Define $u_t$ as the utility received in period $t$. The discounted utility of a utility flow ${\mathbf{u}=(u_0,u_1,\ldots,u_t,\ldots)}$ is: $$U(\mathbf{u})=\sum_{t=0}^{+\infty}\delta^{t}\cdot u_t$$
The discount factor $\delta$ is an exogenous parameter that reflects the agent's time preferences: the lower $\delta$, the more impatient the agent. Alternatively, $\delta$ can be interpreted as the probability that the agent survives to the next period.
Let $D(t)$ be a function of $t$ such that ${D(0)=1}$ and ${0\lt D(t+1)\lt D(t)}$. We call $D(\cdot)$ the discount factor function, and the corresponding discounted utility is: $$U(\mathbf{u})=\sum_{t=0}^{+\infty}D(t)\cdot u_t$$
Notice that the standard model—also called exponential discounting—is the case in which ${D(t)=\delta^t}$ for ${0\lt\delta\lt 1}$.
In the standard model (exponential utility), the discount function is: $$D(t)=\delta^t = \exp \left(-k\cdot t\right),\quad\text{where } k=-\log(\delta).$$
The $(\beta,\delta)$-model uses a different discount function: $$D(t)=\begin{cases} 1 &\text{if } t=0,\\ \beta\cdot \delta^t &\text{if } t\gt 0.\end{cases}$$
The hyperbolic discounting function is: $$D(t)=\bigl(1+\alpha\cdot t \bigr)^{-\gamma/\alpha}$$ where ${\alpha\gt 0}$ and ${\gamma\gt 0}$. Notice that as ${\alpha\searrow 0}$, the discount factor converges to ${D(t)=\exp(-\gamma\cdot t)=\delta^{t}}$, where ${\delta=\exp(-\gamma)}$. Thus, exponential discounting is a limit case of hyperbolic discounting.
Notice that with hyperbolic discounting or with $(\beta,\delta)$ discounting, individual choices might not be dynamically consistent. That is, as time elapses, an individual might change his/her ranking over two flows of utilities even if during the elapsed time both flows delivered the same utility.
When time preferences might exhibit dynamically inconsistent behavior, we need to specify how individuals will react. There are three possible alternatives:
A $(\beta,\delta)$-individual lives for three periods $t=0,1,2$ and starts with wealth ${W_{0}=100}$. In each period $t$, the individual consumes $c_{t}$ and saves the rest, ${s_{t}=W_{t}-c_{t}}$. Savings are invested at a 100% rate of return, so next period's wealth is ${W_{t+1}=2\cdot s_{t}}$.
Consuming $c_{t}$ yields utility $\ln c_{t}$.
Notice that in the last period $t=2$ there is no point in saving, so ${c_{2}=W_{2}}$ and ${s_{2}=0}$. Thus, we only need to find $c_{0}$ and $c_{1}$, where $c_{0}$ can be any amount between $0$ and $W_{0}$, and $c_{1}$ any amount between $0$ and $W_{1}$.
Earlier selves of the consumer anticipate the decisions of the future selves.
If the naive solution were implemented, the agent would save ${s_{0}=\frac{100\,\beta\delta(1+\delta)}{1+\beta\delta(1+\delta)}}$ in period $0$ and ${s_{1}=\frac{200\,\beta\delta^{2}}{1+\beta\delta(1+\delta)}}$ in period $1$, and the utility would be: $$U_{0}^{\text{ideal}}=\ln\left(\frac{100}{1+\beta\delta(1+\delta)}\right)+\beta\delta\ln\left(\frac{200\,\beta\delta}{1+\beta\delta(1+\delta)}\right)+\beta\delta^{2}\ln\left(\frac{400\,\beta\delta^{2}}{1+\beta\delta(1+\delta)}\right)$$ But in period $t=1$, the agent would not choose according to the naive solution; the agent changes its mind and chooses: $$s_{1}=\frac{\beta+\beta\delta}{1+\beta\delta}\cdot\frac{200\,\beta\delta^{2}}{1+\beta\delta(1+\delta)}$$ So the actual utility of the naive agent ends up being: $$U_{0}^{\text{actual}}=\ln\left(\frac{100}{1+\beta\delta(1+\delta)}\right)+\beta\delta\ln\left(\frac{200\,\beta\delta(1+\delta)}{(1+\beta\delta)\bigl(1+\beta\delta(1+\delta)\bigr)}\right)+\beta\delta^{2}\ln\left(\frac{400\,\beta^{2}\delta^{2}(1+\delta)}{(1+\beta\delta)\bigl(1+\beta\delta(1+\delta)\bigr)}\right)$$ which is lower than the utility of the naive solution: $$U_{0}^{\text{ideal}}-U_{0}^{\text{actual}}=\beta\delta\left[(1+\delta)\ln\left(\frac{1+\beta\delta}{1+\delta}\right)-\delta\ln\beta\right]\gt 0$$
If the agent has a commitment device that forces the naive solution to be chosen in period $t=1$, then, as long as the cost of the commitment device is not too high, the agent will commit to the naive solution.
