Changes

Recall that <math>M=e(p,u)\,</math>,

so that <math>v(e(p,u),p)=u\,</math> when the expenditure function is evaluated at <math>p\,</math> and

<math>u\,</math>.

<center><math>\mathbb{E}(C) = K \mathbb{E}(p_{min}^{(n)}) + cn\,</math> </center>

<center>where <math>\mathbb{E}(p_{min}^{(n)}) = \mathbb{E}(min\{p_1,p_2,\ldots,p_n\}) \,</math> </center>

The distribution of the lowest <math>n\,</math> draws is: <center><math>F_{min}^{(n)}(p) = 1 - (1-F(p))^n\,</math> </center>

<center><math>\therefore \mathbb{E}(C) = K \int_{\underline{p}}^{\overline{p}} p \; dF_{min}^{(n)}(p) + cn\,</math></center>

<center><math>\therefore \mathbb{E}(C) = K \left [ \underline{p} + \int_{\underline{p}}^{\overline{p}} (1-F(p))^n \; dp \right ] + cn\,</math></center>

To see this, first recall that <center><math>\mathbb{E}(X) = \int_{\underline{x}}^{\overline{x}} x f(x) dx \,</math></center>

Then use the [http://en.wikipedia.org/wiki/Integration_by_parts integration by parts] formula <center><math>\int u\, \frac{dv}{dx}\; dx=uv-\int v\, \frac{du}{dx} \; dx\!</math></center>