Maintaining cost and ruin probability

Abstract

Specialized funds such as charitable trusts do not attach much value to consumption, instead, they pursue to maintain a satisfactory level of spending and avoid ruin to achieve their managerial goals. We employ an objective function tailored for studying ruin probability of a specialized fund, which implies simple analytical conditions to judge if the fund can be operable permanently. We analytically show that even if the fund has fixed portfolio weights and faces both fixed and proportional maintaining cost, there may still exist a positive probability for the fund to maintain operability permanently. Since if the stock is profitable enough, the wealth process has a large positive drift to offset effects of the fixed cost and downside risk. We extend the benchmark model to a case allowing portfolio rebalance between risky assets, and also obtain analytical expressions for optimal portfolio choices and ruin probability. Allowing portfolio selection potentially improves survival probability. Finally, we provide conditions needed to enjoy a positive probability of permanent survival when the fund can invest in a short bond (potentially with a risky asset) with stochastic nominal riskless interest rate.

Appendix

1.1 Derivation of Bellman equation for benchmark model

We now solve for the ruin probability using the martingale technique pioneer by Fleming and Rishel (1975). Define

$$\begin{aligned} M_{t}\equiv \int _{s=0}^{t}d\iota _{s}+P\left( w_{t},\iota _{t}\right) , \end{aligned}$$

for $t>0$, where $P(\cdot ,\cdot )$ is the proposed value function defined as $1-P^{s}$ and $P^{s}$ is given in (8) in the statement of Proposition 1. This denotes the probability of subsequent termination in the future after date t. Note that by the dynamics of wealth (4) and by Itô’s Lemma (letting $dt^{2}=dtdZ=0$, $dZ^{2}=dt)$, we have

$$\begin{aligned} dM_{t}&=d\iota _{t}+P_{w}dw_{t}+\frac{1}{2}P_{ww}\left( dw_{t}\right) ^{2}+\left( P\left( w_{t},1\right) -P\left( w_{t},0\right) \right) d\iota _{t}\\&=d\iota _{t}+\left( 1-\iota _{t}\right) P_{w}\left( \mu w_{t}dt+\sigma wdZ_{t}-k_{0}dt\right) +\left( P\left( w_{t},1\right) -P\left( w_{t},0\right) \right) d\iota _{t}. \end{aligned}$$

Moreover, the ruin probability as a conditional expectation should be a martingale. Because, a martingale do not have tendency to increase or decrease on average, and it well fit the role that it is bounded between 0 and 1 as a probability. Hence, the drift of M should be zero, gives rise to

$$\begin{aligned} 0=\frac{{\mathrm {E}}\left[ dM\right] }{dt}=\frac{{\mathrm {E}}\left[ \left( \mu w_{t}dt+\sigma wdZ_{t}-k_{0}dt\right) P_{w}+\frac{1}{2}\sigma ^{2}w^{2} P_{ww}dt\right] }{dt}, \end{aligned}$$

which yields the following ODE for ruin probability at any finite time,

$$\begin{aligned} \left( \mu w_{t}-k_{0}\right) P_{w}+\frac{1}{2}\sigma ^{2}w^{2}P_{ww}=0. \end{aligned}$$

1.2 Verifying solution and derivation of ruin probability

To verify

$$\begin{aligned} P=C_{1}\int _{0}^{w}\exp \left( -\frac{2k_{0}}{\sigma ^{2}\varpi }\right) \varpi ^{-\frac{2\mu }{\sigma ^{2}}}d\varpi +C_{2} \end{aligned}$$

is the solution of the ODE, we just need to derive

$$\begin{aligned} P_{w}&=C_{1}\exp \left( -\frac{2k_{0}}{\sigma ^{2}w}\right) w^{-\frac{2\mu }{\sigma ^{2}}},\\ P_{ww}&=C_{1}\left[ \frac{2k_{0}}{\sigma ^{2}w^{2}}\exp \left( -\frac{2k_{0}}{\sigma ^{2}w}\right) w^{-\frac{2\mu }{\sigma ^{2}}}-\frac{2\mu }{\sigma ^{2}}\exp \left( -\frac{2k_{0}}{\sigma ^{2}w}\right) w^{-\frac{2\mu }{\sigma ^{2}}-1}\right] , \end{aligned}$$

therefore,

$$\begin{aligned} \frac{P_{ww}}{P_{w}}=\frac{2\left( k_{0}-\mu w_{t}\right) }{\sigma ^{2}w^{2} }, \end{aligned}$$

which is just another form of the ODE (5). This verifies the solution.

