Semi-analytical solution for consumption and investment problem under quadratic security market model with inflation risk

Abstract

There exists strong empirical evidence that all inflation rates, interest rates, market price of risk, and return volatilities of assets are stochastic, which is now a stylized fact. However, to the best of our knowledge, existing models providing solutions to consumption–investment problems do not consider all of the aforementioned stochastic processes. We consider a consumption–investment problem for a long-term investor with constant relative risk aversion utility under a quadratic security market model in which all of the processes are stochastic. We solve a nonhomogeneous linear partial differential equation for the indirect utility function and derive a semi-analytical solution. This study obtains an optimal portfolio decomposed into myopic demand, intertemporal hedging demand, and “inflation hedging demand,” and presents that all three types of demand are nonlinear functions of the state vector. Our numerical analysis presents the nonlinearity and significance of the market timing effect. The cause of this result lies mainly with inflation hedging demand in addition to myopic demand. This result highlights the importance of the market timing effect and inflation hedging demand.

Appendix

Proof of Lemma 1

As there is no arbitrage, there exists a unique risk-neutral measure $\mathrm {P}^*$. It follows from Girsanov’s theorem that process $B^*_t$ defined by

$$\begin{aligned} B^*_t = B_t + \int _0^t \varLambda _s\,ds \end{aligned}$$

(95)

is a standard Brownian motion under $\mathrm {P}^*$. Then, the SDE for $X_t$ under $\mathrm {P}^*$ is rewritten as

$$\begin{aligned} dX_t= & {} - ( \mathscr {K} X_t + \varLambda _t )\,dt + I_N \,dB^*_t \\= & {} - \bigl ( \lambda + ( \mathscr {K} + \varLambda ) X_t \bigr )\,dt + I_N\,dB^*_t. \end{aligned}$$

First, we consider the case of nominal bond $P^T_t$. As $P^T_T = 1$, $P^T_t$ is written as

$$\begin{aligned} P^T_t = \mathrm {E}^*_t \left[ \exp \left( - \int ^T_{t} r_s\,ds \right) \right] , \end{aligned}$$

(96)

where $\mathrm {E}^*$ is the expectation operator under $\mathrm {P}^*$.

Because $r_t$ is a function of $X_t$, $P^T_t$ is expressed as a smooth function f of $(X_t, t)$.

$$\begin{aligned} P^T_t = f(X_t,t), \end{aligned}$$

(97)

and it follows from Feynman–Kac’s formula that f is a solution to the following PDE:

$$\begin{aligned}&f_t + \bigl ( - \lambda - ( \mathscr {K} + \varLambda ) X_t \bigr )' f_X+ \frac{1}{2} \mathrm {tr} [ f_{XX} ] - \left( \rho _0 + \rho ' X_t + \frac{1}{2} X'_t \mathscr {R} X_t \right) f = 0, \nonumber \\&f(X_T,T) = 1. \end{aligned}$$

(98)

As the PDE above is a second-order linear equation, the solution f is expressed as

$$\begin{aligned}&f(X_t, t) = \exp \left( \sigma _0(\tau ) + \sigma (\tau )' X_t + \frac{1}{2} X'_t \varSigma (\tau ) X_t \right) , \nonumber \\&(\sigma _0(0), \sigma (0), \varSigma (\tau )) = (0, 0, 0), \end{aligned}$$

(99)

where $\sigma _0(\tau ), \sigma (\tau )$, and $\varSigma (\tau )$ are smooth functions of $\tau = T-t$ and $\varSigma (\tau )$ is a symmetric matrix. It should be noted that $\varSigma (\tau )' = \varSigma (\tau )$ and

$$\begin{aligned} X'_t ( \mathscr {K} + \varLambda )' \varSigma (\tau ) X_t = X'_t \varSigma (\tau ) ( \mathscr {K} + \varLambda ) X_t. \end{aligned}$$

