) we have
$$\begin{aligned} {}^{\{PGF\}}\Phi (T)&= \Phi _{PF}(T) + \Phi _{PGF}(T) = {\left\{ \begin{array}{ll} L_{T} + \beta \alpha \left( X_{T}- \frac{L_{T}}{\alpha } \right) ^+, &{}\quad X_{T} \ge L_{T} \\ L_{T}, &{}\quad L_{T} - \bar{L}_{T} < X_{T} < L_{T} \\ X_{T} + \bar{L}_{T}, &{}\quad X_{T} \le L_{T} - \bar{L}_{T} \end{array}\right. } \\&=X_{T} + \beta \alpha \left( X_{T}- \frac{L_{T}}{\alpha } \right) ^+ +\min \{L_{T} - X_{T}, \bar{L}_{T} \} \end{aligned}$$
For \(t^*=\tau\) this reduces to
$$\begin{aligned} {}^{\{PGF\}}\Phi (\tau )= \eta L_{\tau } + \bar{L}_{\tau } . \end{aligned}$$
1.3 Payoff with support by sponsor and PGF
If there exists both the sponsor and PGF guarantee, the payoff to the beneficiary has the following form for \(t^*=T\):
$$\begin{aligned} {}^{\{S,PGF\}}\Phi (T)&= {\left\{ \begin{array}{ll} L_{T} + \beta \alpha \left( X_{T}- \frac{L_{T}}{\alpha } \right) ^+,&{}\quad X_{T} \ge L_{T} \\ L_{T} ,&{}\quad X_{T} < L_{T}\quad {\text {and}}\\ &{}\quad C_{{T}}> \psi C_0 e^{g {T}} +(L_{T} -X_{{T}}) \\ X_{T} + (C_{T}-\psi C_0 e^{g {T}}) &{} \\ \qquad + \min \{L_{T}-X_{T}-(C_{T}-\psi C_0 e^{g {T}}),\bar{L}_{T}\}, &{} \quad X_{T} < L_{T} \quad { \text {and}}\quad \\ &{}\quad \psi C_0 e^{g {T}} <C_{{T}}< \psi C_0 e^{g {T}}\\ &{}\qquad + (L_{T} -X_{{T}})\\ X_{T} + \min \{ L_{T}-X_{T}, \bar{L}_{T}\}, &{}\quad X_{T} < L_{T} \quad {\text {and}}\quad C_{{T}}<\psi C_0 e^{g {T}} \end{array}\right. } \\&= X_{T}+\beta \alpha \left( X_{T}- \frac{L_{T}}{\alpha } \right) ^+ + \min \{L_{T}-X_{T},\bar{L}_{T}+(C_{T}-\psi C_0 e^{g {T}})^+\}\\ \end{aligned}$$
and for the case of \(t^*=\tau\) the payoff can be reduced to
$$\begin{aligned} {}^{\{S,PGF\}}\Phi (\tau )&= {\left\{ \begin{array}{ll} L_{{\tau }}, &{}\quad C_{{\tau }} > \psi C_0 e^{g {\tau }} + (1-\eta )L_{\tau } \\ \eta L_{{\tau }} + (C_{{\tau }}-\psi C_0 e^{g {\tau }}) &{}\\ \qquad +\min \{(1-\eta )L_{{\tau }}-(C_{{\tau }}-\psi C_0 e^{g {\tau }}),\bar{L}_{{\tau }} \}, &{}\quad \psi C_0 e^{g {\tau }}<C_{{\tau }}< \psi C_0 e^{g {\tau }} \\ &{} \qquad + (1-\eta )L_{\tau }\\ \eta L_{{\tau }}+\bar{L}_{{\tau }}, &{}\quad C_{{\tau }}<\psi C_0 e^{g {\tau }}\end{array}\right. }\\&= \eta L_{{\tau }} + \min \{(1-\eta )L_{{\tau }},\bar{L}_{{\tau }}+(C_{{\tau }}-\psi C_0 e^{g {\tau }})^+\} \end{aligned}$$
Shortfall probability
Note that the shortfall probability is zero for \(\eta =1\). Moreover, for \(\eta =0\) all terms of the form \(E[1_{\{\tau \le T\}} \ldots ]\) in “Appendix Shortfall probability” have to be replaced by zero.
