Risk-Sensitive Asset Management with Lognormal Interest Rates

Abstract

Risk-sensitive asset management on both finite and infinite time horizons are treated on a market with a bank account and a risky stock. The risk-free interest rate is formulated as a geometric Brownian motion, and affects the return of the risky stock. The problems become standard risk-sensitive control problems. We derive the Hamilton–Jacobi–Bellman equations and study these solutions. Using solutions, we construct optimal strategies and optimal values.

Appendices

Appendix 1: Formal Derivation of HJB Equation (1.5)

We give a formal derivation of (1.5). Consider

$$\begin{aligned} V(t, x, y):=\inf _{\pi \in {\mathcal {A}}_{t,T}}E_{t,x,y} \left[ \left( X_{t,T}^\pi \right) ^\gamma \right] , \end{aligned}$$

where $X_{t,T}^\pi :=X_{T}^\pi /X_{t}^\pi$.

Let $0\le t \le T$ and $\delta >0$ such that $t+\delta <T$. By dynamic programming principle (III.7 of Fleming and Soner 2006), we have

$$\begin{aligned} V(t, x, y)&= \inf _{\pi \in {\mathcal {A}}_{t, t+\delta }} E_{t, x, y}\left[ V(t+\delta , X^\pi _{t+\delta }, Y_{t+\delta })\right] . \end{aligned}$$

(A.1)

Here, we recall

$$\begin{aligned} dX_{s}^\pi&=X_{s}^\pi \{r(Y_s)+\pi \lambda (Y_s) \}ds+X^\pi _s\pi _s e dw(s), \quad X^\pi _t=x, \\ dY_s&=b(Y_s)ds+c(Y_s)dw_2(s), \quad Y_t=y. \end{aligned}$$

Assume $V(t, x, y) \in {\mathcal {C}}^{1,2,2}([0,T]\times (0, \infty ) \times (0, \infty ))$. Applying It$\hat{\text {o}}$’s formula, we have

$$\begin{aligned} V(t+\delta , X^\pi _{t+\delta }, Y_{t+\delta })&= V(t, x, y)+\int ^{t+\delta }_t \left[ \partial _s V(s, X^\pi _{s}, Y_{s}) \right. \\&\quad \left. + \frac{c(Y_s)^2}{2} \partial _{yy}V(s, X^\pi _{s}, Y_{s})+b(Y_s)\partial _yV(s, X^\pi _{s}, Y_{s})\right. \\&\quad \left. +\frac{1}{2}(X_s^\pi )^2 \pi _s^2 \partial _{xx}V(s, X^\pi _{s}, Y_{s}) \right. \\&\quad \left. +X_s^\pi \{ r(Y_s)+\pi _s \lambda (Y_s)\}\partial _xV(s, X^\pi _{s}, Y_{s}) \right. \\&\left. +\,X_s^\pi \pi _s \rho c(Y_s) \partial _{xy}V(s, X^\pi _{s}, Y_{s}) \right] ds \\&\qquad + \int ^{t+\delta }_t \partial _xV(s, X^\pi _{s}, Y_{s})X_s^\pi \pi _s e dw(s)\\&+\int ^{t+\delta }_t \partial _{y}V(s, X^\pi _{s}, Y_{s}) c(Y_s)dw_2(s). \end{aligned}$$

Assuming some suitable condition on the derivatives of V and also $\pi$ such that

$$\begin{aligned} E_{t,x,y}\left[ \int ^{t+\delta }_t \partial _xV(s, X^\pi _{s}, Y_{s})X_s^\pi \pi _s e dw(s) \right] =0, \end{aligned}$$

and

$$\begin{aligned} E_{t,x,y}\left[ \int ^{t+\delta }_t \partial _{y}V(s, X^\pi _{s}, Y_{s}) c(Y_s)dw_2(s) \right] =0, \end{aligned}$$