Now we derive the analytical expression of ruin probability when $\mu /\sigma ^{2}=1$. We have

$$\begin{aligned} P=C_{1}\int _{0}^{w}\exp \left( -\frac{2k_{0}}{\sigma ^{2}\varpi }\right) \varpi ^{-2}d\varpi +C_{2}. \end{aligned}$$

Note that $-\varpi ^{-2}d\varpi =d\varpi ^{-1}$, thus

$$\begin{aligned} P&=C_{1}\frac{\sigma ^{2}}{2k_{0}}\int _{0}^{w}\exp \left( -\frac{2k_{0} }{\sigma ^{2}}\varpi ^{-1}\right) d\left( -\frac{2k_{0}}{\sigma ^{2}} \varpi ^{-1}\right) +C_{2}\\&=C_{1}\frac{\sigma ^{2}}{2k_{0}}\exp \left( -\frac{2k_{0}}{\sigma ^{2}\varpi }\right) +C_{2}. \end{aligned}$$

Substitute the boundary conditions of

$$\begin{aligned} P\left( w=0\right) =1\,\,{\text { and }}\,\,P\left( w=\infty \right) =0, \end{aligned}$$

we have

$$\begin{aligned} 1=C_{2}+0\,\,{\text { and }}\,\,0=C_{2}+\frac{\sigma ^{2}}{2k_{0}}C_{1}, \end{aligned}$$

hence, we obtain

$$\begin{aligned} C_{1}=-\frac{2k_{0}}{\sigma ^{2}}\,\,{\text { and }}\,\,C_{2}=1, \end{aligned}$$

and therefore

$$\begin{aligned} P=1-\exp \left( -\frac{2k_{0}}{\sigma ^{2}w}\right) . \end{aligned}$$

□

1.3 Proof of Proposition 2

1.3.1 Derivation of ruin probability

We now derive the solution of the following Bellman equation which is an ODE,

$$\begin{aligned} \min _{\theta _{t}}\left[ \left( w\left( \mu +\theta _{t}{\hat{\Delta }}\right) -k_{0}\right) P_{w}+\frac{1}{2}\left( \sigma ^{2}+2\theta \sigma \left( \sigma _{1}-\sigma \right) +\theta _{t}^{2}{\hat{\sigma }}^{2}\right) w^{2} P_{ww}\right] =0. \end{aligned}$$

(21)

The first-order condition with respect to $\theta$ gives

$$\begin{aligned} {\hat{\Delta }}wP_{w}+\sigma \left( \sigma _{1}-\sigma \right) w^{2}P_{ww} +\theta _{t}{\hat{\sigma }}^{2}w^{2}P_{ww}=0\Longrightarrow \theta _{t}=-\frac{{\hat{\Delta }}P_{w}}{{\hat{\sigma }}^{2}wP_{ww}}-\frac{\sigma \left( \sigma _{1}-\sigma \right) }{{\hat{\sigma }}^{2}}. \end{aligned}$$

Substitute it back to ODE (21) yields

$$\begin{aligned} awP_{w}-k_{0}+bw^{2}\frac{P_{ww}}{P_{w}}-\frac{{\hat{\Delta }}^{2}P_{w}}{2{\hat{\sigma }}^{2}P_{ww}}=0 \end{aligned}$$

Let

$$\begin{aligned} g\left( w\right) \equiv \frac{P_{ww}}{P_{w}}=\frac{\partial \ln \left( -P_{w}\right) }{\partial w}, \end{aligned}$$

(22)

we can rewrite the ODE as

$$\begin{aligned} aw-k_{0}+bw^{2}g-\frac{{\hat{\Delta }}^{2}}{2{\hat{\sigma }}^{2}}\frac{1}{g}=0\Leftrightarrow bw^{2}g^{2}+\left( aw-k_{0}\right) g-\frac{\hat{\Delta }^{2}}{2{\hat{\sigma }}^{2}}=0. \end{aligned}$$

Let

$$\begin{aligned} {\bar{a}}=bw^{2}, {\bar{b}}=aw-k_{0},\,\,{\bar{c}}=-\frac{{\hat{\Delta }}^{2}}{2{\hat{\sigma }}^{2}}, \end{aligned}$$

we have

$$\begin{aligned} g_{1}\left( w\right)&=\frac{-{\bar{b}}+\sqrt{{\bar{b}}^{2}-4{\bar{a}}{\bar{c}}} }{2{\bar{a}}}=\frac{k_{0}-aw+\sqrt{\left( aw-k_{0}\right) ^{2}+\frac{2bw^{2}{\hat{\Delta }}^{2}}{{\hat{\sigma }}^{2}}}}{2bw^{2}}>0,\\ {\text {and }}\,\,g_{2}\left( w\right)&=\frac{-{\bar{b}}-\sqrt{{\bar{b}} ^{2}-4{\bar{a}}{\bar{c}}}}{2{\bar{a}}}=\frac{k_{0}-aw-\sqrt{\left( aw-k_{0}\right) ^{2}+\frac{2bw^{2}{\hat{\Delta }}^{2}}{{\hat{\sigma }}^{2}}}}{2bw^{2}}<0. \end{aligned}$$

Note that the ruin probability decreases with wealth, therefore, we have $P_{w}<0$. Since we need to assume that $P_{ww}>0$ to ensure a minimization problem, by the expression of optimal portfolio, we have $g=P_{ww}/P_{w}<0$, therefore, we have $g=g_{2}\left( w\right)$. Now to obtain the ruin probability, we only need to solve Eq. (22) with $g=g_{2}\left( w\right)$.

