Note: These explanations are in the process of being adapted from my textbook.
I'm trying to make them each a "standalone" treatment of a concept, but there may still
be references to the narrative flow of the book that I have yet to remove.
This work is under development and has not yet been professionally edited.
If you catch a typo or error, or just have a suggestion, please submit a note here. Thanks!

# First-Price Sealed-Bid Private-Value Auctions

In a private-value auction each bidder $i$ has some valuation $v_i$ of an object, which they know and nobody else does.

In a first-price sealed-bid auction, each bidder $i$ submits some bid $b_i$ in secret (hence the term “sealed bid”); the highest bidder wins the object and pays their own bid.

Whoever wins the auction gets payoff $v_i - b_i$; everyone else gets a payoff of zero.

Let’s make the following simplifying assumptions:

• there are only two players, player 1 and player 2.
• it is common knowledge that each player’s valuation is drawn from a uniform distribution along the interval $[0,100]$
• player 2’s strategy is to bid some fraction $a$ of their true valuation: $b_2 = av_2$.

What is player 1’s best response to this strategy? Well, they want to choose the bid $b_1$ that maximizes their expected payoff:

$u(b_1 | b_2) = \underbrace{(v_1 - b_1)}_\text{Payoff if win} \times \overbrace{\Pr\{b_2 < b_1\}}^{\text{Prob of winning}}$

To visualize this, let’s first realize that the uniform distribution means that the probability that $v_2$ is less than some number $\hat v$ is given by $\hat v/100$:

If the other player is bidding $b_2 = av_2$, this means that $b_2$ is distributed uniformly along the interval $[0,100a]$. Therefore, for any bid $b_1 < 100a$, the probability that $b_2 < b_1$ is given by $b_1/100a$. Try changing the value of $a$ in the diagram below to see how this affects the probability distribution of bids:

Now suppose player 1’s valuation is $v_1$, and they know player 2 is playing according to $b_2 = av_2$. Then player 1’s expected utility from bidding $b_1$, $u_1(b_1)$, is given by their payoff if they win $(v_1 - b_1)$ times the probability that they win. This is shown by the blue area in the graph below:

Try dragging the valuation $v$ around in the graph above, and finding the utility-maximizing bid. You’ll find that in this case, the bid is always half the valuation v. Why is this? Well, using the fact from above that $$\Pr\{b_2 < b_1\} = \frac{b_1}{100a}$$ We can write player 1’s expected utility from bidding $b$ to be $$u_1(b_1) = (v_1 - b_1) \times \frac{b_1}{100a} = \frac{v_1b_1 - b_1^2}{100a}$$ Taking the derivative of this with respect to $b$ and setting it equal to zero gives us $$u_1^\prime(b_1) = \frac{v_1 - 2b_1}{100a} = 0 \Rightarrow b_1^* = \frac{v_1}{2}$$ Therefore a Bayes Nash Equilibrium of this game is for each player to bid half their valuation.

Copyright (c) Christopher Makler / econgraphs.org