we have

$$\begin{aligned}&E_{t, x, y}\left[ V(t+\delta , X^\pi _{t+\delta }, Y_{t+\delta }) \right] =V(t, x, y) + E_{t,x,y} \left[ \int ^{t+\delta }_t \left[ \partial _s V(s, X^\pi _{s}, Y_{s}) \right. \right. \\&\qquad + \frac{c(Y_s)^2}{2} \partial _{yy}V(s, X^\pi _{s}, Y_{s})+b(Y_s)\partial _yV(s, X^\pi _{s}, Y_{s})+\frac{1}{2}(X_s^\pi )^2 \pi _s^2 \partial _{xx}V(s, X^\pi _{s}, Y_{s}) \\&\quad \left. \left. +X_s^\pi \{ r(Y_s)+\pi _s \lambda (Y_s)\}\partial _xV(s, X^\pi _{s}, Y_{s}) +X_s^\pi \pi _s \rho c(Y_s) \partial _{xy}V(s, X^\pi _{s}, Y_{s}) \right] ds \right] . \end{aligned}$$

Hence, (A.1) becomes

$$\begin{aligned}&\inf _{ \pi \in {\mathcal {A}}_{t, t+\delta }} E_{t,x, y}\left[ \int ^{t+\delta }_t \left[ \partial _s V(s, X^\pi _{s}, Y_{s})+ \frac{c(Y_s)^2}{2}\partial _{yy}V(s, X^\pi _{s}, Y_{s}) \right. \right. \\&\left. \left. +b(Y_s)\partial _yV(s, X^\pi _{s}, Y_{s})+\frac{1}{2}(X_s^\pi )^2 \pi _s^2 \partial _{xx}V(s, X^\pi _{s}, Y_{s}) \right. \right. \\&\left. \left. X_s^\pi \{ r(Y_s)+\pi _s \lambda (Y_s)\}\partial _xV(s, X^\pi _{s}, Y_{s}) +X_s^\pi \pi _s \rho c(Y_s) \partial _{xy}V(s, X^\pi _{s}, Y_{s})\right] ds \right] =0. \end{aligned}$$

Dividing by $\delta$ and letting $\delta \rightarrow 0$, we can formally derive

$$\begin{aligned} \begin{aligned}&-\partial _{t}V= \frac{c(y)^2}{2} \partial _{yy}V +\inf _{\pi \in {{\mathbb {R}}}} \left[ b(y)\partial _{y}V + \rho c(y) \pi x \partial _{xy}V -\frac{1}{2}\pi ^2 x^2 \partial _{xx}V \right. \\&\quad \left. + \{ r(y)+\pi \lambda (y)\}x\partial _xV \right] ,\\&V(T, x, y)=x^\gamma . \end{aligned} \end{aligned}$$

Setting $V(t,x,y)=x^\gamma {{\mathrm {e}}}^{\gamma v(t,y)}$, we obtain (1.5).

Appendix 2: Preliminaries

In this section, we introduce the following result, which will be used several times in the proofs of our theorems.

Lemma B.1

(Hata and Sekine (2006) : Theorem 3.1.) For $f:=(f_1,f_2): (0,\infty ) \mapsto {{\mathbb {R}}}^2$, denote $f(Y):=\left( f(Y_t)\right) _{t\in [0,T]}$. Suppose f(Y) is progressively measurable such that $\int _0^T |f(Y_t)|^2 dt<\infty$ a.e.. Then the martingale property of $(M_t)_{t\in [0,T]}$ is equivalent to that of $(M_{2,t})_{t\in [0,T]}$. Here, $(M_t)_{t\in [0,T]}$ and $(M_{2,t})_{t\in [0,T]}$ are defined as follows :

$$\begin{aligned} M_t:&={{\mathrm {e}}}^{\int ^t_0 f(Y_s)dw(s)-\frac{1}{2}\int ^t_0 |f(Y_s)|^2 ds },\\ M_{2,t}&:={{\mathrm {e}}}^{\int ^t_0 f_2(Y_s)dw_2(s)-\frac{1}{2}\int ^t_0 |f_2(Y_s)|^2 ds }. \end{aligned}$$

