Informal Introduction to Auction Theory

Setup

There is a single indivisible object and \(n\) risk-neutral bidders indexed by \(i\).

Each bidder \(i\) has valuation \(v_i\) drawn i.i.d. from a distribution \(F\) with support \[ v \in [\underline v, \overline v], \] where \(F\) is strictly increasing and admits density \(f\).

The type of bidder \(i\) is \[ t_i := v_i \]

A bidding strategy is a function \[ b : [\underline v, \overline v] \to \mathbb{R} \]

Payoff and Expected Surplus

Consider a first-price auction. The realised surplus of bidder \(i\) is \[ S(v_i) = \begin{cases} v_i - b(v_i), & \text{if bidder $i$ wins},\\ 0, & \text{otherwise}. \end{cases} \]

The expected surplus is \[ S(v_i) = (v_i - b(v_i)) F(v_i)^{n-1} \]

Deviation and Incentive Compatibility

Consider a bidder with true valuation \(v_i\) who mimics a generic type \(\tilde v\). The expected surplus from mimicking is \[ S(v_i;\tilde v) = (v_i - b(\tilde v)) F(\tilde v)^{n-1} \]

Maximising with respect to \(\tilde v\), the first-order condition is \[ \frac{\partial S(v_i;\tilde v)}{\partial \tilde v} = (v_i - b(\tilde v))(n-1)F(\tilde v)^{n-2}f(\tilde v) - b'(\tilde v)F(\tilde v)^{n-1} \]

Setting this equal to zero and imposing symmetry (\(\tilde v = v_i\)) yields \[ b'(v_i) = (v_i - b(v_i))(n-1)\frac{f(v_i)}{F(v_i)} \]

To solve the differential equation define: \[ x(v) := v - b(v), \quad\text{so that}\quad b'(v) = 1 - x'(v) \]

Substituting gives \[ 1 - x'(v) = x(v)(n-1)\frac{f(v)}{F(v)} \]

Rearranging, \[ x'(v) + x(v)(n-1)\frac{f(v)}{F(v)} = 1 \]

Multiplying by \(F(v)^{n-1}\), \[ \frac{d}{dv}\bigl[x(v)F(v)^{n-1}\bigr] = F(v)^{n-1} \]

Integrating, \[ x(v)F(v)^{n-1} = \int_{\underline v}^v F(t)^{n-1}\,dt + C \]

Using the boundary condition \(b(\underline v)=\underline v\), we have \(x(\underline v)=0\), hence \(C=0\).

Therefore, \[ b(v) = v - \frac{\displaystyle\int_{\underline v}^v F(t)^{n-1}\,dt}{F(v)^{n-1}} \]

Or, even better: \[ b(v) = v - \displaystyle\int_{\underline v}^v \left(\frac{F(t)}{F(v)}\right)^{n-1}\,dt \]

It now is useful to recall that \[ S(v_i;\tilde v) = (v_i - b(\tilde v))F(\tilde v)^{n-1} \]

Using the expression for the bidding function, we obtain \[ S(v_i;\tilde v) = \left( v_i - \tilde v + \int_{\underline v}^{\tilde v} \left\{ \frac{F(t)}{F(\tilde v)} \right\}^{n-1} dt \right) F(\tilde v)^{n-1} \]

Rearranging, \[ S(v_i;\tilde v) = (v_i - \tilde v)F(\tilde v)^{n-1} + \int_{\underline v}^{\tilde v} F(t)^{n-1}\,dt \]

Notice that \[ S(v_i) = (v_i - b(v_i))F(v_i)^{n-1} = \int_{\underline v}^{v_i} F(t)^{n-1}\,dt \]

Hence, \[ S(v_i) - S(v_i;\tilde v) = (\tilde v - v_i)F(\tilde v)^{n-1} - \int_{v_i}^{\tilde v} F(t)^{n-1}\,dt \ge 0 \]

Since \(F(t)\) is strictly increasing, the right-hand side is always non-negative, and therefore deviations from one’s own type are never profitable (one can also convince themselves of this statement by drawing a chart).

Revenue Equivalence Theorem

Now, note that the expected surplus of player of type \(\tilde v\) is: \[ S(\tilde v) = \int_{\underline v}^{\tilde v} F(t)^{n-1}\,dt \]

This allows us to rewrite \[ S(v_i) - S(v_i;\tilde v) = S(v_i) - \left[ (v_i - \tilde v)F(\tilde v)^{n-1} + S(\tilde v) \right] \ge 0 \]

Hence, \[ S(v_i) \ge S(\tilde v) + (v_i - \tilde v)F(\tilde v)^{n-1} \]

If we drop the indices and let \(P(v)\) denote the probability of receiving the object in equilibrium given type \(v\), we obtain the more general statement: \[ S(v) \ge S(\tilde v) + (v - \tilde v)P(\tilde v) \]

Let \(\tilde v = v + dv\). Then is must be that both \[ \frac{S(v+dv)-S(v)}{dv} \ge P(v), \quad \frac{S(v+dv)-S(v)}{dv} \le P(v+dv) \]

As \(dv \to 0\), \[ \frac{dS(v)}{dv} = P(v) \]

Integrating, \[ S(v) = \int_{\underline v}^v P(t)\,dt + C \]

Since clearly \(C = S(\underline v)\), we obtain \[ S(v) = S(\underline v) + \int_{\underline v}^v P(t)\,dt \]

Therefore, any two mechanisms with the same allocation rule \(P(v)\) and the same S() generate the same surplus function.