Best-responding to a stochastically higher distribution of bids

Question

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)?

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).