Integration of both side of the Eq. (22) form 1 to w yields

$$\begin{aligned} \int _{1}^{w}g\left( \varpi \right) d\varpi =\ln \left( -P_{w}\right) +\bar{C}_{1}. \end{aligned}$$

Integrating again from 0 to w yields the general solution of the ruin probability,

$$\begin{aligned} P\left( w\right) ={\hat{C}}_{1}\int _{0}^{w}\exp \left( \int _{1}^{h}g\left( \varpi \right) d\varpi \right) dh+{\hat{C}}_{2}, \end{aligned}$$

(23)

where constant ${\hat{C}}_{1}$ and ${\hat{C}}_{2}$ is to be determined by boundary conditions. Similar to the discussion in the previous section, we also have the two boundary conditions as

$$\begin{aligned} P\left( w=0\right) =1\,\,{\text { and }}\,\,P\left( w=\infty \right) =0. \end{aligned}$$

Imposing the two boundary conditions to (23) gives

$$\begin{aligned} {\hat{C}}_{1}=-\frac{1}{\int _{0}^{\infty }\left[ \exp \left( \int _{1} ^{h}g\left( \varpi \right) d\varpi \right) \right] dh}\ \ \,\,{\text { and }}\,\,{\hat{C}}_{2}=1. \end{aligned}$$

Hence, to calculate ${\hat{C}}_{1}$ and, thus, the general solution, we need to first solve for function $\int _{1}^{h}g\left( \varpi \right) d\varpi$. The following subsection provides an analytical expression of the indefinite integral $\int g\left( \varpi \right) d\varpi$.

1.3.2 Calculate integral

By the expression of $g_{2}\left( \varpi \right)$, we have

$$\begin{aligned} \int g_{2}\left( \varpi \right) d\varpi&=\int \frac{k_{0}-a\varpi -\sqrt{\left( a\varpi -k_{0}\right) ^{2}+2b\varpi ^{2}{\hat{\Delta }}^{2} /{\hat{\sigma }}^{2}}}{2b\varpi ^{2}}d\varpi \\&=\frac{1}{2b}\int \frac{k_{0}-a\varpi }{\varpi ^{2}}d\varpi -\frac{1}{2b} \int \frac{\sqrt{\left( a\varpi -k_{0}\right) ^{2}+2b\varpi ^{2}\hat{\Delta }^{2}/{\hat{\sigma }}^{2}}}{\varpi ^{2}}d\varpi . \end{aligned}$$

Rearranging the first integral, we have

$$\begin{aligned} \int \frac{k_{0}-a\varpi }{\varpi ^{2}}d\varpi =k_{0}\int \frac{1}{\varpi ^{2} }d\varpi -a\int \frac{1}{\varpi }d\varpi =-k_{0}\frac{1}{\varpi }-a\ln \varpi . \end{aligned}$$

Rearranging the second integral, and let ${\bar{\mu }}^{2}=a^{2}+2b\hat{\Delta }^{2}/{\hat{\sigma }}^{2}$, we have

$$\begin{aligned} SI\equiv \int \frac{\sqrt{\left( a\varpi -k_{0}\right) ^{2}+2b\varpi ^{2} {\hat{\Delta }}^{2}/{\hat{\sigma }}^{2}}}{\varpi ^{2}}d\varpi =\int \frac{1}{\varpi }\sqrt{{\bar{\mu }}^{2}-\frac{2ak_{0}}{\varpi }+\frac{k_{0}^{2}}{\varpi ^{2}} }d\varpi . \end{aligned}$$

Let $t=1/\varpi \Longrightarrow 1/t=\varpi$ and $-\left( 1/t^{2}\right) dt=d\varpi$, hence the integral becomes

$$\begin{aligned} SI=\int \frac{1}{\varpi }\sqrt{{\bar{\mu }}^{2}-\frac{2ak_{0}}{\varpi }+\frac{k_{0}^{2}}{\varpi ^{2}}}d\varpi =-\int \frac{1}{t}\sqrt{k_{0}^{2}t^{2} -2ak_{0}t+{\bar{\mu }}^{2}}dt. \end{aligned}$$

Let

$$\begin{aligned} \sqrt{k_{0}^{2}t^{2}-2ak_{0}t+{\bar{\mu }}^{2}}=k_{0}t-x, \end{aligned}$$

we have

$$\begin{aligned}&k_{0}^{2}t^{2}-2ak_{0}t+{\bar{\mu }}^{2}=k_{0}^{2}t^{2}+x^{2}-2k_{0} tx \\&\quad \Longrightarrow 2k_{0}tx-2ak_{0}t=x^{2}-{\bar{\mu }}^{2}\,\,{\text { and }}\,\,t=\frac{x^{2}-{\bar{\mu }}^{2} }{2k_{0}x-2k_{0}a}. \end{aligned}$$