By differentiating Eq. (99) and by inserting the result into Eq. (98), we have

$$\begin{aligned}&- \frac{d\sigma _0(\tau )}{d\tau } - X'_t \frac{d\sigma (\tau )}{d\tau } - \frac{1}{2} X'_t \frac{d\varSigma (\tau )}{d\tau } X_t - \lambda ' ( \sigma (\tau ) + \varSigma (\tau ) X_t) - X'_t ( \mathscr {K} + \varLambda )' \sigma (\tau )\nonumber \\&\quad -\frac{1}{2} X'_t ( \mathscr {K} + \varLambda )' \varSigma (\tau ) X_t -\frac{1}{2} X'_t \varSigma (\tau ) ( \mathscr {K} + \varLambda ) X_t + \frac{1}{2} \bigl ( \sigma (\tau )' \sigma (\tau ) + \mathrm {tr} [\varSigma (\tau )] \bigr ) \nonumber \\&\quad + X'_t \varSigma (\tau ) \sigma (\tau ) + \frac{1}{2} X'_t \varSigma (\tau )^2 X_t - \left( \rho _0 + \rho ' X_t + \frac{1}{2} X'_t \mathscr {R} X_t \right) = 0. \end{aligned}$$

(100)

Because the equation above is identical on $X_t$, Eqs. (21)–(23) are obtained. By differentiating Eq. (99), we obtain SDE (33).

In the case of inflation-indexed bond $P^T_{It}$, we define an equivalent probability measure $\bar{\mathrm {P}}$ by the following Radon-Nikodym derivative with respect to $\mathrm {P}^*$:

$$\begin{aligned} \frac{d\bar{\mathrm {P}}\,\,}{d\mathrm {P}^*} = \exp \left( - \frac{1}{2} \int ^{T^*}_0 (-\varLambda _{Is})'(-\varLambda _{Is})\,ds - \int ^{T^*}_0 (-\varLambda '_{Is})\,dB^*_s \right) . \end{aligned}$$

Then, it follows by Girsanov’s theorem that a process ${\bar{B}}_t$ defined by

$$\begin{aligned} {\bar{B}}_t = B^*_t - \int _0^t \varLambda _{Is}\,ds \end{aligned}$$

(101)

is a standard Brownian motion under $\bar{\mathrm {P}}$, and the SDE for $X_t$ under $\bar{\mathrm {P}}$ is rewritten as

$$\begin{aligned} dX_t= & {} - ( \mathscr {K} X_t + \varLambda _t - \varLambda _{It} )\,dt + I_N \,d{\bar{B}}_t \\= & {} - \bigl ( {\bar{\lambda }} + ( \mathscr {K} + {\bar{\varLambda }} \bigr ) X_t )\,dt + I_N\,d{\bar{B}}_t. \end{aligned}$$

Thus, $P^T_{It}$ is calculated as

$$\begin{aligned} \begin{aligned} P^T_{It}&= E^*_t \left[ \exp \left( - \int ^T_{t} r_s \,ds\right) p_T \right] \\&= p_t \mathrm {E}^*_t \left[ \exp \left( - \int ^T_{t} \left( r_s - i_s + \frac{1}{2} \varLambda '_{Is} \varLambda _{Is} \right) \,ds + \int ^T_{t} \varLambda '_{Is} dB_s \right) \right] \\&= p_t \mathrm {E}^*_t \left[ \exp \left( - \int ^T_{t} \left( r_s - i_s + \frac{1}{2} \varLambda '_{Is}\varLambda _{Is} + \varLambda '_{Is} \varLambda _s \right) \,ds + \int ^T_{t} \varLambda '_{Is} dB^*_s \right) \right] \\&= p_t \mathrm {E}^*_t \left[ \exp \left( - \int ^T_{t} \left( r_s - i_s + \varLambda '_{Is} \varLambda _s \right) \,ds \right) \left( \frac{d\bar{\mathrm {P}}\,\,}{d\mathrm {P}^*} \right) _t \right] \\&= p_t \bar{\mathrm {E}}_t \left[ \exp \left( - \int ^T_{t} {\bar{r}}_s \,ds \right) \right] , \end{aligned}\qquad \end{aligned}$$