Lemma B.2

Assume $(\mathbf{A1})$-$(\mathbf{A3})$. Define $(M_t)_{t\in [0,T]}$ as

$$\begin{aligned} M_t:&={{\mathrm {e}}}^{\int ^t_0 (\beta _0+\beta _1Y_s)dw_2(s)-\frac{1}{2}\int ^t_0 (\beta _0+\beta _1Y_s)^2 ds }. \end{aligned}$$

If $\beta _1<0$ and $\beta _0 \in {\mathbb {R}}$, then we have

$$\begin{aligned} E\left[ M_t \right] =1, \ t\in [0,T]. \end{aligned}$$

(B.1)

Proof

We apply the idea of Lemma 4.1.1 in Bensoussan (1992). Recall that

$$\begin{aligned} dM_t =M_t(\beta _0+\beta _1 Y_t)dw_2(t). \end{aligned}$$

(B.2)

Let $\epsilon >0$ be arbitrary. We apply It$\hat{\text {o}}$’s formula to $\frac{M_t}{1+\epsilon M_t}$ and have

$$\begin{aligned}&d\left( \frac{M_t}{1+\epsilon M_t} \right) = dN_t-A_t dt, \end{aligned}$$

(B.3)

where $N_t$ and $A_t$ are defined by

$$\begin{aligned} N_t:=\int ^{t}_{0} \frac{M_s(\beta _0+\beta _1 Y_s)}{\left( 1+\epsilon M_s\right) ^2} dw_2(u),\ \mathrm{and} \ A_t:=\frac{\epsilon M_t^2(\beta _0+\beta _1 Y_t)^2}{\left( 1+\epsilon {M_t}\right) ^3}. \end{aligned}$$

If we can check that there is $K_{1,T, y}>0$ such that

$$\begin{aligned} E\left[ \int ^{t}_{0} M_s(1+Y^2_s)ds\right] \le K_{1,T, y}, \end{aligned}$$

(B.4)

then we see that

$$\begin{aligned} E\left[ |N_t|^2 \right]&\le \overline{C}_{\epsilon } E\left[ \int ^{t}_{0} M_s(1+Y^2_s) ds \right] \\&\le \overline{C}_{\epsilon }K_{1,T,y}, \end{aligned}$$

and that $N_t$ is a square-integrate martingale. Therefore, integrating (B.3) on [0, t] and taking expectation for both sides, we have

$$\begin{aligned} E\left[ \frac{M_t}{1+\epsilon M_t} \right] =\frac{1}{1+\epsilon }-E\left[ \int ^t_0 A_s ds \right] . \end{aligned}$$

(B.5)

Here, we observe the following:

• $\displaystyle A_s \rightarrow 0 \ a.e.\ a.s.$ as $\epsilon \rightarrow 0$.

• $\displaystyle A_s \le K_2 M_s(1+Y^2_s), \ \exists K_2>0$.

Hence, from (B.4) and the dominated convergence theorem, we have

$$\begin{aligned} E\left[ \int ^t_0 A_s ds \right] \rightarrow 0 \ \text {as} \ \epsilon \rightarrow 0. \end{aligned}$$

Meanwhile, since $E\left[ M_t \right] \le 1$,

$$\begin{aligned} E\left[ \frac{M_t}{1+\epsilon M_t} \right] \rightarrow E \left[ M_t \right] \ \text {as} \ \epsilon \rightarrow 0. \end{aligned}$$

Hence, letting $\epsilon \rightarrow 0$ in (B.5), we obtain $E\left[ M_t \right] = 1$.