Note that

$$\begin{aligned} \left( \frac{x^{2}-{\bar{\mu }}^{2}}{2k_{0}\left( x-a\right) }\right) ^{\prime }=-\frac{\left( x^{2}-{\bar{\mu }}^{2}\right) }{2k_{0}\left( x-a\right) ^{2}}+\frac{2x\left( x-a\right) }{2k_{0}\left( x-a\right) ^{2}}=\frac{x^{2}-2xa+{\bar{\mu }}^{2}}{2k_{0}\left( x-a\right) ^{2}}, \end{aligned}$$

therefore

$$\begin{aligned} dt=\frac{x^{2}-2xa+{\bar{\mu }}^{2}}{2k_{0}\left( x-a\right) ^{2}}dx. \end{aligned}$$

Moreover, we have

$$\begin{aligned}&\sqrt{k_{0}^{2}t^{2}-2ak_{0}t+{\bar{\mu }}^{2}}=k_{0}t-x\\&\quad =\frac{k_{0}x^{2}-k_{0}{\bar{\mu }}^{2}}{2k_{0}x-2k_{0}a}-\frac{2k_{0} x^{2}-2k_{0}ax}{2k_{0}x-2k_{0}a}=\frac{-k_{0}{\bar{\mu }}^{2}-k_{0}x^{2} +2k_{0}ax}{2k_{0}x-2k_{0}a}. \end{aligned}$$

Therefore,

$$\begin{aligned} SI&=-\int \frac{1}{t}\sqrt{k_{0}^{2}t^{2}-2ak_{0}t+{\bar{\mu }}^{2}}dt\\&=-\int \frac{2k_{0}x-2k_{0}a}{x^{2}-{\bar{\mu }}^{2}}\frac{-k_{0}{\bar{\mu }} ^{2}-k_{0}x^{2}+2k_{0}ax}{2k_{0}x-2k_{0}a}\frac{x^{2}-2xa+{\bar{\mu }}^{2} }{2k_{0}\left( x-a\right) ^{2}}dx\\&=\frac{1}{2}\int \frac{\left( {\bar{\mu }}^{2}+x^{2}-2ax\right) ^{2}}{\left( x-a\right) ^{2}\left( x^{2}-{\bar{\mu }}^{2}\right) }dx. \end{aligned}$$

We can express the above integrand as a sum of easily integrable functions

$$\begin{aligned} \frac{\left( {\bar{\mu }}^{2}+x^{2}-2ax\right) ^{2}}{\left( x-a\right) ^{2}\left( x^{2}-{\bar{\mu }}^{2}\right) }=\frac{x}{x-a}+\frac{A}{x-a}+\frac{B}{\left( x-a\right) ^{2}}+\frac{C}{x-\sqrt{{\bar{\mu }}^{2}}}+\frac{D}{x+\sqrt{{\bar{\mu }}^{2}}}, \end{aligned}$$

where constants A, B, C, and D are to determined. Thus, we have

$$\begin{aligned} \left( {\bar{\mu }}^{2}+x^{2}-2ax\right) ^{2}&=x\left( x-a\right) \left( x^{2}-{\bar{\mu }}^{2}\right) +A\left( x-a\right) \left( x^{2}-{\bar{\mu }} ^{2}\right) +B\left( x^{2}-{\bar{\mu }}^{2}\right) \\&\quad +C\left( x-a\right) ^{2}\left( x+\sqrt{{\bar{\mu }}^{2}}\right) +D\left( x-a\right) ^{2}\left( x-\sqrt{{\bar{\mu }}^{2}}\right) . \end{aligned}$$

Let $x=a$, we have

$$\begin{aligned} \left( {\bar{\mu }}^{2}+a^{2}-2a^{2}\right) ^{2}=B\left( a^{2}-{\bar{\mu }} ^{2}\right) \Longrightarrow B=a^{2}-{\bar{\mu }}^{2}. \end{aligned}$$

Let $x=\sqrt{{\bar{\mu }}^{2}}$, we have

$$\begin{aligned} \left( {\bar{\mu }}^{2}+x^{2}-2ax\right) ^{2}&= C\left( x-a\right) ^{2}\left( x+\sqrt{{\bar{\mu }}^{2}}\right) \Longleftrightarrow 2\left( {\bar{\mu }} ^{2}-a\sqrt{{\bar{\mu }}^{2}}\right) ^{2}=C\sqrt{{\bar{\mu }}^{2}}\left( \sqrt{{\bar{\mu }}^{2}}-a\right) ^{2} \\&\Longrightarrow C=2\sqrt{{\bar{\mu }}^{2}}. \end{aligned}$$

