# discount factor, function, and rate

Consider an exponential discount factor $$\delta\in(0,1)$$.

Similarly, consider an exponential discount $$\textit{function}$$: $$g(t)=\delta^t$$.

Then, is defining the discount $$\textit{rate}$$ as below a correct way of defining one? My intuition is that the discount rate represents the rate at which the discount function decays:

$$\rho=\frac{\frac{-dg(t)}{dt}}{g(t)}.$$

## 1 Answer

Given a discount function $$g(t)$$, the discount rate is the rate at which the discount function declines over time. If time is discrete, then the (possibly time-varying) discount rate is $$\begin{equation} \rho(t)=-\frac{g(t)-g(t-1)}{g(t)}. \end{equation}$$ If we divide each period into $$n$$ equal intervals and let $$n\to\infty$$, then we get the discount rate in continuous time as $$\begin{equation} \rho(t)=-\frac{g'(t)}{g(t)}, \end{equation}$$ which is the same as your expression.

The (possibly time-varying) discount factor is defined as $$\begin{equation} \delta(t)=\frac{g(t)}{g(t-1)}. \end{equation}$$

With exponential discounting, the discount rate is constant, i.e. $$\rho(t)=\rho$$ for all $$t$$, and the (discrete time) discount function is given by $$\begin{equation} g(t)=\frac{1}{(1+\rho)^t}=\delta^t \end{equation}$$ where $$\delta=\frac{1}{1+\rho}$$ is the constant discount factor.

The continuous time analogs are \begin{align} g(t)&=\mathrm e^{-\rho t}\\ \delta(t)=\delta&=\mathrm e^{-\rho} \end{align} Again, these are obtained by dividing each time period into $$n$$ equal intervals and observing that $$\lim_{n\to\infty}(1+\frac{\rho}{n})^{-nt}=\mathrm e^{-\rho t}$$.

• Hey Herr, long time no see. Hope all is well, and thanks for the response! – Frank Swanton Jun 13 at 4:32
• @FrankSwanton: Glad to help 😃 – Herr K. Jun 13 at 4:43
• Hi Herr, in your second equation where you let $n\rightarrow\infty$ and obtain the discount rate in continuous time, can you provide the detail with the limit sign step by step? – Frank Swanton Jun 13 at 11:39
• @FrankSwanton: The numerator of the first equation becomes $\lim_{n\to\infty}\frac{g(t)-g(t-1/n)}{1/n}=g'(t)$. – Herr K. Jun 13 at 16:27