Finally, we prove (B.4). Applying It$\hat{\text {o}}$’s formula to $Y_t^2$, we have

$$\begin{aligned} dY_t^2=(2b+c^2)Y^2_tdt+2cY^2_t dw_2(t). \end{aligned}$$

(B.6)

Using (B.2) and (B.6), we have

$$\begin{aligned} d\left\{ M_t Y^2_t\right\}&=M_t dY^2_t+Y^2_t dM_t + dM_t \cdot dY^2_t \\&=M_t Y^2_t (2c\beta _1 Y_t +2c\beta _0+2b+c^2)dt+M_t Y^2_t(\beta _1 Y_t+\beta _0+2c)dw_2(t). \end{aligned}$$

Setting $\tau _n:=\inf \{t>0 ; Y_t<1/n, n<Y_t\}$, we have

$$\begin{aligned} E\left[ M_{T \wedge \tau _n} Y^2_{t \wedge \tau _n} \right] -y^2=E\left[ \int ^{T \wedge \tau _n} _0 M_s Y^2_s (2c\beta _1 Y_s +2c\beta _0+2b+c^2)ds \right] . \end{aligned}$$

Observing that

$$\begin{aligned}&M_s Y^2_s (2c\beta _1 Y_s +2c\beta _0+2b+c^2)\\&\quad =M_s Y^2_s \left\{ 2c\beta _1 Y_s +(2c\beta _0+2b+c^2+1)-1\right\} \\&\quad \le K_3 M_s-M_s Y^2_s, \quad \exists K_3>0, \end{aligned}$$

we have

$$\begin{aligned} E\left[ \int ^{T \wedge \tau _n}_0 M_s Y^2_s ds \right] \le y^2+K_3T. \end{aligned}$$

As $n\rightarrow \infty$, we obtain (B.4) with $K_{1,T,y}=y^2+K_3T$. $\square$

Appendix 3: The Smoothness of the Function of ${\widehat{v}}(t,y)$

From (2.11) we recall that

$$\begin{aligned} {\widehat{v}}(t,y)&=\frac{1}{k}\log E\left[ F_{T-t}\right] \\&=\frac{1}{k}\log \tilde{E}\left[ {{\mathrm {e}}}^{k\int ^{T-t}_0 \{r(Y_s)+\frac{\lambda (Y_s)^2}{2(1-\gamma )} \} ds } \right] , \end{aligned}$$

where $\tilde{E}[\cdot ]$ denotes the expectation with respect to the probability measure $\tilde{P}$ on $(\Omega , {\mathcal {F}})$ defined by

$$\begin{aligned} \frac{d\tilde{P}}{dP}\biggr |_{{\mathcal {F}}_{t}}:={{\mathrm {e}}}^{\frac{\gamma \rho }{1-\gamma }\int ^t_0 \lambda (Y_{s})dw_{2}(s)-\frac{1}{2}(\frac{\gamma \rho }{1-\gamma })^2 \int ^t_0 \lambda (Y_{s})^2 ds}. \end{aligned}$$

(C.1)

From Lemma B.2$\tilde{P}$ is well-defined. Under $\tilde{P}$ $Y_t$ solves

$$\begin{aligned} dY_t=\left\{ \left( b+\frac{\gamma c\rho \lambda _0}{1-\gamma } \right) Y_t+\frac{\gamma c\rho \lambda _1}{1-\gamma }Y_t^2 \right\} dt+cY_t d\tilde{w}_2(t), \quad Y_0=y, \end{aligned}$$

(C.2)

where $\tilde{w}_2(t)$ is a Brownian motion under $\tilde{P}$ :

$$\begin{aligned} \tilde{w}_2(t):=w_2(t)-\int ^t_0 \left( \frac{\gamma \rho \lambda _0}{1-\gamma }+\frac{\gamma \rho \lambda _1}{1-\gamma } Y_s \right) ds. \end{aligned}$$

To show the smoothness of ${\widehat{v}}$ we show the smoothness of $\phi$ :

$$\begin{aligned} \phi (t,y):=\tilde{E}\left[ {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s)ds } \right] , \end{aligned}$$

(C.3)

where $\theta (y)$ is defined by

$$\begin{aligned} \theta (y):=k\left\{ r(y)+\frac{\lambda (y)^2}{2(1-\gamma )} \right\} . \end{aligned}$$