Let $x=-\sqrt{{\bar{\mu }}^{2}}$, we have

$$\begin{aligned} \left( {\bar{\mu }}^{2}+x^{2}-2ax\right) ^{2}&= D\left( x-a\right) ^{2}\left( x-\sqrt{{\bar{\mu }}^{2}}\right) \\&\Longrightarrow D=-\frac{\left( {\bar{\mu }}^{2}+{\bar{\mu }}^{2}+2a\sqrt{\bar{\mu }^{2}}\right) ^{2}}{\left( \sqrt{{\bar{\mu }}^{2}}+a\right) ^{2}2\sqrt{{\bar{\mu }}^{2}}}=-2\sqrt{{\bar{\mu }}^{2}}. \end{aligned}$$

Let $x=0$, we have

$$\begin{aligned}{\bar{\mu }}^{4}&=Aa{\bar{\mu }}^{2}-B{\bar{\mu }}^{2}+Ca^{2}\sqrt{{\bar{\mu }}^{2}} -Da^{2}\sqrt{{\bar{\mu }}^{2}}=Aa{\bar{\mu }}^{2}-\left( a^{2}-{\bar{\mu }} ^{2}\right) {\bar{\mu }}^{2}+4{\bar{\mu }}^{2}a^{2} \\ &\Longrightarrow {\bar{\mu }}^{2}=Aa-\left( a^{2}-{\bar{\mu }}^{2}\right) +4a^{2}=A+3a\Longrightarrow A=-3a. \end{aligned}$$

Thus, we have

$$\begin{aligned}&\int \frac{\left( {\bar{\mu }}^{2}+x^{2}-2ax\right) ^{2}}{\left( x-a\right) ^{2}\left( x^{2}-{\bar{\mu }}^{2}\right) }dx\\&\quad =\int \left( \frac{x}{x-a}+\frac{A}{x-a}+\frac{B}{\left( x-a\right) ^{2} }+\frac{C}{x-\sqrt{{\bar{\mu }}^{2}}}+\frac{D}{x+\sqrt{{\bar{\mu }}^{2}}}\right) dx\\&\quad =x+\left( a+A\right) \ln \left| x-a\right| -B\frac{1}{x-a} +C\ln \left| x-{\bar{\mu }}\right| +D\ln \left| x+{\bar{\mu }}\right| . \end{aligned}$$

Therefore, eventually, we have

$$\begin{aligned} \int g\left( \varpi \right) d\varpi&=\frac{1}{2b}\int \frac{k_{0}-a\varpi }{\varpi ^{2}}d\varpi -\frac{1}{2b}\int \frac{\sqrt{\left( a\varpi -k_{0}\right) ^{2}+2b\varpi ^{2}{\hat{\Delta }}^{2}/{\hat{\sigma }}^{2}}}{\varpi ^{2}}d\varpi \\&=\frac{1}{2b}\left( -k_{0}\frac{1}{\varpi }-a\ln \varpi \right) -\frac{1}{4b}\left( \begin{array}[c]{c} x+\left( a+A\right) \ln \left| a-x\right| -B\frac{1}{x-a}\\ +C\ln \left| {\bar{\mu }}-x\right| +D\ln \left| x+{\bar{\mu }}\right| \end{array} \right) , \end{aligned}$$

where

$$\begin{aligned} x=k_{0}t-\sqrt{k_{0}^{2}t^{2}-2ak_{0}t+{\bar{\mu }}^{2}}=\frac{k_{0}}{\varpi }-\sqrt{\frac{k_{0}^{2}}{\varpi ^{2}}-\frac{2ak_{0}}{\varpi }+{\bar{\mu }}^{2}}. \end{aligned}$$

1.3.3 Simplified expression for ruin probability

With the expression of $\int g\left( \varpi \right) d\varpi$, we have

$$\begin{aligned}&\exp \left( \int g\left( \varpi \right) d\varpi \right) =\exp \left( -\frac{k_{0}}{2b}\frac{1}{\varpi }+\ln \varpi ^{-\frac{a}{2b}}\right) \\&\qquad \times \exp \left( -\frac{x}{4b}+\ln \left| a-x\right| ^{-\frac{a+A}{4b} }+\frac{B}{4b}\frac{1}{\left( x-a\right) }+\ln \left| \bar{\mu }-x\right| ^{-\frac{C}{4b}}+\ln \left| x+{\bar{\mu }}\right| ^{-\frac{D}{4b}}\right) \\&\quad =\varpi ^{-\frac{a}{2b}}\left( \left| a-x\right| \right) ^{-\frac{a+A}{4b}}\left( \left| \frac{x+{\bar{\mu }}}{{\bar{\mu }} -x}\right| \right) ^{\frac{C}{4b}}\exp \left( \frac{B}{4b}\frac{1}{\left( x-a\right) }-\frac{k_{0}}{2b}\frac{1}{\varpi }-\frac{x}{4b}\right) \\&\quad =\left( \left| \frac{a-x}{\varpi }\right| \right) ^{\frac{a}{2b} }\left( \left| \frac{{\bar{\mu }}+x}{{\bar{\mu }}-x}\right| \right) ^{\frac{{\bar{\mu }}}{2b}}\exp \left( \frac{a^{2}-{\bar{\mu }}^{2}}{4b\left( x-a\right) }-\frac{k_{0}}{2b\varpi }-\frac{x}{4b}\right) . \end{aligned}$$