1.1 Shortfall probability with support by sponsor
The shortfall probability under sponsor guarantee can be further determined as follows:
$$\begin{aligned} {}^{\{S\}}SP&= E\left[ 1_{\{\tau >T\}} 1_{\{X_T < L_T\}} 1_{\{C_{T}< \psi C_0e^{g T} + (L_T -X_{T})\}} \right] \nonumber \\&\quad + E[ 1_{\{\tau \le T\}} 1_{\{ C_{\tau }< \psi C_0e^{g \tau } + (1-\eta )L_{\tau } \}} ]. \end{aligned}$$
(5)
The first component of RHS can be further split into two terms:
$$\begin{aligned}&E\left[ 1_{\{\tau >T\}} 1_{\{X_T < L_T\}} 1_{\{C_{T}< \psi C_0 e^{g T} + (L_T -X_{T})\}} \right] \\&\quad = E\left[ 1_{\{X_T < L_T\}} 1_{\{C_{T}< \psi C_0 e^{g T} + (L_T -X_{T})\}} \right] -E\left[ 1_{\{\tau \le T\}} 1_{\{X_T < L_T\}} 1_{\{C_{T}< \psi C_0 e^{g T} + (L_T -X_{T})\}} \right] \end{aligned}$$
Relying on the joint distribution of \((\ln X_T, \ln C_T)\) (bivariate normally distributed log-returns) and the density function of the first hitting time \(\tau\), we obtain
$$\begin{aligned}&E\left[ 1_{\{\tau >T\}} 1_{\{X_T < L_T\}} 1_{\{C_{T}< \psi C_0 e^{g T} + (L_T -X_{T})\}} \right] \\&\quad = E\left[ 1_{\{X_T < L_T\}} 1_{\{C_{T}< \psi C_0 e^{g T} + (L_T -X_{T})\}} \right] \\&\qquad - \int _0^T E\left[ 1_{\{X_T < L_T\}} 1_{\{C_{T}< \psi C_0 e^{g T} + (L_T -X_{T})\}}|\tau =t \right] P(\tau \in d t )\\&\quad = \int _{-\infty }^{d_{x1}} \int _{-\infty }^{d_{y1}(x)} n(x)\, n(y)\,dy\,dx -\int _0^T \int _{-\infty }^{d_{x2}(t)} \int _{-\infty }^{d_{y2}(x,t)}n(x)\,n(y) \,f(t)\,dy\,dx\,dt\\&\quad = \int _{-\infty }^{d_{x1}} n(x)\, N(d_{y1}(x))\,dx -\int _0^T \int _{-\infty }^{d_{x2}(t)} n(x)\,N(d_{y2}(x,t)) \,f(t)\,dx\, dt \end{aligned}$$
with
$$\begin{aligned} d_{x1}&= \frac{\ln \frac{L_T }{X_0}-(\hat{\mu }-\frac{1}{2} \theta ^2\sigma ^2)T}{\theta \sigma \sqrt{T}}\end{aligned}$$
(6)
$$\begin{aligned} d_{x2}(t)&= \frac{\ln \frac{L_T }{\eta L_t}-(\hat{\mu }-\frac{1}{2} \theta ^2\sigma ^2)(T-t)}{\theta \sigma \sqrt{T-t}}\end{aligned}$$
(7)
$$\begin{aligned} d y_1(x)&= \frac{\ln \frac{\psi C_0 e^{g T}+L_T- X_0\exp \left\{ \hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)T + \theta \sigma \sqrt{T} x\right\} }{C_0}-(\mu _c-\frac{1}{2} \sigma _c^2)T-\sigma _c\rho \sqrt{T}x}{\sigma _c\sqrt{T}\sqrt{1-\rho ^2}}\end{aligned}$$
(8)
$$\begin{aligned} d y_2(x,t)&= \frac{\ln \frac{\psi C_0 e^{g T}+L_T- \eta L_t \exp \left\{ \hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)(T-t) + \theta \sigma \sqrt{T-t} x\right\} }{C_0}-(\mu _c-\frac{1}{2} \sigma _c^2)T}{\sigma _c\sqrt{T}\sqrt{1-\rho ^2}}\nonumber \\&\quad -\frac{\sigma _c\rho ( \frac{\ln \frac{\eta L_t}{X_0}-(\hat{\mu }-\frac{1}{2} \sigma ^2 \theta ^2)t}{\theta \sigma }+\sqrt{T-t}x)}{\sigma _c\sqrt{T}\sqrt{1-\rho ^2}} \end{aligned}$$
(9)
where \(n(\cdot )\) and \(N(\cdot )\) are the density function and the cumulative distribution function of a standard normal distribution, \(\hat{\mu }:=r+\theta (\mu -r)\) and \(f(\cdot )\) is the density function of the first hitting time \(\tau\):
$$\begin{aligned} f(t)&= \frac{-\ln \frac{\eta L_0}{X_0}}{\theta \sigma t^{3/2}} n\left( \frac{\ln \frac{\eta L_0}{X_0} -\tilde{\mu } \sigma \theta t}{\theta \sigma \sqrt{t}} \right) \end{aligned}$$
(10)
with \(\tilde{\mu }:= (r - \delta +\theta (\mu -r) -\frac{\theta ^2 \sigma ^2}{2} )/(\theta \sigma )\).
