Optimal Consumption Under Deterministic Income

Abstract

We consider an individual or household endowed with an initial wealth, having an income and consuming goods and services. The wealth development rate is assumed to be a deterministic continuous function of time. The objective is to maximize the discounted consumption over a finite time horizon. Via the Hamilton–Jacobi–Bellman approach, we prove the existence and the uniqueness of the solution to the considered problem in the viscosity sense. Furthermore, we derive an algorithm for explicit calculation of the value function and optimal strategy. It turns out that the value function is in general not continuous. The method is illustrated by two examples.

Appendix

1.1 HJB Equation—Viscosity Solution

Proof of Theorem 3.1

The proof of Theorem 3.1 consists of two steps: the super- and subsolution proofs. The following lemma presents the subsolution direction; the supersolution proof does not include subtleties and follows standard techniques, see Mnif and Sulem [39]. □

Lemma A.1

The function V(t,x) given by (1) is a viscosity subsolution to (13).

Proof

We need to show, that for every $\bar{\psi}\in C^{(1,1)}[0,T]\times[0,\infty[$:

at every $(\bar{t},\bar{x})\in\,]0,T[\times]0,\infty[$ which is a (strict) maximizer of $V^{*}-\bar{\psi}$ on [0,T]×[0,∞[ with $V^{*}(\bar {t},\bar{x})=\bar{\psi}(\bar{t},\bar{x})$.

As usual, the subsolution proof is done by way of contradiction. Suppose there are some $(\bar{t},\bar{x})$ and $\bar{\psi}$ with the properties stated before but with

(14)

for some ξ>0. Consider the function $\psi(t,x)=\bar{\psi}(t,x)+\frac{(x-\bar{x})^{2}+(t-\bar {t})^{2}}{\bar{t}^{2}+\bar{x}^{2}}\xi$.

Then it holds that $V^{*}(\bar{t},\bar{x})=\bar{\psi}(\bar{t},\bar {x})=\psi(\bar{t},\bar{x}),\bar{\psi}_{x}(\bar{t},\bar{x})=\psi _{x}(\bar{t},\bar{x})$ and $\bar{\psi}_{t}(\bar{t},\bar{x})=\psi _{t}(\bar{t},\bar{x})$, which gives

Since ψ(t,x) is continuously differentiable in both t and x and μ _t is continuous, there is $\delta\in\,]0,\frac{\sqrt{\bar{t}^{2}+\bar{x}^{2}}}{2}[$ such that

for $(t,x)\in B_{\delta}(\bar{t},\bar{x})$. We obtain $V^{*}(t,x)\leq\bar{\psi}(t,x)=\psi(t,x)-\frac{\delta^{2}}{\bar {t}^{2}+\bar{x}^{2}}\xi$ for $(t,x)\in\partial B_{\delta}(\bar{t},\bar {x})$.

Let now $\varepsilon=\frac{1}{2}\frac{\delta^{2}}{\bar{t}^{2}+\bar {x}^{2}}\xi$, then on $B_{\delta}(\bar{t},\bar{x})$ we have

(15)

while for $(t,x)\notin B_{\delta}(\bar{t},\bar{x})$ we have V ^∗(t,x)≤ψ(t,x)−2ε.

Now let $(t_{n},x_{n})\to(\bar{t},\bar{x})$ such that $V(t_{n},x_{n})\to V^{*}(\bar{t},\bar{x})$ and assume (w.l.o.g.) that $(t_{n},x_{n})\in B_{\delta}(\bar{t},\bar{x})$ for all $n\in\mathbb{N}$.

Let $C^{n}\in\mathcal{C}(t_{n},x_{n})$, $X^{C^{n}}$ be the corresponding wealth starting in (t _n ,x _n ) and $\tau^{*}=\tau_{n}\wedge\bar{T}$ (with some $\bar{T}$ such that $t_{n}<\bar{T}\leq T$ for all n) where

At first we observe that $X^{C^{n}}$ can only have downward jumps in the x direction and that V(t,x) is increasing in x. Because of continuity of μ _t , jumps in the wealth process are due to jumps in the consumption process, and we have $X_{s}^{C^{n}}-X_{s-}^{C^{n}}=-\Delta C^{n}_{s}$.