(102)

where $\bar{\mathrm {E}}$ is the expectation operator under $\bar{\mathrm {P}}$. Because all of the processes $r_t, i_t, \varLambda '_{It}$, and $\varLambda _t$ are functions of $X_t$, the real price of $P^T_{It}$ is expressed as a smooth function $f(X_t,t)$

$$\begin{aligned} \frac{P^T_{It}}{p_t} = f(X_t,t), \end{aligned}$$

(103)

and it follows from Feynman–Kac’s formula that f is a solution to the following PDE:

$$\begin{aligned}&f_t - \big ( {\bar{\lambda }} + ( \mathscr {K} + {\bar{\varLambda }} ) X_t \bigr )' f_X+ \frac{1}{2} \mathrm {tr} [ f_{XX} ] - \left( {\bar{\rho }}_0 + {\bar{\rho }}' X_t + \frac{1}{2} X'_t \bar{\mathscr {R}} X_t \right) f = 0,\nonumber \\&f(X_T,T) = 1. \end{aligned}$$

(104)

Hence, f is expressed as

$$\begin{aligned}&f(X_t,t) = \exp \left( \sigma _{I0}(\tau ) + \sigma _I(\tau )' X_t + \frac{1}{2} X'_t \varSigma _I(\tau ) X_t \right) ,\nonumber \\&(\sigma _{I0}(0), \sigma _I(0), \varSigma _I(\tau ) ) = (0, 0, 0), \end{aligned}$$

(105)

where $\sigma _{I0}(\tau ), \sigma _I(\tau )$, and $\varSigma _I(\tau )$ are smooth functions and $\varSigma _I(\tau )$ is a symmetric matrix. It should be noted that $\varSigma _I(\tau )' = \varSigma _I(\tau )$ and

$$\begin{aligned} X'_t ( \mathscr {K} + {\bar{\varLambda }})' \varSigma _I(\tau ) X_t = X'_t \varSigma _I(\tau ) ( \mathscr {K} + {\bar{\varLambda }}) X_t. \end{aligned}$$

By differentiating Eq.(105) and by inserting the result into Eq.(104), the following equation is obtained:

$$\begin{aligned}&- \frac{d\sigma _{I0}(\tau )}{d\tau } - X'_t \frac{d\sigma _I(\tau )}{d\tau } - \frac{1}{2} X'_t \frac{d\varSigma _I(\tau )}{d\tau } X_t - {\bar{\lambda }}' ( \sigma _I(\tau ) + \varSigma _I(\tau ) X_t) - X'_t ( \mathscr {K} + {\bar{\varLambda }})' \sigma _I(\tau )\nonumber \\&\quad -\frac{1}{2} X'_t ( \mathscr {K} + {\bar{\varLambda }})' \varSigma _I(\tau ) X_t -\frac{1}{2} X'_t \varSigma _I(\tau ) ( \mathscr {K} + {\bar{\varLambda }}) X_t + \frac{1}{2} \bigl ( \sigma _I(\tau )' \sigma _I(\tau ) + \mathrm {tr} [\varSigma _I(\tau )] \bigr ) \nonumber \\&\quad + X'_t \varSigma _I(\tau ) \sigma _I(\tau ) + \frac{1}{2} X'_t \varSigma _I(\tau )^2 X_t - \left( {\bar{\rho }}_0 + {\bar{\rho }}' X_t + \frac{1}{2} X'_t \bar{\mathscr {R}} X_t \right) = 0. \end{aligned}$$

(106)

Note the following equation:

$$\begin{aligned} \frac{dP^T_{It}}{P^T_{It}} = \frac{dp_t}{p_t} + \frac{df(X_t,t)}{f(X_t,t)} + \frac{dp_t}{p_t} \frac{df(X_t,t)}{f(X_t,t)}. \end{aligned}$$

(107)

Therefore, we obtain Eqs. (25)–(27) and (34).

On the j-th index, Kikuchi [26] proves that $S^j_t$ is given by Eq. (28). Hence, the instantaneous dividend rate process is

$$\begin{aligned} \frac{D^j_{t}}{S^j_t} = \delta _{0j} + \delta '_{j} X_t + \frac{1}{2} X'_t \varDelta _j X_t. \end{aligned}$$

(108)

In a similar way, the following identical equation on $X_t$ is obtained from Eqs. (28) and (108).

