<p>Actually it was easy: they summed the revenue over time.</p>
<p><span class="math-container">$\sum_{i\ge0} \frac1{(1+r)^i} = \frac{1}{1 - \frac{1}{1+r}} = \frac{1+r}r$</span></p>

CC BY-SA 4.0
2022-12-10T17:05:21.117Z
2022-12-10T17:05:21.117Z
Labo
<p>In Shapiro, C. (1983). Premiums for High Quality Products as Returns to Reputations. The Quarterly Journal of Economics, 98(4), 659. doi:10.2307/1881782, they assume the one-period interest rate is <span class="math-container">$r=\exp(iT) - 1$</span>. <span class="math-container">$p(q)$</span> and <span class="math-container">$c(q)$</span> are the price and cost of producing a unit good of quality <span class="math-container">$q$</span>.</p>
<p>They write on page 8:</p>
<blockquote>
<p>The alternative strategy of maintaining quality forever yields
a constant flow of &quot;profits&quot; of <span class="math-container">$p(q) - c(q)$</span>, which has a present value of <span class="math-container">$(p(q) - c(q))(1 + r)/r$</span></p>
</blockquote>
<p>I am very confused as I would think the present value of profits would be <span class="math-container">$\frac{p(q) - c(q)}{1+r}$</span></p>
<p>Can someone explain me why?</p>

<interest-rate>
Present value of profits
Labo
# Present value of profits

(Asked by: username_0 on 2022-12-10)

In Shapiro, C. (1983). Premiums for High Quality Products as Returns to Reputations. The Quarterly Journal of Economics, 98(4), 659. doi:10.2307/1881782, they assume the one-period interest rate is $r=\exp(iT) - 1$. $p(q)$ and $c(q)$ are the price and cost of producing a unit good of quality $q$.

They write on page 8:

> The alternative strategy of maintaining quality forever yields
> a constant flow of "profits" of $p(q) - c(q)$, which has a present value of $(p(q) - c(q))(1 + r)/r$

I am very confused as I would think the present value of profits would be $\frac{p(q) - c(q)}{1+r}$

Can someone explain me why?

**Tags:** interest-rate

---

## 1 answers

### Answer by username_0 on 2022-12-10. Score: 2 | Accepted answer

Actually it was easy: they summed the revenue over time.

$\sum\_{i\ge0} \frac1{(1+r)^i} = \frac{1}{1 - \frac{1}{1+r}} = \frac{1+r}r$


---

