2
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

The CDF satisfies $G(0) = 0$, $G(1) = 1$ and $G'(\theta) = g(\theta)$.

Integrating by parts, with $u = 1- G(\theta)$ and $dv = d\theta$, we have $du = -G'(\theta)\,d\theta = - g(\theta)\,d\theta$, $v = \theta$, and

$$\int_{\bar{R}}^1(1- G(\theta)) \, d\theta = \left.(1- G(\theta))\theta\right|_{\bar{R}}^1 + \int_{\bar{R}}^1 \theta\,g(\theta) \, d\theta = -(1- G(\bar{R}))\bar{R} + \int_{\bar{R}}^1 \theta\, g(\theta) \, d\theta\\=-\bar{R} \int_{\bar{R}}^1 g(\theta)\, d\theta+ \int_{\bar{R}}^1 \theta\, g(\theta) \, d\theta = \int_{\bar{R}}^1 (\theta-\bar{R}) \, g(\theta) \, d\theta$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .