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

<p>The (possibly time-varying) <strong>discount factor</strong> is defined as 
<span class="math-container">\begin{equation}
\delta(t)=\frac{g(t)}{g(t-1)}.
\end{equation}</span></p>

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

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

CC BY-SA 4.0
2019-06-13T04:32:14.297Z
Hey Herr, long time no see. Hope all is well, and thanks for the response!
Frank Swanton
CC BY-SA 4.0
2019-06-13T04:43:00.723Z
@FrankSwanton: Glad to help 
Herr K.
CC BY-SA 4.0
2019-06-13T11:39:01.363Z
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
CC BY-SA 4.0
2019-06-13T16:27:40.703Z
@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.
CC BY-SA 4.0
2019-06-12T07:28:14.043Z
2019-06-12T07:28:14.043Z
Herr K.
<p>Consider an exponential discount factor <span class="math-container">$\delta\in(0,1)$</span>.</p>

<p>Similarly, consider an exponential discount <span class="math-container">$\textit{function}$</span>: <span class="math-container">$g(t)=\delta^t$</span>.</p>

<p>Then, is defining the discount <span class="math-container">$\textit{rate}$</span> as below a correct way of defining one? My intuition is that the discount rate represents the rate at which the discount function decays:</p>

<p><span class="math-container">$\rho=\frac{\frac{-dg(t)}{dt}}{g(t)}.$</span></p>

<macroeconomics><microeconomics><discount-rate>
discount factor, function, and rate
Frank Swanton
# discount factor, function, and rate

(Asked by: username_0 on 2019-06-12)

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:

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

**Tags:** macroeconomics, microeconomics, discount-rate

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## 1 answers

### Answer by username_1 on 2019-06-12. Score: 4 | Accepted 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 $\lim\_{n\to\infty}(1+\frac{\rho}{n})^{-nt}=\mathrm e^{-\rho t}$.

> **Comments:**
> **username_0**: Hey Herr, long time no see. Hope all is well, and thanks for the response!  
> **username_1**: @FrankSwanton: Glad to help   
> **username_0**: 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?  
> **username_1**: @FrankSwanton: The numerator of the first equation becomes $\lim_{n\to\infty}\frac{g(t)-g(t-1/n)}{1/n}=g'(t)$.  

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