We try to prove that $\phi (t,y)$ is differentiable with respect to t. We observe

$$\begin{aligned} \begin{aligned} \frac{\phi (t+h,y)-\phi (t,y)}{h}&=\tilde{E}\left[ \frac{1}{h}\left( {{\mathrm {e}}}^{\int ^{t+h}_0 \theta (Y_s)ds } - {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s)ds } \right) \right] \\&=\tilde{E}\left[ \frac{1}{h} \int ^h_0 \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t+\epsilon }_0 \theta (Y_s)ds } \right) d\epsilon \right] . \end{aligned} \end{aligned}$$

(C.4)

We also observe

$$\begin{aligned} \begin{aligned} \left| \frac{1}{h} \int ^h_0 \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t+\epsilon }_0 \theta (Y_s)ds } \right) d\epsilon \right|&\le \frac{1}{h} \int ^h_0|\theta (Y_{t+\epsilon })|d\epsilon \\&\le K \left( \sup _{t\in [0,T]}|Y_t|^2+1 \right) . \end{aligned} \end{aligned}$$

(C.5)

Now, we recall, for $p>1$

$$\begin{aligned} \begin{aligned}&dY_t^p=pY_t^{p-1}dY_t+\frac{p(p-1)}{2}Y_t^{p-2}d\langle Y_{\cdot } \rangle _t\\&=p\left[ \frac{\gamma c\rho \lambda _1}{1-\gamma }Y_t^{p+1} +\left( b+\frac{\gamma c\rho \lambda _0}{1-\gamma } +\frac{c^2(p-1)}{2} \right) Y_t^p\right] dt+pcY_t^p d\tilde{w}_2(t) \end{aligned} \end{aligned}$$

(C.6)

Using (C.6) with $p=2$, we have

$$\begin{aligned} dY_t^2=2\left[ \frac{\gamma c\rho \lambda _1}{1-\gamma }Y_t^{3} +\left( b+\frac{\gamma c\rho \lambda _0}{1-\gamma } +\frac{c^2}{2} \right) Y_t^2\right] dt+2cY_t^2 d\tilde{w}_2(t). \end{aligned}$$

Define

$$\begin{aligned} \tilde{M}_p:=\sup _{y>0} \left[ p\left\{ \frac{\gamma c\rho \lambda _1}{1-\gamma }y^{p+1} +\left( b+\frac{\gamma c\rho \lambda _0}{1-\gamma } +\frac{c^2(p-1)}{2} \right) y^p \right\} \right] . \end{aligned}$$

(C.7)

Then, we have

$$\begin{aligned} \tilde{E}\left[ \sup _{t\in [0,T]}Y_t^2 \right]&\le y^2+\tilde{M}_2 T+2c \tilde{E}\left[ \sup _{t\in [0,T]} \int _0 ^t Y_s^2 d\tilde{w}_2(s) \right] \nonumber \\&\le y^2+\tilde{M}_2 T+2c \tilde{E}\left[ \left( \sup _{t\in [0,T]} \int _0 ^t Y_s^2 d\tilde{w}_2(s) \right) ^2 \right] ^{1/2} \nonumber \\&\le y^2+\tilde{M}_2 T+8c \tilde{E}\left[ \langle \int _0 ^{(\cdot )} Y_s^2 d\tilde{w}_2(s) \rangle _T \right] ^{1/2}\nonumber \\&\le y^2+\tilde{M}_2 T+8c \left( \int _0^T \tilde{E}\left[ Y_t^4\right] dt\right) ^{1/2}. \end{aligned}$$

(C.8)