$$\begin{aligned}&\sigma _{0j} - \lambda ' (\sigma _j + \varSigma _j X_t) - X'_t (\mathscr {K} + \varLambda )' \sigma _j - \frac{1}{2} X'_t (\mathscr {K} + \varLambda )' \varSigma _j X_t - \frac{1}{2} X'_t \varSigma _j (\mathscr {K} + \varLambda ) X_t\nonumber \\&\quad + \frac{1}{2} \bigl (\sigma '_j \sigma _j + \mathrm {tr}[\varSigma _j]\bigr ) + X'_t \varSigma _j \sigma _j + \frac{1}{2} X'_t \varSigma ^2_j X_t \nonumber \\&\quad +\left( \delta _{0j} - \rho _0 + (\delta _j - \rho )' X_t + \frac{1}{2} X'_t (\varDelta _j - \mathscr {R})X_t \right) = 0. \end{aligned}$$

(109)

Therefore, we obtain Eqs. (29)–(31) and (35).

Proof of Lemma 2

Let $(\vartheta , (\vartheta (\tau )), (\vartheta ^I(\tau )), (\vartheta ^{j}))$ denote a portfolio. The nominal wealth $p_t W_t$ is given by

$$\begin{aligned} p_t W_t = \vartheta _t P_t + \int _0^{\tau ^*} \bigl (\vartheta _t(\tau ) P_t(\tau ) + \vartheta ^I_t(\tau ) P_{It}(\tau )\bigr ) d\tau + \sum _{j=1}^J \vartheta ^{j}_t S^{j}_t. \end{aligned}$$

Then, given $c_t$, the self-financing portfolio $(\vartheta , (\vartheta (\tau )), (\vartheta ^I(\tau )), (\vartheta ^{j}))$ satisfies

$$\begin{aligned} \begin{aligned} \frac{d(p_tW_t)}{p_tW_t}&= \frac{1}{p_tW_t} \Biggl \{ \vartheta _t dP_t + \int _0^{\tau ^*} \bigl ( \vartheta _t(\tau )dP_t(\tau ) + \vartheta ^I_t(\tau )dP_{It}(\tau ) \bigr )d\tau \\&\quad + \sum _{j=1}^J \vartheta ^{j}_t \Bigl ( dS^{j}_t + D^{j}_t dt \Bigr ) - p_t c_t dt \Biggr \} \\&= \frac{\vartheta _t P_t}{p_tW_t} \frac{dP_t}{P_t} + \int _0^{\tau ^*} \left( \frac{\vartheta _t(\tau )P_t(\tau )}{p_tW_t} \frac{dP_t(\tau )}{P_t(\tau )} + \frac{\vartheta ^I_t(\tau )P_{It}(\tau )}{p_tW_t} \frac{dP_{It}(\tau )}{P_{It}(\tau )} \right) d\tau \\&\quad + \sum _{j=1}^J \frac{\vartheta ^{j}_t S^{j}_t}{p_tW_t} \frac{dS^{j}_t + D^{j}_t dt}{S^{j}_t} - \frac{c_t}{W_t} dt \\&= \Biggl ( 1 - \int _0^{\tau ^*}\bigl ( \varphi _t(\tau ) + \varphi ^I_t(\tau ) \bigr )d\tau - \sum _{j=1}^J \varPhi ^j_t \Biggr ) \frac{dP_t}{P_t}\\&\quad + \int _0^{\tau ^*} \left( \varphi _t(\tau ) \frac{dP_t(\tau )}{P_t(\tau )} + \varphi ^I_{t}(\tau ) \frac{dP_{It}(\tau )}{P_{It}(\tau )} \right) d\tau \\&\quad + \sum _{j=1}^J \varPhi ^j_t \frac{dS^{j}_t + D^{j}_t dt}{S^{j}_t} - \frac{c_t}{W_t} dt. \end{aligned} \end{aligned}$$

Substituting Eqs. (32)–(36) into the equation above yields

$$\begin{aligned} \frac{d(p_tW_t)}{p_tW_t} = \left( r_t + (\varPsi _t + \varLambda _{It})'\varLambda _t - \frac{c_t}{W_t} \right) \,dt + (\varPsi _t + \varLambda _{It})'\,dB_t. \end{aligned}$$

(110)

Noting that

$$\begin{aligned} \frac{d(p_tW_t)}{p_tW_t} = \frac{dW_t}{W_t} + \frac{dp_t}{p_t} + \frac{dW_t}{W_t} \frac{dp_t}{p_t}, \end{aligned}$$

and that the volatility of $W_t$ is equal to $\varPsi _t$, we get

$$\begin{aligned} \frac{dW_t}{W_t} = \frac{d(p_tW_t)}{p_tW_t} - i_t dt - \varLambda '_{It} dB_t - \varPsi '_t \varLambda _{It}\,dt. \end{aligned}$$

By inserting Eq. (110) into the equation above, we obtain Eq. (37).