The last term of (5) can be computed by using the density of \(\tau\):
$$\begin{aligned} E[ 1_{\{\tau \le T\}} 1_{\{C_{\tau }< \psi C_0e^{g \tau } + (1-\eta )L_{\tau } \}} ]&= \int _0^T N({d_{y3}(t)}) f(t) dt, \end{aligned}$$
with
$$\begin{aligned} d_{y3}(t)&= \frac{\ln \frac{\psi C_0 e^{g t} + (1-\eta )L_{t} }{C_0 } -(\mu _c -\frac{1}{2} \sigma _c^2) t -\frac{\rho \sigma _c}{\sigma \theta } (\ln \frac{\eta L_{t} }{X_0} - (\hat{\mu }-\frac{1}{2} \sigma ^2 \theta ^2) t) }{\sigma _c \sqrt{t}\sqrt{1-\rho ^2} }. \end{aligned}$$
(11)
1.2 Shortfall probability with support by sponsor and PGF
In this case, the shortfall probability can be determined very similarly as in 7.1. Note that we have to adjust the limits for the integration:
$$\begin{aligned} \widetilde{d_{x1}}&= \frac{\ln \frac{L_T-\bar{L}_T}{X_0}-(\hat{\mu }-\frac{1}{2} \theta ^2\sigma ^2)T}{\theta \sigma \sqrt{T}}\end{aligned}$$
(12)
$$\begin{aligned} \widetilde{d_{x2}}(t)&= \frac{\ln \frac{L_T-\bar{L}_T }{\eta L_t}-(\hat{\mu }-\frac{1}{2} \theta ^2\sigma ^2)(T-t)}{\theta \sigma \sqrt{T-t}}\end{aligned}$$
(13)
$$\begin{aligned} \widetilde{d_{y1}}(x)&= \frac{\ln \frac{\psi C_0 e^{g T}+L_T-\bar{L}_T- X_0\exp \left\{ \hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)T + \theta \sigma \sqrt{T} x\right\} }{C_0}-(\mu _c-\frac{1}{2} \sigma _c^2)T-\sigma _c\rho \sqrt{T}x}{\sigma _c\sqrt{T}\sqrt{1-\rho ^2}} \end{aligned}$$
(14)
$$\begin{aligned} \widetilde{ d_{y2}}(x,t)&=\frac{\ln \frac{\psi C_0 e^{g T}+L_T-\bar{L}_T- \eta L_t \exp \left\{ \hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)(T-t) + \theta \sigma \sqrt{T-t} x\right\} }{C_0}-(\mu _c-\frac{1}{2} \sigma _c^2)T}{\sigma _c\sqrt{T}\sqrt{1-\rho ^2}} \end{aligned}$$
(15)
$$\begin{aligned}&\quad -\frac{\sigma _c\rho ( \frac{\ln \frac{\eta L_t}{X_0}-(\hat{\mu }-\frac{1}{2} \sigma ^2 \theta ^2)t}{\theta \sigma }+\sqrt{T-t}x)}{\sigma _c\sqrt{T}\sqrt{1-\rho ^2}} \end{aligned}$$
(16)
$$\begin{aligned} \widetilde{d_{y3}}(t)&= \frac{\ln \frac{\psi C_0 e^{g t} + (1-\eta )L_t - \bar{L}_t }{C_0 } -(\mu _c -\frac{1}{2} \sigma _c^2) t -\frac{\rho \sigma _c}{\sigma \theta } \ln \frac{\eta L_t }{X_0} - (\hat{\mu }-\frac{1}{2} \sigma ^2 \theta ^2) t }{\sigma _c \sqrt{t}\sqrt{1-\rho ^2} }. \end{aligned}$$
