In Auction Theory, Krishna writes that:

a bidder who faces a stochastically higher distribution of bids–in the sense of reverse hazard rate dominance–will bid higher

(This follows the proof of proposition 4.4 in the section 'Weakness Leads to Aggression'.)

Supposedly, this is 'easy to see' - however, I am having trouble seeing it! I would be grateful if someone could explain why this is true.

My (unsuccessful) efforts so far:

If I bid $b$ when my valuation is $v$, then (under risk neutrality) my expected utility is

$$(v - b)F(b)$$

where $F(b)$ is the cumulative distribution function of all the bids apart from mine (and thus the probability that I win the auction).

Maximising my expected utility with respect to $b$, we obtain the first order condition

$$(v - b)f(b) - F(b) = 0$$

Which implies that

$$b = v - \frac{F(b)}{f(b)}$$

A distribution $G$ is stochastically higher than a distribution $F$ in the sense of reverse hazard rate dominance when, for every $b$

$$\frac{g(b)}{G(b)} > \frac{f(b)}{F(b)}$$

Now, if $f(b)/F(b)$ and $g(b)/G(b)$ do not depend on $b$, it is obvious from the first order condition that I will bid higher when facing a stochastically higher distribution of bids. However, it is not clear to me what happens in the general case where the 'hazard rates' may depend on $b$.

Many thanks in advance!

Bonus question: When my bids increase, will my new distribution of bids 'stochastically dominate' my old distribution (in the sense of the reverse hazard rate)?

  • $\begingroup$ I'm not sure if your first order conditions are correct. The expected profit is $(v - b)F(\beta^{-1}(x))$, where $b=\beta (x)$. You can further consider $\beta^{-1} = \phi$, therefore the expected profit can be rewritten as $(v - b)F(\phi (b))$. $\endgroup$
    – superhulk
    Commented Dec 26, 2018 at 15:50
  • $\begingroup$ Your distribution F is the distribution of the highest valuation (apart from my own). However, my distribution F is the distribution of the highest bid (apart from my own). Hence, both of our expressions are correct - we are just using different notation. $\endgroup$
    – user17900
    Commented Dec 26, 2018 at 19:19
  • $\begingroup$ Is your last sentence correct? Do you mean to say when this does or does not depend on $v$ rather than when these functions depend on (b)? $\endgroup$
    – 123
    Commented Dec 27, 2018 at 10:49
  • $\begingroup$ No, I do mean $b$. If the hazard rate does not depend on $b$, then we have an explicit formula for the optimal bid. However, if the hazard rate does depend on $b$, we only have an equation which defines $b$ implicitly, which makes things harder. $\endgroup$
    – user17900
    Commented Dec 27, 2018 at 11:50

1 Answer 1


Here is a sketch of an answer.

As noted in the question, the optimal $b$ must satisfy

$$b = v - \frac{F(b)}{f(b)}$$

Let us define the function $h(b) = v - F(b)/f(b)$. Our problem is to find the the fixed points of the function $h(b)$, i.e. the $b$ such that

$$b = h(b)$$

What can we say about $h(b)$? Assuming there is a unique optimal bid, it crosses the 45 degree line exactly once. Moreover, assuming that bids cannot be negative, i.e. $F(b) = 0$, it follows that $h(0) = v$. Thus, assuming that $v > 0$, the function $h(b)$ starts off above the 45 degree line.

Now, when the opponent bids are 'stochastically higher', the function $h(b)$ moves upwards in the sense that the 'new' function lies everywhere above $h(b)$. It thus seems reasonably clear that the unique fixed point must increase, though admittedly I lack a conclusive proof of this fact.

I am also unsure about the bonus question (whether my new bids seem stochastically higher in the sense of the reverse hazard rate).


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