<p>When integrals look different than what pops into your head, often the reason is <a href="https://en.wikipedia.org/wiki/Integration_by_parts" rel="noreferrer">integration by parts</a>. For your example note that
<span class="math-container">$$\int_R^1 (\theta -R) g(\theta) d \theta + \int_R^1 G(\theta) d \theta = (1-R) - 0,$$</span>
where the right-hand side is equivalent to <span class="math-container">$\int^1_R 1 d\theta$</span>.
Hence, the two expressions you consider are equivalent.</p>
<p>It's of the form
<span class="math-container">$$\int u(x) v'(x) dx + \int u'(x) v(x) dx = \int u(x) v(x) dx.$$</span></p>

CC BY-SA 4.0
2021-01-18T15:44:41.640Z
2021-01-18T15:44:41.640Z
Bayesian
<p>In period 1 the consumer of type <span class="math-container">$\theta$</span> selects an option contract consisting of an up-front fee, <span class="math-container">$B&gt;0$</span>, and exercise price, <span class="math-container">$\bar{R}$</span>. The consumer pays <span class="math-container">$B$</span> at the end of the first period. In period 2, he realises his valuation, <span class="math-container">$\theta$</span>, distributed on <span class="math-container">$[0,1]$</span> by CDF <span class="math-container">$G(\theta)$</span> with density <span class="math-container">$g&gt;0$</span>. His expected payoff, from choosing contract <span class="math-container">$(B, \bar{R})$</span>, should thus be <span class="math-container">$-B+\int_{\bar{R}}^{1}(\theta-\bar{R})g(\theta)d(\theta).$</span> However, in the paper I am told that it is instead: <span class="math-container">$-B+\int_{\bar{R}}^{1}(1-G(\theta))d\theta.$</span> Are these two expressions equivalent? What does it mean to integrate a CDF in this fashion? Thank you.</p>

<microeconomics><game-theory><mechanism-design><contract-theory>
Payoff from an option contract
Charles
# Payoff from an option contract

(Asked by: username_0 on 2021-01-18)

In period 1 the consumer of type $\theta$ selects an option contract consisting of an up-front fee, $B>0$, and exercise price, $\bar{R}$. The consumer pays $B$ at the end of the first period. In period 2, he realises his valuation, $\theta$, distributed on $[0,1]$ by CDF $G(\theta)$ with density $g>0$. His expected payoff, from choosing contract $(B, \bar{R})$, should thus be $-B+\int\_{\bar{R}}^{1}(\theta-\bar{R})g(\theta)d(\theta).$ However, in the paper I am told that it is instead: $-B+\int\_{\bar{R}}^{1}(1-G(\theta))d\theta.$ Are these two expressions equivalent? What does it mean to integrate a CDF in this fashion? Thank you.

**Tags:** microeconomics, game-theory, mechanism-design, contract-theory


---

## 1 answers

### Answer by username_1 on 2021-01-18. Score: 5

When integrals look different than what pops into your head, often the reason is [integration by parts](https://en.wikipedia.org/wiki/Integration_by_parts). For your example note that
$$\int\_R^1 (\theta -R) g(\theta) d \theta + \int\_R^1 G(\theta) d \theta = (1-R) - 0,$$
where the right-hand side is equivalent to $\int^1\_R 1 d\theta$.
Hence, the two expressions you consider are equivalent.

It's of the form
$$\int u(x) v'(x) dx + \int u'(x) v(x) dx = \int u(x) v(x) dx.$$


---