Suppose τ ^∗=τ _n , i.e. stopping because of leaving $B_{\delta }(\bar{t},\bar{x})$. Then either we hit the boundary continuously or leave the ball due to a jump at time τ ^∗ in which case $X_{\tau ^{*}-}\in B_{\delta}(\bar{t},\bar{x})$. From the above estimates we get $V(\tau^{*},X_{\tau^{*}-}^{C_{n}})\leq\psi (\tau^{*},X_{\tau^{*}-}^{C_{n}})-2\varepsilon I_{\{X_{\tau^{*}-}=X_{\tau ^{*}}\}}$.

If $\tau^{*}=\bar{T}$ then $X_{\tau^{*}}^{C_{n}}$, as well as $X_{\tau ^{*}-}^{C_{n}}$, is still inside the ball and we have $V(\tau^{*},X_{\tau^{*}-}^{C_{n}})\leq\psi(\tau^{*},X_{\tau^{*}-}^{C_{n}})$. In total, we arrive at

In the above formula C ^n,c denotes the continuous part of strategy C ⁿ.

If $X_{s}^{C_{n}}\neq X_{s-}^{C^{n}}$, we have that $X_{s}^{C^{n}}- X_{s-}^{C^{n}}=-\Delta C_{s}^{n}$ and we can write

Combining the last expression with the continuous part of C ⁿ and using e ^−βt≤ψ _x (t,x) on $B_{\delta}(\bar{t},\bar {x})$, we arrive at

Finally, using $\psi_{t}(s,X_{s}^{C^{n}})+\mu_{s} \psi_{x}(s,X_{s}^{C^{n}})\leq -\varepsilon$ from (15) we get

which is the same as

Since also $\psi(t_{n},x_{n})\to V^{*}(\bar{t},\bar{x})$ if n→∞, we can choose n large enough such that $0\leq\psi(t_{n},x_{n})-V(t_{n},x_{n})\leq\frac{(\tau^{*}-t_{n}) \varepsilon +2\varepsilon I_{\{\tau_{n}=\tau^{*}\wedge X_{\tau^{*}-}=X_{\tau^{*}}\}}}{2}$. We get

(16)

Now, before we can state that (16) is a contradiction to (12), we have to discuss the case τ ^∗=τ _n =t _n , where an immediate lump-sum consumption leads to

$$x_n=X_{\tau^*-}^{C^n}>X_{\tau^*}^{C^n} \notin B_\delta(\bar{t},\bar {x}). $$

We notice that property (P6) and the construction of the value function show that in this case V is continuously differentiable around the point $(\bar{t},\bar{x})$ and $V_{x}(\bar{t},\bar{x})=e^{-\beta \bar{t}}$. Therefore V ^∗=V around $(\bar{t},\bar{x})$, which furthermore yields that $\psi_{x}(\bar{t},\bar{x})=e^{-\beta \bar{t}}$. Since the test function ψ is continuously differentiable in x, inequality (15) cannot be true and states a contradiction to (14).

Thus we have, when stating (14), that there exists an area around $(\bar{t},\bar{x})$ inside which it is not optimal to consume a lump sum from the wealth. Consequently a strategy C ⁿ, for playing a role in the dynamic programming principle for n large enough such that (t _n ,x _n ) are inside this non-paying area, has the feature that $X_{t}^{C^{n}}$ can leave $B_{\delta}(\bar{t},\bar {x})$ through a jump not before leaving the non-paying area continuously.

Therefore τ ^∗>t _n , which completes the proof and we can conclude that V(t,x) is a viscosity solution to (13). □

Proof of Lemma 3.1

Since V(t,x)≥e ^−βt x for (t,x)∈[0,T]×[0,∞[ (you can always pay out everything and quit by consuming a small constant rate such that X _t+<0), we also have V ^∗(t,x)≥e ^−βt x and V _∗(t,x)≥e ^−βt x. From e ^−βT x=V(T,x)≥V _∗(T,x) we get V _∗(T,x)=e ^−βT x.

Assume that V ^∗(T,x)>e ^−βT x, then there exists η>0 with V ^∗(T,x)≥2η+e ^−βT x.