In the third inequality we use the Burkholder-Davis-Gundy inequality. Using (C.6) with $p=2$ and (C.7) with $p=4$, we have

$$\begin{aligned} \tilde{E}\left[ Y_{t \wedge \tau _n}^4 \right] \le y^4 +\tilde{M}_4 T, \end{aligned}$$

where $\tau _n:=\inf \{t>0 ; Y_t<1/n, n<Y_t \}$. As $n \rightarrow \infty$, we have

$$\begin{aligned} \tilde{E}\left[ Y_{t}^4 \right] \le y^4 +\tilde{M}_4 T. \end{aligned}$$

Hence, we have

$$\begin{aligned} \tilde{E}\left[ \sup _{t\in [0,T]}Y_t^2 \right] \le y^2+\tilde{M}_2 T+8c\sqrt{(y^4 +\tilde{M}_4 T)T}. \end{aligned}$$

Using (C.5) and the dominated convergence theorem, we have

$$\begin{aligned} \frac{\partial \phi }{\partial t}(t,y)&=\lim _{h\rightarrow 0}\tilde{E}\left[ \frac{1}{h} \int ^h_0 \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t+\epsilon }_0 \theta (Y_s)ds } \right) d\epsilon \right] \\&=\tilde{E}\left[ \theta (Y_t) {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s)ds } \right] . \end{aligned}$$

Next, we try to show that $\phi (t,y)$ is differentiable with respect to y. We observe

$$\begin{aligned} \begin{aligned} \frac{\phi (t,y+h)-\phi (t,y)}{h}&=\tilde{E}\left[ \frac{1}{h}\left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+h})ds } - {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^y)ds } \right) \right] \\&=\tilde{E}\left[ \frac{1}{h} \int ^h_0 \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) d\epsilon \right] . \end{aligned} \end{aligned}$$

(C.9)

Here, $Y_s^{z}$ solves (C.2) with $Y_0^{z}=z$. We also observe

$$\begin{aligned} \begin{aligned} \left| \frac{1}{h} \int ^h_0 \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) d\epsilon \right|&= \frac{1}{h} \int ^h_0 \left| \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) \right| d\epsilon . \end{aligned} \end{aligned}$$

(C.10)

We also have

$$\begin{aligned} \begin{aligned} \left| \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) \right|&\le \left| k \int ^t_0 \left\{ 1+\frac{\lambda (Y^{y+\epsilon }_s)}{1-\gamma } \right\} \frac{\partial Y^{y+\epsilon }_s}{\partial \epsilon } ds \right| \\&\le K_T \int ^t_0 (1+Y^{y+\epsilon }_s) \left| \frac{\partial Y^{y+\epsilon }_s}{\partial \epsilon }\right| ds. \end{aligned} \end{aligned}$$

(C.11)

Following the arguments of Lemma 2.2, we see that (C.2) has a unique solution :

$$\begin{aligned} Y^{y+\epsilon }_{t}=\frac{\displaystyle (y+\epsilon ) {{\mathrm {e}}}^{c\tilde{w}_2(t)+c^{2}\tilde{\delta } t}}{\displaystyle 1-\frac{\gamma c\rho \lambda _1}{1-\gamma }(y+\epsilon ) \int _{0}^{t} {{\mathrm {e}}}^{c\tilde{w}_2(s)+c^{2}\tilde{\delta } s}ds}, \end{aligned}$$

where

$$\begin{aligned} \tilde{\delta }:=\frac{1}{c^2}\left( b+\frac{\gamma c \rho \lambda _0}{1-\gamma }-\frac{c^2}{2} \right) . \end{aligned}$$

By a direct calculation we have

$$\begin{aligned} \frac{\partial Y^{y+\epsilon }_{t}}{\partial \epsilon }=\frac{\displaystyle {{\mathrm {e}}}^{c\tilde{w}_2(t)+c^{2}\tilde{\delta } t}}{\displaystyle \left\{ 1-\frac{\gamma c\rho \lambda _1}{1-\gamma }(y+\epsilon ) \int _{0}^{t} {{\mathrm {e}}}^{c\tilde{w}_2(s)+c^{2}\tilde{\delta } s}ds\right\} ^2 }. \end{aligned}$$