Note that in a form of definite integral, we have

$$\begin{aligned} \exp \left( \int _{1}^{h}g\left( \varpi \right) d\varpi \right)&=\exp \left( \int g\left( \varpi \right) d\varpi \left| _{_{\varpi =h} }\right. -\int g\left( \varpi \right) d\varpi \left| _{_{\varpi =1} }\right. \right) \\&=\exp \left( -\int g\left( \varpi \right) d\varpi \left| _{_{\varpi =1} }\right. \right) \exp \left( \int g\left( \varpi \right) d\varpi \left| _{_{\varpi =h}}\right. \right) , \end{aligned}$$

therefore,

$$\begin{aligned} \int _{0}^{w}\exp \left( \int _{1}^{h}g\left( \varpi \right) ds\right) dh=\exp \left( -\int g\left( \varpi \right) d\varpi \left| _{_{\varpi =1} }\right. \right) \int _{0}^{w}\exp \left( \int g\left( \varpi \right) d\varpi \left| _{_{\varpi =h}}\right. \right) dh. \end{aligned}$$

Substitute it back to (23), we obtain the expression for ruin probability.

1.3.4 Derivation of optimal portfolio weight

With the expression of optimal portfolio weight in asset B, when $P_{ww}>0,$ we have

$$\begin{aligned} \theta&=-\frac{{\hat{\Delta }}P_{w}}{{\hat{\sigma }}^{2}wP_{ww}}-\frac{\sigma \left( \sigma _{1}-\sigma \right) }{{\hat{\sigma }}^{2}}\\&=-\frac{{\hat{\Delta }}}{{\hat{\sigma }}^{2}w}\frac{2bw^{2}}{k_{0}-aw-\sqrt{\left( aw-k_{0}\right) ^{2}+2bw^{2}{\hat{\Delta }}^{2}/{\hat{\sigma }}^{2}} }-\frac{\sigma \left( \sigma _{1}-\sigma \right) }{{\hat{\sigma }}^{2}}\\&=\frac{1}{{\hat{\Delta }}}\left( \frac{k_{0}}{w}-a+\sqrt{\left( a-\frac{k_{0}}{w}\right) ^{2}+\frac{2b{\hat{\Delta }}^{2}}{{\hat{\sigma }}^{2}}}\right) -\frac{\sigma \left( \sigma _{1}-\sigma \right) }{{\hat{\sigma }}^{2}}. \end{aligned}$$

Moreover, the absolute portfolio weight in asset B is given as $\theta w$, hence, we have $\lim _{w\rightarrow 0}\theta w=2k_{0}/{\hat{\Delta }}$.

1.4 Lemma 2 and Proof

Lemma 5

Given $x=k_{0}/\varpi -\sqrt{k_{0}^{2}/\varpi ^{2}-2ak_{0}/\varpi +{\bar{\mu }}^{2} }$, and ${\bar{\mu }}^{2}=a^{2}+2b{\hat{\Delta }}^{2}/{\hat{\sigma }}^{2}$, we have $\left( {\bar{\mu }}+x\right) /\left( {\bar{\mu }}-x\right) \sim 1/\varpi$.

Proof

We now prove $\left( {\bar{\mu }}+x\right) /\left( {\bar{\mu }}-x\right) \sim 1/\varpi$ as $\varpi \rightarrow \infty$ and $x-a\sim \varpi$ as $\varpi \rightarrow 0$. Since $\lim _{\varpi \rightarrow \infty }x=-{\bar{\mu }}$, hence $\lim _{\varpi \rightarrow \infty }{\bar{\mu }}-x=-2{\bar{\mu }}$. Therefore, to proof $\frac{{\bar{\mu }}+x}{{\bar{\mu }}-x}\sim \frac{1}{\varpi }$, we only need to proof ${\bar{\mu }}+x\sim \frac{1}{\varpi }$, which is easy to see by

$$\begin{aligned} {\bar{\mu }}+x={\bar{\mu }}+\frac{k_{0}}{\varpi }-\sqrt{\frac{k_{0}^{2}}{\varpi ^{2} }-\frac{2ak_{0}}{\varpi }+{\bar{\mu }}^{2}}=\frac{1}{\varpi }\frac{2k_{0}\left( {\bar{\mu }}+a\right) }{{\bar{\mu }}+\frac{k_{0}}{\varpi }+\sqrt{\frac{k_{0}^{2} }{\varpi ^{2}}-\frac{2ak_{0}}{\varpi }+{\bar{\mu }}^{2}}}, \end{aligned}$$

where

$$\begin{aligned} \lim _{\varpi \rightarrow \infty }\frac{2k_{0}\left( {\bar{\mu }}+a\right) }{{\bar{\mu }}+\frac{k_{0}}{\varpi }+\sqrt{\frac{k_{0}^{2}}{\varpi ^{2}} -\frac{2ak_{0}}{\varpi }+{\bar{\mu }}^{2}}}=\frac{k_{0}\left( {\bar{\mu }}+a\right) }{{\bar{\mu }}}. \end{aligned}$$

