<p>The value <span class="math-container">$e^{-\rho t}$</span> is the discount rate that you need to apply if the per period interest is <span class="math-container">$\rho$</span>, but interest is accumulated continuously. The value of <span class="math-container">$\rho$</span> can be interpreted as the per period interest rate, so it usually takes values in the interval <span class="math-container">$(0,1)$</span>.</p>
<p>To see how to get to this value, assume that you invest an amount of <span class="math-container">$a$</span> at an interest rate of <span class="math-container">$\rho$</span> per time period. After <span class="math-container">$t$</span> periods, you will have an amount equal to:
<span class="math-container">$$
a\underbrace{(1+\rho)(1+ \rho)\ldots (1+\rho)}_{t \text{ periods }} = a (1+\rho)^t.
$$</span>
Now assume that instead you accumulate interest not every period but ever (1/2) period. In particular, the bank gives you an interest rate of <span class="math-container">$\rho/2$</span> every (1/2) period. Then after 1 period, the amount of money will be:
<span class="math-container">$$
a \left(1+\frac{\rho}{2}\right)\left(1+ \frac{\rho}{2}\right) = a \left(1+\frac{\rho}{2}\right)^2.
$$</span><br />
So after <span class="math-container">$t$</span> periods, you will have:
<span class="math-container">$$
a \left(1 + \frac{\rho}{2}\right)^{2t}.
$$</span>
We can generalize this. Assume that that every period is split into <span class="math-container">$n$</span> subperiods and you receive an interest of <span class="math-container">$\rho/n$</span> every subperiod, i.e. every <span class="math-container">$(1/n)$</span> period. Then after <span class="math-container">$t$</span> periods, you'll have:
<span class="math-container">$$
a \left(1 + \frac{\rho}{n}\right)^{n t}.
$$</span>
If interest is gathered continuously, this means that we take this value for  <span class="math-container">$n$</span> going to infinity. This gives:
<span class="math-container">$$
\lim_{n \to \infty} a \left(1 + \frac{\rho}{n}\right)^{n t} = a e^{\rho t}.
$$</span>
So if the per period interest rate is <span class="math-container">$\rho$</span> but interest is accumulated continuously, then you will receive an amount of <span class="math-container">$a e^{\rho t}$</span> after <span class="math-container">$t$</span> periods.</p>
<p>In order to have accumulated an amount of, say <span class="math-container">$A$</span> after <span class="math-container">$t$</span> periods (i.e. <span class="math-container">$A = a e^{\rho t}$</span>), you need to invest an amount equal to:
<span class="math-container">$$
a = A e^{-\rho t},
$$</span>
This means that <span class="math-container">$e^{-\rho t}$</span> is the period <span class="math-container">$t$</span> discount rate that you need to apply if interest is accumulated continuously.</p>

CC BY-SA 4.0
2024-02-03T10:32:05.253Z
Just to add that in real world calculations, the internet rate can be negative as was the case for a long time in Europe and Japan. In this case, the discount factor is actually >1.
AKdemy
CC BY-SA 4.0
2024-02-05T10:14:36.010Z
Thank you for your clear explanation!   
I understand that the discount rate is the inverse of the interest rate, and $e^{\rho t}$ represents the interest accumulated over t periods in continuous time.

@AKdemy also thank you for your insightful comment!
Yoji Yamamoto
CC BY-SA 4.0
2024-02-03T09:00:24.350Z
2024-02-03T09:00:24.350Z
tdm
<p>In game theory at the undergraduate level, the discount factor is denoted as <span class="math-container">$\delta \in [0,1]$</span>.<br />
On the other hand, in macroeconomics at the graduate level, the notation <span class="math-container">$e^{-\rho t}$</span> is used(<span class="math-container">$t$</span> is time).<br />
I am unclear as to why the second notation is introduced.</p>
<p>Additionally, I would like to confirm if the range of <span class="math-container">$\rho$</span> is <span class="math-container">$\rho \in [0, \infty]$</span>.</p>
<p>Thank you for any assistance you can provide.</p>

