Consider the following variant of a 1st price auction in an IPV setup. Sealed bids are collected. The highest bidder pays his bid, but receives the object only if the outcome of a toss of a fair coin is heads. If the outcome is tails, the seller keeps the object and the high bidders bid. Assume bidder symmetry.

My attempt was as follows:

To find the symmetric equilibrium strategy in a first price auction we need to first find our optimal bidding strategy and that of all other $N-1$ bidders. We write out our strategy in terms of the expectation of winning the auction as $\beta^\textbf{I}(x)=E[Y_1|Y_1 < x]$ where $Y_1$ is the second highest bidder. Due to the context of our problem we must generate our payoff statement as $$G(\beta^{-1}(b)) \cdot \frac{(x-b)}{2}$$ where $G(\beta^{-1}(b)$ is the CDF with our bid to the inverse of our strategy which yields our value. We now maximize by setting the equation to $0$ with respect to our bid $b$. $$\frac{g(\beta^{-1}(b)}{\beta^{\prime}(\beta^{-1}(b))} \cdot \frac{(x-b)}{2}-G(\beta^{-1}(b))=0$$ Substituting $\beta(x)$ for $b$ yields $$\frac{g(x)}{\beta^{\prime}(x)}\frac{(x-\beta(x))}{2}-G(x)=0$$ $$\frac{g(x)x-\beta(x)g(x)}{2\beta^{\prime}(x)}-\frac{2\beta^{\prime}(x)G(x)}{2\beta^{\prime}(x)}=0$$ $$\beta(x)g(x)+2\beta^{\prime}(x)G(x)=g(x)x$$ $$2\beta(x)g(x)+2\beta^{\prime}(x)G(x)=g(x)x+\beta(x)g(x)$$

However, I am stuck in trying to simplify this final equation. Normally, without halved probability, you are able to solve the differential equation on the left-hand side and then integrate both sides.

Is my initial equation or approach to the problem?

  • 1
    $\begingroup$ Shouldn't the payoff be $$ G(\beta^{-1}(b))\left(\frac{x}{2}-b\right) $$ since the auctioneer keeps the the high bid even if the coin toss reveals tails? $\endgroup$ Nov 5, 2019 at 14:57
  • $\begingroup$ @TheoreticalEconomist So the value that you receive through signal x in a normal 1PA is halved and your bid should be adjusted accordingly? $\endgroup$
    – sardinsky
    Nov 5, 2019 at 15:04
  • $\begingroup$ Yes. When you bid $b$, your bid is the highest with probability $G(\beta^{-1}(b))$. Whenever you make the highest bid, you receive the object with probability $1/2$, but always pay the bid $b$, so your payoff whenever you bid the highest is $(x/2 -b)$. $\endgroup$ Nov 5, 2019 at 15:32

1 Answer 1


I'm going to assume that the bidder valuations are iid with distribution $F$ supported on $\left[ \underline{v} ,\overline{v}\right]\subset \mathbb R_+$. Denote $G(v) = (F(v))^{N-1}$, which gives the probability that the $N-1$ other bidders have value less than $v$.

We fix one bidder, call them $i$, and conjecture that the $N-1$ other bidders employ a symmetric equilibrium strategy $\beta\colon \left[ \underline{v} ,\overline{v}\right]\to \mathbb R$. We further assume that $\beta$ is strictly increasing (and thus invertible) and differentiable.

Considering the perspective of the bidder $i$, we know that the probability that $i$ wins the auction after bidding $b \ge 0$ is $G(\beta^{-1}(b))$. Suppose that $i$'s valuation for the object is $x$. Whenver $i$ wins the auction, they always pay the bid $b$, but receive the good with probability $1/2$. Thus, their payoff from winning the auction while bidding $b$ and having valuation $x$ must be $x/2 - b$. In particular, note that the bid is not halved since the winning bidder will pay the bid $b$ regardless of the outcome of the coin toss.

Thus, $i$'s expected payoff from bidding $b$ must be $$ G(\beta^{-1}(b))\left(\frac{x}{2}-b\right). $$

To find the optimal bid, we can differentiate the expression above with respect to $b$ to find the first-order condition $$ \frac{g(\beta^{-1}(b^*))}{\beta^{\prime}(\beta^{-1}(b^*))} \left( \frac x2 - b^* \right) - G(\beta^{-1}(b^*)) = 0. $$

From our conjecture that $\beta$ is a symmetric equilibrium strategy, we must have that $b^* = \beta(x)$. This gives us the ODE $$ \frac{g(x)x}{2} = G(x)\beta^\prime(x) + g(x) \beta(x). \label{1}\tag{1} $$

The right-hand side can be written as $$ G(x)\beta^\prime(x) + g(x) \beta(x) = \frac{\mathrm d }{\mathrm d x} (G(x)\beta(x)). $$

Integrating \eqref{1}, we find that $$ G(x) \beta(x) = \frac12 \int_\underline{v}^x tg(t)\,\mathrm dt $$ so that $$ \beta(x) = \frac{1}{2G(x)} \int_\underline{v}^x tg(t)\,\mathrm dt, $$ which means that each bidder bids half as much as they would in the standard first-price auction, which is pretty much what you'd expect given the form of the auction.

  • $\begingroup$ This is a superb explanation! $\endgroup$
    – sardinsky
    Nov 5, 2019 at 15:49
  • $\begingroup$ Just a bid (I mean bit :)) of an aside here but what is the intuition behind calculating the sellers expected revenues. I understand one multiplies N times the ex ante equilibrium bid but I fail to see how that bid can be equal to $$y(1-F(y))g(y)$$ in $$N\int^{w}_{0}y(1-F(y))g(y)dy$$ $\endgroup$
    – sardinsky
    Nov 5, 2019 at 16:05
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    $\begingroup$ @sardinsky a bidder with valuation $v$ expects to pay (in equilibrium) $G(v) \beta (v)$, since they win with probability $G(v)$, and pay $\beta (v)$ whenever they win. Thus, before they learn their valuation, they expect to pay $\int G(v) \beta (v) f(v) \,\mathrm d v$ (limits of integration omitted). Multiply this by $N$ to get the auctioneer’s expected revenue. Not really sure where you get the expression you give — but you might get there with a little manipulation of the expression I give in this comment. Perhaps try integration by parts? $\endgroup$ Nov 6, 2019 at 5:51
  • $\begingroup$ The expression I have is a result of changing the limits of integration where x is our value and the inner integral of the expected revenue calculation is from 0 to y with respect to x and the outer integral was from 0 to omega with respect to y. But thank you for your response! You've been really helpful. $\endgroup$
    – sardinsky
    Nov 6, 2019 at 13:36

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