Raising the bar Consumption-Savings Problem Archives

Nov 17

Consumption-Savings Problem

Consumption-Savings Problem

Here we look at the solution of

Example 7.5 (mt3042 Optimization Guide by M. Baltovic)

Suppose a consumer, let's call her Ms Thrifty, has an initial wealth of $w$ . In any period $t=1,...,T$ , denote the wealth she has at the start of the period by $w_t$ , so that $w_1=w.$ She can choose to spend $c_t\in[0,w_t]$ , giving her a utility of $u(c_t)=c_t$ . However, the wealth $w_t-c_t$ she does not consume in that period will be stored in a bank account where it earns an interest rate of $r>0$ by the beginning of the next period. Thus, she will enter the next time period $t+1$ with wealth $w_{t+1}=(1+r)(w_t-c_t)$ .

Actions of Ms Thrifty can be identified with her spendings $\{c_1,...,c_T\}$ and the same spendings are her rewards. Thus, a strategy is a sequence of spendings $\{c_1,...,c_T\}$ and the value of that strategy is the sum of rewards $V(\{c_1,...,c_T\})=c_1+...+c_T.$ To obtain the solution, we use backwards induction.

Period $T$ . The reward $c_T\in[0,w_T]$ is obviously maximized at $c_T=w_T$ so that

(1) $c_T=w_T=(1+r)(w_{T-1}-c_{T-1}).$

Period $T-1$ . According to the Bellman equations, at each state $s$ we need to select an action $a$ that maximizes the sum

(2) $r(s,a)+V(remainder).$

I deliberately don't use the full notation because I prefer the following verbal description: $r(s,a)$ is the immediate reward for taking action $a$ at state $s$ (for taking one step leading us to the next state). Whatever is the next state, $remainder$ means an optimal strategy leading us from that state to the final destination. For period $T-1$ we use (1) to write (2) as

(3) $c_{T-1}+c_T=c_{T-1}+(1+r)(w_{T-1}-c_{T-1})=-rc_{T-1}+(1+r)w_{T-1}.$

Since $c_{T-1}\in[0,w_{T-1}],$ (3) is maximized at $c_{T-1}=0$ , giving

(4) $V(\{0,c_{T}\})=(1+r)w_{T-1}$ .

Period $T-2$ . Here (2) is

$c_{T-2}+V(\{0,c_{T}\})$ (using (4))

$=c_{T-2}+(1+r)w_{T-1}$ (using $w_{T-1}=(1+r)(w_{T-2}-c_{T-2})$ )

$=c_{T-2}+(1+r)^{2}(w_{T-2}-c_{T-2})=[ 1-(1+r)^2]c_{T-2}+(1+r)^2w_{T-2}.$

This is maximized at $c_{T-2}=0$ and the value function for three last actions is $V(\{0,0,c_T\})=(1+r)^{2}w_{T-2}$ . This easily generalizes leading to the solution $V(\{0,...,0,c_T\})=(1+r)^{T-1}w_1=(1+r)^{T-1}w.$