# Can't Derive Expectation in Second Price Auction

I'm working on a SPA where we solve:

$\max_{\beta \ge 0} Pr\{Winning\}[v - \mathbb{E}(b^{} \ | \ b^{} \le \beta)]$

We assume all beliefs on bids are independently and identically distributed on $[0,1]$ given by CDF $F(b)$.

$b^{}$ 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:

$\mathbb{E}(b^{} \ | \ b^{} \le \beta) = \frac{1}{Pr(b^{} \le beta)} \int_0^{\beta} bdF(b)^N$

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.

• Not sure I understand you right, but I think you may be confusing conditional probability with conditional expectation?
– BCLC
Jul 16, 2015 at 18:42
• Also I think dF(x) where F is a cdf just means f(x)dx. Perhaps $dF(b)^N$ means $Nf(b)^{N-1}db$?
– BCLC
Jul 16, 2015 at 18:44
• @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 $F(b)^{N-1}$? Jul 16, 2015 at 19:01

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:

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

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

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