<p>The <span class="math-container">$r$</span>, or rather <span class="math-container">$\frac{1}{1/r}$</span>, is the &quot;average&quot; part of the average discounted payoff. As you note, <span class="math-container">$\mathrm e^{-rs}$</span> is the discount factor. Observe that
<span class="math-container">\begin{equation}
\int_0^\infty \mathrm e^{rs}\,\mathrm ds = \frac1r.
\end{equation}</span>
If the payoff is <span class="math-container">$p$</span> in every instant of time, the discounted payoff (with no averaging) is <span class="math-container">$p/r$</span>. But if we want to get <span class="math-container">$p$</span> as the <em>average</em> discounted payoff, we would weight the discount factor by <span class="math-container">$\frac{1}{1/r}$</span> so that
<span class="math-container">\begin{equation}
\int_0^\infty \frac{\mathrm e^{rs}}{1/r}p\,\mathrm ds = r\int_0^\infty \mathrm e^{rs}p\,\mathrm ds = p.
\end{equation}</span></p>
<p>In the paper you referred to, the average discounted payoff on page 10 comes from:
<span class="math-container">\begin{equation}
\int_t^\infty (1)\frac{\mathrm e^{-rs}}{1/r}\,\mathrm ds - \int_0^t \frac{\mathrm e^{-rs}}{1/r} c_i(u_{i,s})\,\mathrm ds
\end{equation}</span>
where the discount factor is weighted by <span class="math-container">$\frac{1}{1/r}$</span>.</p>

CC BY-SA 4.0
2022-06-13T18:31:22.377Z
Thankyou very much!
nupro
CC BY-SA 4.0
2022-06-13T18:20:09.433Z
2022-06-13T18:40:36.070Z
2022-06-13T18:40:36.070Z
Herr K.
<p>While reading some theory papers that model a continuous time problem, I have noticed the way they represent the expression for average discounted payoff is of this form, where <span class="math-container">$r&gt; 0 $</span> is the discount rate,<span class="math-container">$$r\int_0^\infty e^{-rs}u\left(x_s\right)ds$$</span> Can someone explain to me where does the <span class="math-container">$r$</span> in this expression come from? I understand the use of <span class="math-container">$e^{-rs}$</span> since it is the discount factor, but can't understand why we also multiply by <span class="math-container">$r$</span></p>
<p>For example see this paper <a href="https://elischolar.library.yale.edu/cgi/viewcontent.cgi?article=3011&amp;context=cowles-discussion-paper-series#page=12" rel="nofollow noreferrer">https://elischolar.library.yale.edu/cgi/viewcontent.cgi?article=3011&amp;context=cowles-discussion-paper-series#page=12</a>. On page 10 (PDF page 12).</p>

CC BY-SA 4.0
2022-06-13T18:32:59.040Z
I changed a few typos (at least I think they were) in your main formula. Feel free to revert back if you think I'm mistaken.
Herr K.
<mathematical-economics><game-theory>
Average Discounted Payoffs in Continuous Time
nupro
# Average Discounted Payoffs in Continuous Time

(Asked by: username_0 on 2022-06-13)

While reading some theory papers that model a continuous time problem, I have noticed the way they represent the expression for average discounted payoff is of this form, where $r> 0 $ is the discount rate,$$r\int\_0^\infty e^{-rs}u\left(x\_s\right)ds$$ Can someone explain to me where does the $r$ in this expression come from? I understand the use of $e^{-rs}$ since it is the discount factor, but can't understand why we also multiply by $r$

For example see this paper <https://elischolar.library.yale.edu/cgi/viewcontent.cgi?article=3011&context=cowles-discussion-paper-series#page=12>. On page 10 (PDF page 12).

**Tags:** mathematical-economics, game-theory

> **Comments:**
> **username_1**: I changed a few typos (at least I think they were) in your main formula. Feel free to revert back if you think I'm mistaken.  

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## 1 answers

### Answer by username_1 on 2022-06-13. Score: 4 | Accepted answer

The $r$, or rather $\frac{1}{1/r}$, is the "average" part of the average discounted payoff. As you note, $\mathrm e^{-rs}$ is the discount factor. Observe that
\begin{equation}
\int\_0^\infty \mathrm e^{rs}\,\mathrm ds = \frac1r.
\end{equation}
If the payoff is $p$ in every instant of time, the discounted payoff (with no averaging) is $p/r$. But if we want to get $p$ as the *average* discounted payoff, we would weight the discount factor by $\frac{1}{1/r}$ so that
\begin{equation}
\int\_0^\infty \frac{\mathrm e^{rs}}{1/r}p\,\mathrm ds = r\int\_0^\infty \mathrm e^{rs}p\,\mathrm ds = p.
\end{equation}

In the paper you referred to, the average discounted payoff on page 10 comes from:
\begin{equation}
\int\_t^\infty (1)\frac{\mathrm e^{-rs}}{1/r}\,\mathrm ds - \int\_0^t \frac{\mathrm e^{-rs}}{1/r} c\_i(u\_{i,s})\,\mathrm ds
\end{equation}
where the discount factor is weighted by $\frac{1}{1/r}$.

> **Comments:**
> **username_0**: Thankyou very much!  

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