Proof of Proposition 1

First, using Eq. (41), optimal consumption control (51) is obtained as follows:

$$\begin{aligned} c^*_t = \alpha ^{\frac{1}{\gamma }} e^{-\frac{\beta }{\gamma } t} J^{-\frac{1}{\gamma }}_W = \alpha ^{\frac{1}{\gamma }} e^{-\frac{\beta }{\gamma }t} \left\{ e^{-\beta t} (W^*_t)^{-\gamma } G^{\gamma } \right\} ^{-\frac{1}{\gamma }} = \alpha ^{\frac{1}{\gamma }} \frac{W^*_t}{G}. \end{aligned}$$

Then, by inserting $c^*_t$ into budget constraint (37) and by solving the SDE, we obtain Eq. (52).

Second, the derivatives of J are given by

$$\begin{aligned} \begin{aligned} J_t&= -\beta J + \gamma J \frac{G_t}{G}, \quad \quad W J_W = (1-\gamma ) J, \quad \quad J_X=\gamma \,J \frac{G_X}{G},\\ W^2 J_{WW}&= -\gamma (1-\gamma )J,\quad \quad W J_{XW} = \gamma (1-\gamma ) J \frac{G_X}{G},\\ J_{XX}&= \gamma \,J \left\{ (\gamma -1) \frac{G_X}{G} \frac{G'_X}{G} + \frac{G_{XX}}{G} \right\} . \end{aligned} \end{aligned}$$

Then, the numerator and the denominator on the right-hand side of Eq. (42) are rewritten as

$$\begin{aligned}&\pi _t = (\gamma -1) J\left( {\bar{\varLambda }}_t + \gamma \frac{G_X}{G} \right) , \end{aligned}$$

(111)

$$\begin{aligned}&W^2_t J_{WW} = \gamma (\gamma -1) J. \end{aligned}$$

(112)

Therefore, by inserting Eqs. (111) and (112) into Eq. (42), we obtain equation (53). The second and third terms in PDE (49) are calculated from Eqs. (111) and (112) as

$$\begin{aligned}&\frac{1}{2} \,\mathrm {tr}\left[ J_{XX} \right] - \frac{\pi '_t \pi _t}{2W^2_t J_{WW}} \nonumber \\&\quad = \frac{\gamma }{2} J\,\mathrm {tr}\left[ \left\{ (\gamma -1) \frac{G_X}{G} \frac{G'_X}{G} + \frac{G_{XX}}{G} \right\} \right] - \frac{\gamma -1}{2\gamma } J \left( {\bar{\varLambda }}_{t} + \gamma \frac{G_X}{G} \right) ' \left( {\bar{\varLambda }}_{t} + \gamma \frac{G_X}{G} \right) \nonumber \\&\quad = J \left\{ \frac{\gamma }{2} \,\mathrm {tr} \left[ \frac{G_{XX}}{G} \right] - \frac{\gamma -1}{2\gamma } {\bar{\varLambda }}'_t {\bar{\varLambda }}_t - (\gamma -1) {\bar{\varLambda }}'_t \frac{G_X}{G} \right\} . \end{aligned}$$

(113)

The sixth term in PDE (49) is calculated from Eq. (51) as

$$\begin{aligned} \frac{\gamma }{1-\gamma } c^*_t J_W = \frac{\gamma }{1-\gamma } \alpha ^{\frac{1}{\gamma }} \frac{W^*_t}{G} (1-\gamma ) \frac{J}{G} = \gamma \alpha ^{\frac{1}{\gamma }} \frac{J}{G}. \end{aligned}$$

(114)

Substituting Eqs. (113) and (114) into Eq. (49) and dividing by $\gamma J/G$ yield Eq. (54).