Moreover, note that

$$\begin{aligned} x-a&=\frac{\left( \frac{k_{0}}{\varpi }-a-\sqrt{\frac{k_{0}^{2}}{\varpi ^{2}}-\frac{2ak_{0}}{\varpi }+{\bar{\mu }}^{2}}\right) \left( \frac{k_{0} }{\varpi }-a+\sqrt{\frac{k_{0}^{2}}{\varpi ^{2}}-\frac{2ak_{0}}{\varpi }+\bar{\mu }^{2}}\right) }{\frac{k_{0}}{\varpi }-a+\sqrt{\frac{k_{0}^{2}}{\varpi ^{2} }-\frac{2ak_{0}}{\varpi }+{\bar{\mu }}^{2}}}\\&=\varpi \frac{a^{2}-{\bar{\mu }}^{2}}{k_{0}-\varpi a+\sqrt{k_{0}^{2} -2ak_{0}\varpi +{\bar{\mu }}^{2}\varpi ^{2}}}. \end{aligned}$$

Hence, we have

$$\begin{aligned} \lim _{\varpi \rightarrow 0}\frac{x-a}{\varpi }=\frac{a^{2}-{\bar{\mu }}^{2}}{k_{0}+k_{0}}=\frac{a^{2}-{\bar{\mu }}^{2}}{2k_{0}}, \end{aligned}$$

and this completes the proof. □

1.5 Derivations of optimal portfolio and Bellman equation

By the dynamics of wealth (10) and by Itô’s Lemma (letting $dt^{2}=dtdZ=0$, $dZ^{2}=dt)$, we have

$$\begin{aligned} dM_{t}&=d\iota _{t}+P_{w}dw_{t}+\frac{1}{2}P_{ww}\left( dw_{t}\right) ^{2}+\left( P\left( w_{t},1\right) -P\left( w_{t},0\right) \right) d\iota _{t}\\&=d\iota _{t}+\left( P\left( w_{t},1\right) -P\left( w_{t},0\right) \right) d\iota _{t},\\&\quad +\left( 1-\iota _{t}\right) P_{w}\left( w\left( \mu dt+\sigma dZ_{t}+\theta _{t}\left( \left( \mu _{1}+\Delta -\mu \right) dt+\left( \sigma _{1}-\sigma \right) dZ_{t}+\sigma _{0}dZ_{t}^{0}\right) \right) -k_{0}dt\right) \\&\quad +\frac{1}{2}\left( 1-\iota _{t}\right) ^{2}P_{ww}w^{2}\left( \sigma ^{2}+2\theta _{t}\sigma \left( \sigma _{1}-\sigma \right) +\theta _{t}^{2}\left( \left( \sigma _{1}-\sigma \right) ^{2}+\sigma _{0}^{2}\right) \right) dt \end{aligned}$$

where we employ

$$\begin{aligned} \left( dw\right) ^{2}&=\left( w\left( \mu dt+\sigma dZ_{t}+\theta _{t}\left( \left( \mu _{1}+\Delta -\mu \right) dt+\left( \sigma _{1} -\sigma \right) dZ_{t}+\sigma _{0}dZ_{t}^{0}\right) \right) -k_{0}dt\right) ^{2}\\&=w^{2}\left( \theta _{t}^{2}\left( \left( \sigma _{1}-\sigma \right) ^{2}+\sigma _{0}^{2}\right) +2\theta _{t}\left( \sigma _{1}-\sigma \right) \sigma +\sigma ^{2}\right) dt. \end{aligned}$$

For the optimal portfolio, M should be a martingale, hence, the drift of M should be zero, which yields the following Bellman equation as an ODE

$$\begin{aligned} \min _{\theta _{t}}\left[ \left( w\left( \mu +\theta _{t}{\hat{\Delta }}\right) -k_{0}\right) P_{w}+\frac{1}{2}\left( \sigma ^{2}+2\theta _{t}\sigma \left( \sigma _{1}-\sigma \right) +\theta _{t}^{2}{\hat{\sigma }}^{2}\right) w^{2} P_{ww}\right] =0. \end{aligned}$$

where ${\hat{\Delta }}=\mu _{1}-\mu +\Delta$, and ${\hat{\sigma }}^{2}=\sigma _{0} ^{2}+\left( \sigma _{1}-\sigma \right) ^{2}$. Rearranging, we have

$$\begin{aligned} \min _{\theta _{t}}\left[ \mu wP_{w}+\theta _{t}{\hat{\Delta }}wP_{w}-k_{0} P_{w}+\frac{1}{2}\sigma ^{2}w^{2}P_{ww}+\theta _{t}\sigma \left( \sigma _{1}-\sigma \right) w^{2}P_{ww}+\frac{1}{2}\theta _{t}^{2}{\hat{\sigma }}^{2} w^{2}P_{ww}=0\right] . \end{aligned}$$