CC BY-SA 4.0
2024-02-03T07:35:15.350Z
I heard that when I consider $p' = e^\rho -1$,
I can derive $e^{-\rho t} = (e^\rho)^{-t} = (\rho'+1)^{-t} = \frac{1}{1+\rho'}^t \equiv \beta^t$.
But it's still unclear what makes this attractive.
Yoji Yamamoto
<macroeconomics><game-theory><discount-rate>
Differences of two discount factors
Yoji Yamamoto
# Differences of two discount factors

(Asked by: username_0 on 2024-02-03)

In game theory at the undergraduate level, the discount factor is denoted as $\delta \in [0,1]$.  
On the other hand, in macroeconomics at the graduate level, the notation $e^{-\rho t}$ is used($t$ is time).  
I am unclear as to why the second notation is introduced.

Additionally, I would like to confirm if the range of $\rho$ is $\rho \in [0, \infty]$.

Thank you for any assistance you can provide.

**Tags:** macroeconomics, game-theory, discount-rate

> **Comments:**
> **username_0**: I heard that when I consider $p' = e^\rho -1$, I can derive $e^{-\rho t} = (e^\rho)^{-t} = (\rho'+1)^{-t} = \frac{1}{1+\rho'}^t \equiv \beta^t$. But it's still unclear what makes this attractive.  

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## 1 answers

### Answer by username_1 on 2024-02-03. Score: 3 | Accepted answer

The value $e^{-\rho t}$ is the discount rate that you need to apply if the per period interest is $\rho$, but interest is accumulated continuously. The value of $\rho$ can be interpreted as the per period interest rate, so it usually takes values in the interval $(0,1)$.

To see how to get to this value, assume that you invest an amount of $a$ at an interest rate of $\rho$ per time period. After $t$ periods, you will have an amount equal to:
$$
a\underbrace{(1+\rho)(1+ \rho)\ldots (1+\rho)}\_{t \text{ periods }} = a (1+\rho)^t.
$$
Now assume that instead you accumulate interest not every period but ever (1/2) period. In particular, the bank gives you an interest rate of $\rho/2$ every (1/2) period. Then after 1 period, the amount of money will be:
$$
a \left(1+\frac{\rho}{2}\right)\left(1+ \frac{\rho}{2}\right) = a \left(1+\frac{\rho}{2}\right)^2.
$$  
So after $t$ periods, you will have:
$$
a \left(1 + \frac{\rho}{2}\right)^{2t}.
$$
We can generalize this. Assume that that every period is split into $n$ subperiods and you receive an interest of $\rho/n$ every subperiod, i.e. every $(1/n)$ period. Then after $t$ periods, you'll have:
$$
a \left(1 + \frac{\rho}{n}\right)^{n t}.
$$
If interest is gathered continuously, this means that we take this value for $n$ going to infinity. This gives:
$$
\lim\_{n \to \infty} a \left(1 + \frac{\rho}{n}\right)^{n t} = a e^{\rho t}.
$$
So if the per period interest rate is $\rho$ but interest is accumulated continuously, then you will receive an amount of $a e^{\rho t}$ after $t$ periods.

In order to have accumulated an amount of, say $A$ after $t$ periods (i.e. $A = a e^{\rho t}$), you need to invest an amount equal to:
$$
a = A e^{-\rho t},
$$
This means that $e^{-\rho t}$ is the period $t$ discount rate that you need to apply if interest is accumulated continuously.

> **Comments:**
> **username_2**: Just to add that in real world calculations, the internet rate can be negative as was the case for a long time in Europe and Japan. In this case, the discount factor is actually >1.  
> **username_0**: Thank you for your clear explanation!    I understand that the discount rate is the inverse of the interest rate, and $e^{\rho t}$ represents the interest accumulated over t periods in continuous time.  @username_2 also thank you for your insightful comment!  

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