Therefore, we have

$$\begin{aligned} 0<Y^{y+\epsilon }_{t}< (y+1) {{\mathrm {e}}}^{c\tilde{w}_2(t)+c^{2}\tilde{\delta } t}, \ \text {and} \ \ 0<\frac{\partial Y^{y+\epsilon }_{t}}{\partial \epsilon }<{{\mathrm {e}}}^{c\tilde{w}_2(t)+c^{2}\tilde{\delta } t}. \end{aligned}$$

(C.12)

Using (C.11) and (C.12), we have

$$\begin{aligned} \sup _{\epsilon \in [0,1]} \left| \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) \right| \le K_T (y+1) \int ^t_0 \left( {{\mathrm {e}}}^{2c\tilde{w}_2(s)+2c^{2}\tilde{\delta } s} +1 \right) ds. \end{aligned}$$

(C.13)

From (C.10) and (C.13) we have

$$\begin{aligned} \begin{aligned} \left| \frac{1}{h} \int ^h_0 \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) d\epsilon \right|&\le K_T (y+1) \int ^t_0 \left( {{\mathrm {e}}}^{2c\tilde{w}_2(s)+2c^{2}\tilde{\delta } s} +1 \right) ds. \end{aligned} \end{aligned}$$

(C.14)

and, we have

$$\begin{aligned} \tilde{E}\left[ \int ^t_0 {{\mathrm {e}}}^{2c\tilde{w}_2(s)+2c^{2}\tilde{\delta } s} ds \right] =\int ^t_0 {{\mathrm {e}}}^{2c^2s+2c^{2}\tilde{\delta } s}ds < \infty . \end{aligned}$$

(C.15)

Using (C.9, (C.14)), (C.15) and the dominated convergence theorem, we have

$$\begin{aligned} \frac{\partial \phi }{\partial y}(t,y)&=\lim _{h\rightarrow 0}\tilde{E}\left[ \frac{1}{h} \int ^h_0 \frac{\partial }{\partial \epsilon } \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) d\epsilon \right] \\&=\tilde{E}\left[ \frac{\partial }{\partial y}\left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^y)ds } \right) \right] . \end{aligned}$$

Finally, we show that $\phi (t,y)$ is twice differentiable with respect to y. We have

$$\begin{aligned} \begin{aligned} \frac{\partial _y \phi (t,y+h)-\partial _y \phi (t,y)}{h}&=\tilde{E}\left[ \frac{1}{h} \int ^h_0 \frac{\partial ^2 }{\partial \epsilon \partial y} \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) d\epsilon \right] . \end{aligned} \end{aligned}$$

(C.16)

Here, we note

$$\begin{aligned} \begin{aligned} \left| \frac{1}{h} \int ^h_0 \frac{\partial ^2 }{\partial \epsilon \partial y} \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) d\epsilon \right|&= \frac{1}{h} \int ^h_0 \left| \frac{\partial ^2 }{\partial \epsilon \partial y} \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) \right| d\epsilon . \end{aligned} \end{aligned}$$

(C.17)

By a direct calculation, we have

$$\begin{aligned} \begin{aligned}&\frac{\partial }{\partial \epsilon } \left\{ \frac{\partial }{\partial y}\left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) \right\} =\int ^t_0 \frac{\partial ^2 }{\partial \epsilon \partial y} \theta (Y^{y+\epsilon }_s)ds {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \\&\quad +\left( \int ^t_0 \frac{\partial }{\partial y} \theta (Y^{y+\epsilon }_s)ds \right) \left( \int ^t_0 \frac{\partial }{\partial \epsilon } \theta (Y^{y+\epsilon }_s)ds \right) {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds }. \end{aligned} \end{aligned}$$

(C.18)