Proof of Proposition 2

By substituting Eqs. (58) and (63) into Eqs. (51) and (53), respectively, we obtain Eqs. (66) and (67). It is straightforward to see that $a_0(\tau )$ and $a(\tau )$ are expressed as Eqs. (68) and (69), respectively.

Following Theorem 5.2 in Arimoto [3], we prove that $A(\tau )$ is a unique symmetric solution to matrix differential Riccati equation (70). We consider the following initial value problem of the linear ODE for the $N \times N$ matrix-value functions $C_1(\tau )$ and $C_2(\tau )$:

$$\begin{aligned} \frac{d}{d\tau } \begin{pmatrix}C_1(\tau ) \\ C_2(\tau ) \end{pmatrix}= \begin{pmatrix}L &{} - I_N \\ -Q &{} -L' \end{pmatrix}\begin{pmatrix}C_1(\tau ) \\ C_2(\tau ) \end{pmatrix}, \quad \quad \quad \begin{pmatrix}C_1(0) \\ C_2(0) \end{pmatrix}= \begin{pmatrix}I_N \\ 0_N \end{pmatrix}, \end{aligned}$$

(115)

where

$$\begin{aligned} L = \mathscr {K} + \frac{\gamma -1}{\gamma } {\bar{\varLambda }}, \quad \quad Q = \frac{\gamma -1}{\gamma } \left( \frac{1}{\gamma } {\bar{\varLambda }}' {\bar{\varLambda }} + \bar{\mathscr {R}} \right) . \end{aligned}$$

A solution to Eq.(115) is given by Eq. (71). As we can prove that $C_1(\tau )$ is regular,¹³ we define $A(\tau )$ by Eq. (70). Subsequently, considering that

$$\begin{aligned} \frac{d}{d\tau } C_1^{-1}(\tau ) = - C_1^{-1}(\tau ) \left\{ \frac{d}{d\tau } C_1(\tau ) \right\} C_1^{-1}(\tau ), \end{aligned}$$

(116)

we can derive

$$\begin{aligned} \frac{d}{d\tau } A(\tau )= & {} \left\{ \frac{d}{d\tau } C_2(\tau ) \right\} C_1^{-1}(\tau ) + C_2(\tau ) \frac{d}{d\tau } C_1^{-1}(\tau ) \\= & {} \left( -Q C_1(\tau ) - L' C_2(\tau ) \right) C^{-1}_1(\tau ) - A(\tau ) \left( L C_1(\tau ) - C_2(\tau ) \right) C^{-1}_1(\tau ) \\= & {} A(\tau )^2 - L' A(\tau ) - A(\tau ) L - Q, \end{aligned}$$

and thus confirm that $A(\tau )$ satisfies matrix differential Riccati equation (62). For the uniqueness of solution to the Riccati equation, see the proof of Theorem 5.2 in Arimoto [3]. Finally, for the symmetry of $A(\tau )$, taking the transposition of Riccati equation (62) for $A (\tau )$ yields the same equation for $A(\tau )'$, which implies that $A(\tau )' = A(\tau )$ due to the uniqueness of the solution to the Riccati equation.

Proof of Lemma 3

First, Eq. (1) can be transformed as follows.

$$\begin{aligned} d \bigl (e^{t\mathscr {K}} X_t\bigr ) = e^{t\mathscr {K}}\, dB_t. \end{aligned}$$

(117)

Integrating the above equation over the interval $[n \varDelta , (n + 1)\varDelta ]$, we obtain the following equation.

$$\begin{aligned} e^{(n+1)\varDelta \mathscr {K}} X_{(n+1)\varDelta } - e^{n \varDelta \mathscr {K}} X_{n \varDelta } = \int _{n \varDelta }^{(n+1)\varDelta } e^{s \mathscr {K}}\,dB_s. \end{aligned}$$

(118)

Dividing both sides of the above equation by $e^{(n+1)\varDelta \mathscr {K}}$, we get Eq.(74). By definition of yield-to-maturity, the following holds:

$$\begin{aligned} \begin{aligned} s_t(\tau )&= - \frac{1}{\tau } \log P_t(\tau ), \\ s^I_t(\tau )&= - \frac{1}{\tau } \log P_{It}(\tau ). \end{aligned} \end{aligned}$$

(119)

Thus, Eq.(75) follows from Eqs.(20), (24), (28), and (119).