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:

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