I am going through this paper https://arxiv.org/pdf/1803.02194.pdf on Bidding Machine. In the section 3.1 Problem definition, I got the part about finding the probability of winning at bidding price $b$ (Equation 1, as it is quite straightforward), but I am kind of stuck on the part where it denotes the expected cost in the second price auction (Equation 2) as:

$$w(b) = \int_{0}^{b} p_z(z) dz$$

$$c(b) = \frac{\int_{0}^{b} z p_z(z) dz}{\int_{0}^{b} p_z(z) dz}$$

where $p_z(z)$ is the PDF of the market price, $w(b)$ is the probability of winning and $c(b)$ denotes the expected cost.

What is the significance of the denominator in the Equation 2 which calculates the expected cost in the second price auction?

Edit: I visited this post

Expected payment in second price seal-bid auction and got some insights about the calculation of the expected value. Shouldn't the probability be as the numerator and outside the integral as the constant term instead of the denominator?

  • $\begingroup$ Your image isn't loading for me. Can you use MathJax to display the equations instead? $\endgroup$
    – Math1000
    Jan 16, 2019 at 19:01
  • $\begingroup$ I have edited the question. $\endgroup$ Jan 16, 2019 at 19:05

1 Answer 1


The quantity $c(b)$ is not just the expected cost but the expected cost of winning with bid $b$. The conditional density $\tilde p_b(z)$ of the winning price given you win with $b$ is

$$\tilde p_b(z) = \frac{p(z)}{\int_0^b p(z) dz}, \ \text{for}\ 0 \leq \ z \leq b. $$

The cost is computed by taking the expectation of $z$ under the truncated distribution.


You must log in to answer this question.

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