(24)

First-order condition w.r.t. $\theta _{t}$ gives

$$\begin{aligned} {\hat{\Delta }}wP_{w}+\sigma \left( \sigma _{1}-\sigma \right) w^{2}P_{ww} +\theta _{t}{\hat{\sigma }}^{2}w^{2}P_{ww}=0. \end{aligned}$$

Thus, we obtain the optimal portfolio as

$$\begin{aligned} \theta _{t}^{*}=-\frac{{\hat{\Delta }}P_{w}+\sigma \left( \sigma _{1} -\sigma \right) wP_{ww}}{{\hat{\sigma }}^{2}wP_{ww}}=-\frac{{\hat{\Delta }}P_{w} }{{\hat{\sigma }}^{2}wP_{ww}}+\phi , \end{aligned}$$

where $\phi =-\frac{\sigma \left( \sigma _{1}-\sigma \right) }{{\hat{\sigma }}^{2} }$. Substitute the optimal portfolio back to the ODE (24), we have

$$\begin{aligned}&\left( w\left( \mu +\left( -\frac{{\hat{\Delta }}P_{w}}{{\hat{\sigma }} ^{2}wP_{ww}}+\phi \right) {\hat{\Delta }}\right) -k_{0}\right) P_{w}\\&\qquad +\frac{1}{2}\left( \sigma ^{2}+2\left( -\frac{{\hat{\Delta }}P_{w}}{{\hat{\sigma }}^{2}wP_{ww}}+\phi \right) \sigma \left( \sigma _{1}-\sigma \right) +\left( -\frac{{\hat{\Delta }}P_{w}}{{\hat{\sigma }}^{2}wP_{ww}}+\phi \right) ^{2}{\hat{\sigma }}^{2}\right) w^{2}P_{ww}=0\\&\quad \Longleftrightarrow \mu wP_{w}+\phi {\hat{\Delta }}wP_{w}-\phi {\hat{\Delta }}wP_{w}-\frac{\hat{\Delta }\sigma \left( \sigma _{1}-\sigma \right) }{{\hat{\sigma }}^{2}}wP_{w}+\frac{{\hat{\Delta }}^{2}P_{w}^{2}}{2{\hat{\sigma }}^{2}P_{ww}}-\frac{{\hat{\Delta }} ^{2}P_{w}^{2}}{{\hat{\sigma }}^{2}P_{ww}}-k_{0}P_{w}\\&\qquad +\phi \sigma \left( \sigma _{1}-\sigma \right) w^{2}P_{ww}+\frac{\sigma ^{2}w^{2}P_{ww}}{2}+\frac{\phi ^{2}{\hat{\sigma }}^{2}w^{2}P_{ww}}{2} =0\\&\quad \Longleftrightarrow \left( \mu -\frac{{\hat{\Delta }}\sigma \left( \sigma _{1}-\sigma \right) }{{\hat{\sigma }}^{2}}\right) wP_{w}-k_{0}P_{w}-\frac{{\hat{\Delta }}^{2}P_{w}^{2} }{2{\hat{\sigma }}^{2}P_{ww}}\\&\qquad +\left( \phi \sigma \left( \sigma _{1}-\sigma \right) +\frac{\sigma ^{2} +\phi ^{2}{\hat{\sigma }}^{2}}{2}\right) w^{2}P_{ww}=0\\&\quad \Longleftrightarrow awP_{w}-k_{0}P_{w}-\frac{{\hat{\Delta }}^{2}P_{w}^{2}}{2{\hat{\sigma }}^{2}P_{ww} }+bw^{2}P_{ww}=0, \end{aligned}$$

where

$$\begin{aligned} a\equiv \mu -\frac{{\hat{\Delta }}\varphi }{{\hat{\sigma }}^{2}},\,\,b\equiv \frac{\sigma ^{2}\sigma _{0}^{2}}{2{\hat{\sigma }}^{2}},\,\,\phi \equiv -\frac{\varphi }{{\hat{\sigma }}^{2}},\,\,{\text {and}}\,\,\varphi \equiv \sigma \left( \sigma _{1}-\sigma \right) . \end{aligned}$$

Note that

$$\begin{aligned} b=\phi \sigma \left( \sigma _{1}-\sigma \right) +\frac{\sigma ^{2}+\phi ^{2} {\hat{\sigma }}^{2}}{2}=\frac{\sigma ^{2}\sigma _{0}^{2}}{2\left( \sigma _{0} ^{2}+\left( \sigma _{1}-\sigma \right) ^{2}\right) }=\frac{\sigma ^{2} \sigma _{0}^{2}}{2{\hat{\sigma }}^{2}}. \end{aligned}$$