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.

  • $\begingroup$ Not sure I understand you right, but I think you may be confusing conditional probability with conditional expectation? $\endgroup$
    – BCLC
    Jul 16, 2015 at 18:42
  • $\begingroup$ Also I think dF(x) where F is a cdf just means f(x)dx. Perhaps $dF(b)^N$ means $Nf(b)^{N-1}db$? $\endgroup$
    – BCLC
    Jul 16, 2015 at 18:44
  • 1
    $\begingroup$ @BCLC This is the case when f(x) exists. I get when trying to find the expectation of the second highest bid we only need to integrate from $[0,\beta]$ because it obviously is not higher than the first best bid, $\beta$. My main concern is why we use $F(b)^N$ which is the probability of $b \ge x \ s.t. $ x respresents all bids (there are N players). Shouldn't it instead by $F(b)^{N-1}$? $\endgroup$ Jul 16, 2015 at 19:01

2 Answers 2


There are $N+1$ bidders. So we only care about the second highest bid. That's the same as finding the highest bid amongst the remaining $N$ bidders. The expected second highest bid is given by:

$$b \cdot Pr[b] \cdot Pr[b \geq b_{-i}] = b \cdot f(b) \cdot F(b)^{N-1} = b \cdot dF(b)^{N}$$

As expected value is an average, we divide out by the probability that $b \leq \beta$, which is $F(\beta)$.

Thus, we have:

$$\mathbb{E}[b^{[2]} | b^{[2]} \leq \beta] = \dfrac{1}{F(\beta)}\int_{0}^{\beta} bdF(b)^{N}$$


Given that $b_i$s are independently and identically distributed on $[0,1]$ with CDF $F$ and PDF $f$, and $b^{[2]}$ denotes the second highest bid, we want to find $\mathbb{E}(b^{[2]}|b^{[2]}\leq \beta)$. To do so, we'll first find the distribution of $b^{[2]}$.

$$\Pr(b^{[2]}\leq x) = N[1-F(x)][F(x)]^{N-1} + [F(x)]^{N}= N[F(x)]^{N-1} - (N-1)[F(x)]^{N}$$ where $N$ denotes the number of bidders. So the density of $b^{[2]}$ is $$N(N-1)\left(1 - F(x)\right)[F(x)]^{N-2}f(x)$$ Consequently, the conditional density of $b^{[2]}$ at $x\in [0, \beta]$ given that $b^{[2]}\leq \beta$ is $$\dfrac{N(N-1)\left(1 - F(x)\right)[F(x)]^{N-2}f(x)}{\Pr(b^{[2]}\leq \beta)} $$ Therefore, $$\mathbb{E}(b^{[2]}|b^{[2]}\leq \beta) = \displaystyle\frac{1}{\Pr(b^{[2]}\leq \beta)}\int_0^\beta xN(N-1)\left(1 - F(x)\right)[F(x)]^{N-2}f(x)dx$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.