(17)
Then
$$\begin{aligned}&E\left[ 1_{\{\tau >T\}} 1_{\{X_T < L_T-\bar{L}_T\}} 1_{\{C_{T}< \psi C_0 e^{g T} + (L_T -X_{T})-\bar{L}_T \}} \right] \\&\quad = \int _{-\infty }^{\widetilde{d_{x1}}} n(x)\, N(\widetilde{d_{y1}}(x)) dx -\int _0^T \int _{-\infty }^{\widetilde{d_{x2}}(t)} n(x)\,N(\widetilde{d_{y2}}(x,t)) \,f(t)\,dx\,dt. \end{aligned}$$
and
$$\begin{aligned} E[ 1_{\{\tau \le T\}} 1_{\{C_{\tau }< \psi C_0e^{g \tau } + (1-\eta )L_{\tau } -\bar{L}_{\tau }\}} ]&= \int _0^T N(\widetilde{d_{y3}}(t)) f(t) \,dt. \end{aligned}$$
Expected shortfall
Note that the expected shortfall is zero for \(\eta =1\). Moreover, for \(\eta =0\) all terms of the form \(E[1_{\{\tau \le T\}} \ldots ]\) in “Appendix Expected shortfall” have to be replaced by zero.
1.1 Expected shortfall without external support
If there is no external support, the expected shortfall can be further determined as follows:
$$\begin{aligned} {}^{\{\}}ES=E[(1-\eta )L_\tau 1_{\{\tau \le T\}}]+E[(L_T-X_T)1_{\{\tau >T\}} 1_{\{L_T>X_T\}}]. \end{aligned}$$
The first term of the RHS is given by
$$\begin{aligned} E[(1-\eta )L_\tau 1_{\{\tau \le T\}}]&=\int _0^T (1-\eta ) L_t f(t)\,dt =(1-\eta ) L_0 \int _0^T e^{\delta t} f(t)\,dt \end{aligned}$$
where \(f(\cdot )\) is defined in (10). The second term of RHS is given by
$$\begin{aligned}&E\left[ (L_T-X_T)1_{\{\tau >T\}} 1_{\{X_T<L_T\}}\right] \\&\quad =\; E[(L_T-X_T)1_{\{X_T<L_T\}}] - \int _0^TE\left[ (L_T-X_T)1_{\{X_T<L_T\}} \mid \tau = t\right] P(\tau \in dt)\\&\quad =\; \int _{-\infty }^{d_{x1}}\left( L_T-X_0\,\exp \left\{ (\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)T+\sigma \theta \sqrt{T}x\right\} \right) n(x)\, dx\\&\qquad -\int _0^T \int _{-\infty }^{d_{x2}(t)} \left( L_T-\eta L_t\,\exp \left\{ (\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)(T-t)+\sigma \theta \sqrt{T-t}x\right\} \right) n(x)\,f(t)\,dx\,dt \end{aligned}$$
where \(d_{x1}\) and \(d_{x2}(t)\) are defined as in (6) and (7) of “Appendix Shortfall probability with support by sponsor”.