Now choose a sequence (t _n ,x _n )→(T,x) such that V(t _n ,x _n )→V ^∗(T,x). There is some n ₀>0 such that for n≥n ₀ we have

(17)

Let $C^{n}\in\mathcal{C}(t_{n},x_{n})$ and define $\tau_{n}=\inf\{t\geq t_{n}\mid X_{t}^{C^{n}}<0\}\wedge T$. Since C ⁿ is admissible, we have $\Delta C_{t_{n}}^{n}\leq x_{n}$, there is no lump-sum payment leading to ruin. Furthermore, by the definition of t _n , we have τ _n −t _n →0 if n→∞. Now, fix some ε>0 and choose n large enough such that

Taking a supremum over strategies C ⁿ we get $V(t_{n},x_{n})\leq e^{-\beta t_{n}} x_{n}+\varepsilon$, which contradicts (17) since ε is arbitrary and if (t _n ,x _n )→(T,x) we have $e^{-\beta t_{n}} x_{n}\to e^{-\beta T} x$. □

Proof of Theorem 3.2

Assume there is $(\hat{t},\hat{x})\in[0,T]\times\mathbb{R}_{+}$ such that $u(\hat{t},\hat{x})-v(\hat{t},\hat{x})>0$. We assume, w.l.o.g., $(\hat{t},\hat{x})\in\{(t,x){:}\ t_{j}< t <t_{j-1}, \gamma_{j,i}(t)< x<\gamma_{j,i+1}(t)\}=:A$ with t _j <t _j−1, γ _j,i <γ _j,i+1<∞ defined as in Sect. 2 and in Remark 2.2. Note that because of (P5) we do not need to consider {(t,γ _j,i (t)): t _j <t<t _j−1}∪{(t _j ,x): γ _j,i (t _j )<x<γ _j,i+1(t _j )}. If u(t,γ _j,i (t))−v(t,γ _j,i (t))>0, then there is h>0 with (t,x+h)∈A with u(t,x+h)−v(t,x+h)>0; the same holds for t=t _j .

We further assume that the comparison principle is already shown for the intervals [t _l ,t _l−1[ with l∈{j−1,…,m}, i.e. u(t,x)≤v(t,x) on $[t_{j-1},T]\times\mathbb{R}_{+}$. Note that Lemma 3.1 yields u(x,T)=v(x,T) for all $x\in \mathbb{R}_{+}$.

Define v ^k=kv for k>1. It is easy to check that $k\tilde{v}$ is still a supersolution with lsc envelope kv. Choose k>1 such that $u(\hat{t},\hat{x})-v^{k}(\hat{t},\hat{x})>0$. Due to Lemma 2.1, we obtain the following inequality:

It is clear that u(t,x)−v ^k(t,x)≤0 for $x\ge\frac{\lambda (0)}{k-1}e^{\beta t_{j-1}}=:\eta$. We can choose k in such a way that η≥γ _j,i+1(t) for all t∈[t _j ,t _j−1[ and define

$$A:=\bigl\{(t,x){:}\ t_j< t< t_{j-1},\gamma_{j,i}(t)< x<\gamma_{j,i+1}(t)\bigr\} . $$

Define further

$$M:=\sup_{(t,x)\in A}\bigl\{u(t,x)-v^k(t,x)\bigr\}. $$

From above we know that M<λ(0)<∞ and obtain $0<u(\hat{t},\hat{x})-v^{k}(\hat{t},\hat{x})\le M$. Since u−v ^k is continuous on A, there is (t ^∗,x ^∗)∈A with $u(t^{*},x^{*})-v^{k}(t^{*},x^{*})>\frac{M}{2}>0$. Further, we let H:={(t,x,s,y): (t,x),(s,y)∈A, y−x≥0, t−s≥0} and for ξ>0:

In the following, we mean by u(t,γ _j,i+1) and v ^k(s,γ _j,i+1) the corresponding lsc envelopes u _∗(t,γ _j,i+1) and kv _∗(s,γ _j,i+1), respectively. Then it holds that

Note that $(t,x,s,\gamma_{j,i}(s)), (t,\gamma_{j,i+1}(t),s,y), (t_{j},x,s,y)\in\bar{H}$ only if t=s and x=γ _j,i (s), t=s and y=γ _j,i+1(t) or s=t _j which yields

$$f_\xi\bigl(t,x,s,\gamma_{j,i}(s)\bigr),f_\xi \bigl(t,\gamma_{j,i+1}(t),s,y\bigr), f_\xi(t_j,x,s,y)<0. $$