Moreover, we have

$$\begin{aligned} \frac{\partial Y^{y+\epsilon }_{t}}{\partial y}=\frac{\displaystyle {{\mathrm {e}}}^{c\tilde{w}_2(t)+c^{2}\tilde{\delta } t}}{\displaystyle \left\{ 1-\frac{\gamma c\rho \lambda _1}{1-\gamma }(y+\epsilon ) \int _{0}^{t} {{\mathrm {e}}}^{c\tilde{w}_2(s)+c^{2}\tilde{\delta } s}ds\right\} ^2 }, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial ^2 Y^{y+\epsilon }_{t}}{\partial y \partial \epsilon }=-2 \frac{\displaystyle -\frac{\gamma c\rho \lambda _1}{1-\gamma } \int _{0}^{t} {{\mathrm {e}}}^{c\tilde{w}_2(s)+c^{2}\tilde{\delta } s}ds \cdot {{\mathrm {e}}}^{c\tilde{w}_2(t)+c^{2}\tilde{\delta } t}}{\displaystyle \left\{ 1-\frac{\gamma c\rho \lambda _1}{1-\gamma }(y+\epsilon ) \int _{0}^{t} {{\mathrm {e}}}^{c\tilde{w}_2(s)+c^{2}\tilde{\delta } s}ds\right\} ^3 }. \end{aligned}$$

Then, we observe

$$\begin{aligned} \begin{aligned}&0<\frac{\partial Y^{y+\epsilon }_{t}}{\partial y}<{{\mathrm {e}}}^{c\tilde{w}_2(t)+c^{2}\tilde{\delta } t}, \\&\quad \left| \frac{\partial ^2 Y^{y+\epsilon }_{t}}{\partial y \partial \epsilon }\right| \le \frac{\displaystyle 2 {{\mathrm {e}}}^{c\tilde{w}_2(t)+c^{2}\tilde{\delta } t}}{\displaystyle y \left\{ 1-\frac{\gamma c\rho \lambda _1}{1-\gamma }(y+\epsilon ) \int _{0}^{t} {{\mathrm {e}}}^{c\tilde{w}_2(s)+c^{2}\tilde{\delta } s}ds\right\} ^2 }. \end{aligned} \end{aligned}$$

(C.19)

Using (C.18) and (C.19), we have

$$\begin{aligned} \sup _{\epsilon \in [0,1]} \left| \frac{\partial ^2 }{\partial \epsilon \partial y} \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) \right| \le K_T (y+1) \int ^t_0 \left( {{\mathrm {e}}}^{4c\tilde{w}_2(s)+4c^{2}\tilde{\delta } s} +1 \right) ds. \end{aligned}$$

(C.20)

From (C.17) and (C.19) we have

$$\begin{aligned} \begin{aligned} \left| \frac{1}{h} \int ^h_0 \frac{\partial ^2 }{\partial \epsilon \partial y} \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) d\epsilon \right|&\le K_T (y+1) \int ^t_0 \left( {{\mathrm {e}}}^{4c\tilde{w}_2(s)+4c^{2}\tilde{\delta } s} +1 \right) ds. \end{aligned} \end{aligned}$$

(C.21)

And, we have

$$\begin{aligned} \tilde{E}\left[ \int ^t_0 {{\mathrm {e}}}^{4c\tilde{w}_2(s)+4c^{2}\tilde{\delta } s} ds \right] =\int ^t_0 {{\mathrm {e}}}^{8c^2s+4c^{2}\tilde{\delta } s}ds < \infty . \end{aligned}$$

(C.22)

Using (C.16, (C.21), (C.22)) and the dominated convergence theorem, we have

$$\begin{aligned} \frac{\partial ^2 \phi }{\partial y^2}(t,y)&=\lim _{h\rightarrow 0}\tilde{E}\left[ \frac{1}{h} \int ^h_0 \frac{\partial ^2 }{\partial \epsilon \partial y} \left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^{y+\epsilon })ds } \right) d\epsilon \right] \\&=\tilde{E}\left[ \frac{\partial ^2 }{\partial y^2}\left( {{\mathrm {e}}}^{\int ^{t}_0 \theta (Y_s^y)ds } \right) \right] . \end{aligned}$$