Derivation of Equilibrium Strategy in 1st-price Auction?

Question

Hey, everyone. I desperately need help understanding some math in auction theory. I have been writing a paper as an undergrad for auction theory, and after all of my research I just cannot understand one step the author of a book uses to derive the equilibrium strategy for first-price auctions under independent private value conditions.
The steps in the derivation process essentially follow this: Let x = the chosen player's personal valuation, G(x) = the probability that the highest opposing bid, an order statistic, Y_1, is < x. \begin{equation} \beta(x) = x - \int_0^x\frac{G(y)}{G(x)} \, dy \\ \end{equation} Then without much explanation, other than "values are uniformly distributed on [0,1]" he just converts this equation to \begin{equation} \beta(x) = \frac{N-1}{N}, \end{equation} Which makes sense and is usable.

The book I am using is Vijay Krishna's Auction Theory, and I would extremely appreciate someone helping me understand this last step.

Answer 1

I remember slaving over the notation in this book when I was a bad undergraduate. It brings up some interesting memories, some which may help you.

• $F(x)$ is the cumulative distribution of a single bidder's valuation.
• $G(x)$ is the cumulative distribution of the highest bidder's valuation, given $N$ bidders.

For the example you are referring to, values being uniformly distributed along $[0,1]$ implies $F(x) = x.$ That's pretty straightforward. The chance of you having a valuation of $\frac{1}{2}$ or less for the object is, well, $\frac{1}{2}$. The chance of you having at least a value of $1$ should be a probability of $1$.

But why is $G(x) = x^{N-1}$ ?

For $n =1$, the chance of your valuation being the largest valuation or less is...well, $1 \ (= x^0)$.

For $n = 2$, the chance of your valuation being the largest valuation or less is just the chance that the other bidder has a valuation lower than you, or exactly $x$.

For $n = 3$, the chance of your valuation being the biggest or less is just the chance that both other bidders value the object less. Since you're working with independent private values, you can think of their valuations as independent (duh) events, so you can just multiply the events together. (More on conditional probability here). So for example the chance of someone having a smaller value than you is $x$, but there is another person who also has to have a smaller value than you, with the same chance $x$. The chance of both of them having smaller values than you is $x \cdot x = x^2$.

So it goes, the more bidders (independent valuations) you have.

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So we have your general formula for an optimal bid under a first-price auction:

$$\beta^1(x) = x - \int^x_0 \frac{G(y)}{G(x)} dy$$

So substitute in $G(x)$:

$$\beta^1(x) = x - \int^x_0 \frac{y^{N-1}}{x^{N-1}} dy$$

Evaluate:

$$= x \ - \ \biggr\lvert \frac{y^N}{Nx^{N-1}} + c\biggr\rvert^x_0$$ $$= x - \frac{x}{N} = \frac{Nx}{N} - \frac{x}{N}$$ $$\boxed{\beta^1(x) = x\frac{N - 1}{N}}$$

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The next example in Krishna's book has an exponential distribution, but only with two bidders. If you try using the same line of reasoning as I used above to find $G(x)$, you'll notice things appear a bit difficult, but the author doesn't explicitly state $G(x)$ in this case in fact. Try seeing for yourself if you understand why Krishna gives the helpful statement:

$$\frac{G(y)}{G(x)} = \left[\frac{F(y)}{F(x)}\right]^{N-1}$$

for the generic case where there is no functional form for the distributions.

Answer 2

I think you have a typo -- the equilibrium bidding strategy in the first-price auction you specify should be $$ \beta(x) = \frac{N-1}{N}x $$

Here's a hint that might help. The CDF of the order statistic, $Y_1$, is $$G(x) = x^{N-1}$$

To see why this is the case, notice that this is exactly the probability that some given quantity $x$ is greater than or equal to $N-1$ independently drawn random variables that are uniformly distributed on $[0,1]$.

I'll write something in more detail when I have more time tomorrow, but I think this should tide you over for now. (At the very least, it'll help you evaluate the integral.)