Suppose you usually go to the movies on Saturdays, and the schedule at the local cinema is Street Fighter this week with utility ${3,}$ Sense and Sensibility next week with utility ${5,}$ Spider-Man: Brand New Day in two weeks with utility ${8,}$ and You Can See Everything in three weeks with utility ${13.}$ Now suppose you must complete a report for work within four weeks, and to do so you must skip the movie on one of the next four Saturdays. When do you complete the report?
Adapted from O’Donoghue & Rabin (1999), p. 109
An individual must complete a task exactly once, and completing it costs $26$. Until the task is completed, the individual bears a loss in every single period: the loss in period $t$ is ${L_{t}=t+1,}$ which grows the longer the task is put off. So if the task is completed in period ${\tau,}$ the individual bears the loss $L_{t}$ in each period ${t=0,1,\ldots,\tau-1,}$ pays $26$ in period ${\tau,}$ and bears nothing afterwards.
Costs are negative utilities: ${u_{t}=-L_{t}}$ for ${t\lt\tau,}$ ${u_{\tau}=-26,}$ and ${u_{t}=0}$ for ${t\gt\tau.}$ Throughout, ${\delta=0.8}$ and ${\beta=0.5.}$
In which period does the individual complete the task?
Seen from period ${t,}$ with the task still pending, the cost of completing it in period ${\tau\ge t}$ is: $$C_{t}(\tau)=\sum_{s=t}^{\tau-1}\delta^{s-t}L_{s}+\delta^{\tau-t}\cdot 26$$ Postponing the task from $\tau$ to $\tau+1$ changes this cost by: $$C_{t}(\tau+1)-C_{t}(\tau)=\delta^{\tau-t}\bigl(L_{\tau}-(1-\delta)\cdot 26\bigr)$$ Postponing adds the loss $L_{\tau}$ but saves ${(1-\delta)\cdot 26=5.2,}$ the gain from paying $26$ one period later. Since $L_{\tau}$ grows over time, the agent completes the task in the first period in which ${L_{\tau}\ge 5.2.}$ This period does not depend on ${t,}$ so the exponential agent never changes its mind.
Since ${L_{4}=5\lt 5.2\lt 6=L_{5},}$ the exponential agent completes the task in period ${\tau=5.}$
For a $(\beta,\delta)$-agent in period ${t,}$ with the task still pending, completing the task now costs ${26,}$ while waiting until period ${\tau\gt t}$ costs: $$C_{t}^{\beta}(\tau)=L_{t}+\beta\Bigl[\sum_{s=t+1}^{\tau-1}\delta^{s-t}L_{s}+\delta^{\tau-t}\cdot 26\Bigr]$$ A sophisticated agent knows that its future selves are also present-biased, so it solves the problem backward. From period $15$ on, every self completes the task, since waiting costs at least ${L_{t}+\beta\delta\cdot 26=(t+1)+10.4\gt 26.}$ Working backward from there, each self waits until the next self that completes the task. For example, self $13$ expects self $15$ to complete it, so waiting costs ${C_{13}^{\beta}(15)=14+0.5\,(0.8\cdot 15+0.64\cdot 26)=28.32\gt 26.}$
| Self | Next self to complete | Cost of waiting | Decision |
|---|---|---|---|
| $14$ | $15$ | $25.4$ | waits |
| $13$ | $15$ | $28.32$ | completes |
| $12$ | $13$ | $23.4$ | waits |
| $11$ | $13$ | $25.52$ | waits |
| $10$ | $13$ | $26.616$ | completes |
| $9$ | $10$ | $20.4$ | waits |
Selves $0$ to $8$ also wait: knowing that self $10$ will complete the task, waiting costs each of them less than $22$. So the sophisticated agent completes the task in period ${\tau=10.}$
A naive agent believes that its future selves will carry out whatever plan the current self makes. Because $\beta$ multiplies all future costs equally, the best plan for later is the exponential agent’s: complete the task in period $5$ if ${t\lt 5,}$ or in period $t+1$ if ${t\ge 5.}$
Before period ${5,}$ waiting is cheaper even for the exponential agent, so the naive agent waits. From period $5$ on, the naive agent completes the task now only if ${26\le L_{t}+\beta\delta\cdot 26,}$ that is, if ${L_{t}\ge(1-\beta\delta)\cdot 26=15.6.}$ Since ${L_{14}=15\lt 15.6\lt 16=L_{15},}$ the naive agent completes the task in period ${\tau=15.}$ In every period from $5$ to ${14,}$ it plans to complete the task next period and then postpones again.
| Agent | Completes the task in period | $C_{0}(\tau)$ |
|---|---|---|
| Exponential | $5$ | $17.14$ |
| Sophisticated | $10$ | $19.74$ |
| Naive | $15$ | $22.40$ |