The Revelation Principle and Optimal Auctions

This scene finds the best rule for selling one item. We skip complicated games and only ask directly.

The revelation principle says honest rules are enough. Ask the buyer what the item is truly worth. Reports below the reserve get nothing. Reports at or above it pay exactly the reserve. A well-chosen reserve maximizes expected revenue. The virtual value in the formula below marks the spot.

For example, you sell one ⚽ soccer ball. Your friend values it at 8,000 won. You set the reserve at 5,000 won. If your friend reports at least 5,000, the ball is theirs. They pay exactly 5,000 💰 won. Lying gains nothing, so honesty wins.

ψ(v)=v−1−F(v)f(v)=2v−1\psi(v) = v - \frac{1-F(v)}{f(v)} = 2v - 1
R(r)=∫r1ψ(v) dv=r−r2,r∗=12R(r) = \int_r^1 \psi(v)\, dv = r - r^2, \qquad r^* = \frac{1}{2}

Part of these learning paths

Examples & Exercises

With buyer value v∼U[0,1]v \sim U[0,1] and reserve rr, show that expected revenue of the direct mechanism is R(r)=r−r2R(r) = r - r^2,

and find the revenue-maximizing r∗r^*.

ψ(v)=v−1−F(v)f(v)=v−(1−v)=2v−1\psi(v) = v - \frac{1 - F(v)}{f(v)} = v - (1 - v) = 2v - 1
R(r)=∫r1ψ(v) dv=r−r2R(r) = \int_r^1 \psi(v)\, dv = r - r^2
r∗=0.5r^* = 0.5, R=0.25R = 0.25 — the optimal reserve.
Show proof

  1. With F(v)=vF(v) = v and density f(v)=1f(v) = 1, the virtual value is ψ(v)=v−(1−v)/1=2v−1\psi(v) = v - (1-v)/1 = 2v - 1. And ψ(v)≥0  ⟺  v≥1/2\psi(v) \ge 0 \iff v \ge 1/2 — the region where selling adds to expected revenue.

  2. Expected revenue is the expected virtual surplus: R(r)=∫r1(2v−1) dv=[v2−v]r1=(1−1)−(r2−r)=r−r2R(r) = \int_r^1 (2v-1)\,dv = [v^2 - v]_r^1 = (1-1) - (r^2-r) = r - r^2.

  3. The first-order condition R′(r)=1−2r=0R'(r) = 1 - 2r = 0 gives r∗=0.5r^* = 0.5, and R′′(r)=−2<0R''(r) = -2 < 0 confirms a maximum.

    R(0.5)=0.5−0.25=0.25R(0.5) = 0.5 - 0.25 = 0.25.

    The intuition: selling at r<0.5r < 0.5 means selling to buyers with negative virtual value, which drags revenue down, while r>0.5r > 0.5 throws away positive virtual surplus — the blue region in the top panel is exactly the expected revenue.

    r∗=12,R(r∗)=12−14=0.25r^* = \frac{1}{2}, \qquad R(r^*) = \frac{1}{2} - \frac{1}{4} = 0.25

With reserve r=0.5r = 0.5, compare the payoff of a buyer with true value v=0.7v = 0.7 who lies with v^=0.3\hat{v} = 0.3 against telling the truth.

What happens when a buyer with v=0.4v = 0.4 lies with v^=0.8\hat{v} = 0.8?

U(v,v^)=(v−r)⋅1{v^≥r}U(v, \hat{v}) = (v - r) \cdot \mathbf{1}\{\hat{v} \ge r\}
U(0.7,0.7)=0.2,U(0.7,0.3)=0U(0.7, 0.7) = 0.2, \qquad U(0.7, 0.3) = 0
v=0.7v = 0.7, v^=0.3\hat{v} = 0.3 — the lie loses the good.
Show proof

  1. Telling the truth, v^=0.7≥0.5\hat{v} = 0.7 \ge 0.5 wins the good, pays r=0.5r = 0.5, and earns 0.7−0.5=0.20.7 - 0.5 = 0.2.

    Lying with v^=0.3<0.5\hat{v} = 0.3 < 0.5 loses the good for a payoff of 0.

    Truth wins by 0.20.2 — under-reporting throws away the whole positive surplus.

  2. The reverse direction is symmetric.

    A buyer with v=0.4v = 0.4 who tells the truth gets 0 since 0.4<0.50.4 < 0.5. Lying with v^=0.8\hat{v} = 0.8 wins the good but costs 0.50.5, for a payoff of 0.4−0.5=−0.10.4 - 0.5 = -0.1 — paying more than the value.

    No lie in either direction beats the truth, so this direct mechanism is incentive compatible.

21. Mechanism Design