Let M _ξ =sup _H f _ξ . Since f _ξ is continuous on H, there is $(t_{\xi},x_{\xi}, s_{\xi},y_{\xi})\in\bar{H}$ such that M _ξ =f _ξ (t _ξ ,x _ξ ,s _ξ ,y _ξ ). Since (t ^∗,x ^∗)∈A, it holds that γ _j,i (t ^∗)<x ^∗<γ _j,i+1(t ^∗), from which it follows that

We obtain directly: M _ξ >0 for

In particular, we have $\liminf_{\xi\to\infty}M_{\xi}\ge \frac{M}{2}>0$. With the usual techniques, see for example [3], one can show that there is ξ ₁>0 such that (t _ξ ,x _ξ ,s _ξ ,y _ξ )∈H∖∂H for ξ>max{ξ ₁,ξ ₀}.

Note that by definition of (t _ξ ,x _ξ ,s _ξ ,y _ξ ) it holds that

$$ f_\xi(t_\xi,x_\xi,s_\xi,x_\xi)+f_\xi(t_\xi,y_\xi,s_\xi,y_\xi )\le2f_\xi(t_\xi,x_\xi,s_\xi,y_\xi). $$

(18)

Choose a sequence ξ _n →∞ such that $(t_{\xi_{n}},x_{\xi _{n}},s_{\xi_{n}},y_{\xi_{n}})\to(\bar{t},\bar{x},\bar{s},\bar{y})$.

Let $z_{n}:= (\xi_{n}(t_{\xi_{n}}-s_{\xi_{n}})+1 ) (\xi_{n}(y_{\xi _{n}}-x_{\xi_{n}}+t_{\xi_{n}}-s_{\xi_{n}})+1 )$. Then using (P2) and (18) we obtain

(19)

Here $\lim_{n\to\infty}\xi_{n}(y_{\xi_{n}}-x_{\xi_{n}}+t_{\xi _{n}}-s_{\xi_{n}})=\infty$ implies $\lim_{n\to\infty}\xi _{n}(y_{\xi_{n}}-x_{\xi_{n}})^{3}<0$, which is a contradiction. Thus, we conclude $\lim_{n\to\infty} (t_{\xi_{n}}-s_{\xi_{n}})= \lim_{n\to\infty} (y_{\xi_{n}}-x_{\xi_{n}})=0$, and consequently $\bar{x}=\bar{y}$, $\bar{t}=\bar{s}$. On the other hand, $\lim_{n\to\infty}\xi_{n}(y_{\xi_{n}}-x_{\xi_{n}}+t_{\xi_{n}}-s_{\xi_{n}})\le \frac{2k}{1+k}$, because otherwise we would again obtain $\lim_{n\to\infty}\xi_{n}(y_{\xi_{n}}-x_{\xi_{n}})^{3}<0$. Note that $\bar{x}$ is bounded away from $\gamma_{j,i+1}(\bar{t})$ and $\gamma_{j,i}(\bar{t})$.

Define the functions

These functions are continuously differentiable in t and in x. Furthermore, $u(t,x)e^{\beta s_{\xi_{n}}}-\psi(t,x)$ attains its maximum at $(t_{\xi_{n}},x_{\xi_{n}})$ and $v^{k}(s,y)e^{\beta t_{\xi_{n}}}-\phi(s,y)$ attains its minimum at $(s_{\xi_{n}},y_{\xi _{n}})$. Thus, $\psi(t,x)e^{-\beta s_{\xi_{n}}}$ and $\phi(s,y)e^{-\beta t_{\xi_{n}}}$ are test functions for u(t,x) and v ^k(s,y), respectively. Due to the arguments after relation (19), we have

$$ \lim_{n\to\infty}\psi_x(t_{\xi_n},x_{\xi_n})= \lim_{n\to\infty}\phi_y(s_{\xi_n},y _{\xi_n})\ge\frac{k+1}{2}>1. $$

Thus, there is $N\in\mathbb{N}$ such that for n>N it holds that

Therefore, it holds by Definition 3.2 of viscosity sub- and super-solutions that:

Subtracting the above inequalities, rearranging the terms and letting ξ _n →∞ yields the following relation:

On the other hand, we know that

which is a contradiction. □