I'm working on a SPA where we solve:
$\max_{\beta \ge 0} Pr\{Winning\}[v - \mathbb{E}(b^{[2]} \ | \ b^{[2]} \le \beta)]$
We assume all beliefs on bids are independently and identically distributed on $[0,1]$ given by CDF $F(b)$.
$b^{[2]}$ denotes the second highest bid.
I've found $Pr\{Winning\} = Pr\{\beta \ge b_1, \ldots , b_n\} = F(b)^N$
In my lecture notes we are given:
$\mathbb{E}(b^{[2]} \ | \ b^{[2]} \le \beta) = \frac{1}{Pr(b^{[2]} \le beta)} \int_0^{\beta} bdF(b)^N$
I'm assuming this comes from $P(A|B) = \frac{P(A \cap B)}{P(B)} $, but can't quite figure it out. Especially the $F(b)^N$ in the integral.