1.2 Expected shortfall with support by PGF
If there exists a PGF, the expected shortfall can be determined very similarly to the case without any support (“Appendix Expected shortfall without external support”). The first term is given by
$$\begin{aligned} E[(1-\eta )(1-\gamma )L_\tau 1_{\{\tau \le T\}}]&=(1-\eta )(1-\gamma ) L_0 \int _0^T e^{\delta t} f(t)\,dt. \end{aligned}$$
The second term is given by
$$\begin{aligned}&E\left[ (L_T(1-\gamma +\gamma \eta )-X_T)1_{\{\tau >T\}} 1_{\{X_T<L_T(1-\gamma +\gamma \eta )\}}\right] \\&\quad = \int _{-\infty }^{\widetilde{d_{x1}}}\left( L_T(1-\gamma +\gamma \eta )-X_0\,\exp \left\{ (\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)T+\sigma \theta \sqrt{T}x\right\} \right) n(x)\, dx\\&\qquad -\! \int _0^T \! \int _{-\infty }^{\widetilde{d_{x2}}(t)} \! \left( L_T(1-\gamma +\gamma \eta )-\eta L_t \exp \left\{ (\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)(T-t)+\sigma \theta \sqrt{T-t}x\right\} \right) \!n(x)\,\!f(t)\,\!dx\,\!dt \end{aligned}$$
where \(\widetilde{d_{x1}}\) and \(\widetilde{d_{x2}}(t)\) are defined as in (12) and (13) of “Appendix Shortfall probability with support by sponsor and PGF”.
1.3 Expected shortfall with support by sponsor
As observed in (4), the expected shortfall under the sponsor guarantee consists of four terms. The determination of these four terms is very similar as in the calculation of the shortfall probability under the sponsor guarantee. Hence, we decide not to write down the detailed derivation and jump to the results immediately. The first term is given by the following double integrals
$$\begin{aligned}&E[((1-\eta )L_\tau - (C_\tau - \psi C_0 e^{g \tau })) 1_{\{\tau \le T\}} 1_{\{C_0 e^{g \tau } < C_{\tau }< \psi C_0 e^{g \tau } + (1-\eta )L_{\tau })\}}]\nonumber \\&\quad = \int _0^T \int _{ d_{y4}(t)}^{d_{y3}(t)} ((1-\eta )L_t - (C_t(y) - \psi C_0 e^{gt}))\, n(y)\, f(t)\, dy\, dt \nonumber \\ {\text {with}} \; \; C_t(y)&= \,C_0\,\exp \left\{ (\mu _c-\frac{1}{2} \sigma _c^2)t+\sigma _c(\rho \frac{ \ln \frac{\eta L_t }{X_0} - (\hat{\mu }-\frac{1}{2} \sigma ^2 \theta ^2) t }{\theta \sigma }+\sqrt{1-\rho ^2}\sqrt{t}y) \right\} \nonumber \\ d_{y4}(t)&:= \frac{\ln \frac{\psi C_0 e^{g t}}{C_0 } -(\mu _c -\frac{1}{2} \sigma _c^2) t -\frac{\rho \sigma _c}{\sigma \theta } (\ln \frac{\eta L_t }{X_0} - (\hat{\mu }-\frac{1}{2} \sigma ^2 \theta ^2) t)}{\sigma _c \sqrt{t}\sqrt{1-\rho ^2} }. \end{aligned}$$
(18)
The second term is determined similarly
$$\begin{aligned}&E[(1-\eta )L_\tau 1_{\{\tau \le T\}} 1_{\{C_{\tau }< \psi C_0 e^{g \tau }\}}]\\&\quad = \int _0^T \int _{-\infty }^{d_{y4}(t)} (1-\eta )L_t\, n(y)\, f(t)\, dy\, dt= \int _0^T (1-\eta )L_t\, N(d_{y4}(t))\, f(t)\, dt \end{aligned}$$
The third term owns the following form:
$$\begin{aligned}&E[(L_T-X_T-(C_T- \psi C_0 e^{g T}))1_{\{\tau >T\}} 1_{\{X_T<L_T\}}1_{\{C_0 e^{g T} < C_{T}< \psi C_0 e^{g T} + (L_{T} -X_{T})\}}]\,\\&\quad = \int _{-\infty }^{d_{x1}} \int _{d_{y5}(x)}^{d_{y1}(x)}(L_T-X_0\,\exp \left\{ (\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)T+\theta \sigma \sqrt{T}x\right\} \\&\qquad -C_0\, \exp \left\{ (\mu _c-\frac{1}{2} \sigma _c^2)T+\sigma _c (\rho \sqrt{T}x+\sqrt{1-\rho ^2} \sqrt{T} y)\right\} +\psi C_0 e^{gT}) n(x)\, n(y)\, dy \,dx\\&\qquad -\int _0^T \int _{-\infty }^{d_{x2}(t)} \int _{d_{y6}(x,t)}^{d_{y2}(x,t)}(L_T-\eta L_t\,\exp \left\{ (\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)(T-t)+\theta \sigma \sqrt{T-t}x\right\} \\&\qquad -C_0\, \exp \left\{ (\mu _c-\frac{1}{2} \sigma _c^2)T+\sigma _c(\rho (\frac{ \ln \frac{\eta L_t }{X_0} - (\hat{\mu }-\frac{1}{2} \sigma ^2 \theta ^2) t }{\theta \sigma }+\sqrt{T-t}x)+\sqrt{1-\rho ^2}\sqrt{T}y) \right\} \\&\qquad +\psi C_0 e^{gT}) n(x)\,n(y)\, f(t)\,dy\, dx\, dt \end{aligned}$$
with
$$\begin{aligned} d_{y5}(x)&:=\; \frac{\ln \frac{\psi C_0 e^{g T}}{C_0}-(\mu _c-\frac{1}{2} \sigma _c^2)T-\sigma _c\rho \sqrt{T}x}{\sigma _c\sqrt{T}\sqrt{1-\rho ^2}}\end{aligned}$$
(19)
$$\begin{aligned} d_{y6}(x,t)&=\; \frac{\ln \frac{\psi C_0 e^{g T}}{C_0}-(\mu _c-\frac{1}{2} \sigma _c^2)T-\sigma _c\rho ( \frac{\ln \frac{\eta L_t}{X_0}-(\hat{\mu }-\frac{1}{2} \sigma ^2 \theta ^2)t}{\theta \sigma }+\sqrt{T-t}x)}{\sigma _c\sqrt{T}\sqrt{1-\rho ^2}} \end{aligned}$$
(20)
Finally, the last term is given by
$$\begin{aligned}&E[(L_T-X_T)1_{\{\tau >T\}} 1_{\{X_T<L_T\}}1_{\{C_{T}< \psi C_0 e^{g T}\}}]\,\\&\quad = \int _{-\infty }^{d_{x1}} \int _{-\infty }^{d_{y5}(x)}(L_T-X_0\,\exp \{(\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)T+\theta \sigma \sqrt{T}x\}) n(y)\,n(x)\, dy dx\\&\qquad -\int _0^T \int _{-\infty }^{d_{x2}(t)} \int _{-\infty }^{d_{y6}(x,t)}\left( L_T-\eta L_t\,\exp \left\{ (\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)(T-t)+\theta \sigma \sqrt{T-t}x\right\} \right) \\&\qquad \times n(y)\,n(x)\,f(t)\,dy\,dx\,dt\\&\quad =\int _{-\infty }^{d_{x1}}(L_T-X_0\,\exp \{(\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)T+\theta \sigma \sqrt{T}x\}) N(d_{y5}(x))\,n(x)\,dx\\&\qquad -\int _0^T \int _{-\infty }^{d_{x2}(t)}\left( L_T-\eta L_t\,\exp \left\{ (\hat{\mu }-\frac{1}{2} \theta ^2 \sigma ^2)(T-t)+\theta \sigma \sqrt{T-t}x\right\} \right) \\&\qquad \times N(d_{y6}(x,t))\,n(x)\,f(t)\,dx\,dt \end{aligned}$$
where \(d_{x1}\), \(d_{y1}(x)\), \(d_{x2}(t)\), \(d_{y2}(x,t)\), \(d_{y5}(x)\), \(d_{y6}(x,t)\) are defined in (6), (8), (7), (9), (19) and (20).
1.4 Expected shortfall with support by sponsor and PGF
The expected shortfall under sponsor guarantee and PGF support can be determined very similarly to the case with only sponsor support (“Appendix Expected shortfall with support by sponsor”). For the first and second term we have to consider the additional factor \((1-\gamma )\) and use the adjusted \(\widetilde{d_{y3}}(t)\) instead of \(d_{y3}(t)\). This yields For the third and fourth term, considering the additional \(\bar{L}_T\) and the adjusted \(\widetilde{d_{x1}}\), \(\widetilde{d_{x2}}(t)\), \(\widetilde{d_{y1}}(x)\) and \(\widetilde{d_{y2}}(x)\), which are defined in (12)–(16). In this place, we leave out the details.
