# Deriving support in bidding strategy

I'm considering some question, and I'm not sure what it asks me to do:

Consider a two-bidder auction with two types of players, high type and low type ($v_h>v_l$). The probability of a low type is $0.3$. The high type plays a continuous mixed strategy with support $s\in[\underline{S},\overline{S}]$. For some bid $s$ the probability of winning is given by the cdf$$\frac{3}{4}\bigg(\frac{S-v_l}{v_h-S}\bigg)$$

What is the value of the upper bound of the support $s$?

I guess the derivation should be roughly as follows. First, it is easy to verify that the chance of high valuation player to win against another high value player is

$$\frac{S-v_l}{v_h-S},$$

Then we have something like this

$$\big(\frac{3}{4}\bigg(\frac{S-v_l}{v_h-S}\bigg)+0.3\big)(v_h-S)=\frac{S-v_l}{v_h-S}$$

Which doesn't quite give the right answer. I suspect I need to weight some arguments by probabilities.

If it helps, the answer is $$\overline{S}=\frac{7v_h+3v_l}{10}$$

• "the upper bound is simply the bid that the highest that yields $0$ payoff, namely, $\overline{S}=v_h$" I cannot make any sense of the either the first or second part of this statement. – Giskard Mar 29 '18 at 20:35
• @denesp I've updated the question. – user526463 Mar 29 '18 at 22:27
• Why is the winning probability a function of capital $S$ while it is qualified by "For some bid [lowercase] $s$"? Is it simply a typo? Also, what is this "probability of winning"? Does it already taken into account the strategy of the other player and is thus an unconditional probability? Or is it conditional on the other player being a certain type (and playing a certain strategy)? – Herr K. Mar 29 '18 at 22:43

I agree with the comments that suggest that the question is almost indecipherable. However, it should be easy, given the cdf, to derive the bounds. Why? Well we know for a cdf $F$, we must have $F(\underline{S}) = 0$ and $F(\overline{S})=1$. We can use the latter to work backwards from the upper bound i.e.
We guess that $F(s)$ is of the form
$$F(s) = k\bigg(\frac{S-v_l}{v_h-S}\bigg)$$
We must have $F(\overline{S}) = 1$ and using your solution, we have
$$\begin{split}F(\overline{S}) = F\bigg(\frac{7v_h+3v_l}{10}\bigg) = k\bigg(\frac{\big(\frac{7v_h+3v_l}{10}\big)-v_l}{v_h-\big(\frac{7v_h+3v_l}{10}\big)}\bigg) &= 1\\ k\bigg(\frac{7v_h+3v_l}{10}\bigg)-kv_l &= v_h-\bigg(\frac{7v_h+3v_l}{10}\bigg)\\ \frac{7kv_h - 7kv_l}{10} &= \frac{3v_h-3v_l}{10}\\ 7kv_h - 7kv_l &= 3v_h-3v_l\\ k &= \frac{3}{7}\end{split}$$
Thus, I suspect that your mixed strategy for the high type is given by the cdf $$F(s) = \frac{3}{7}\bigg(\frac{S-v_l}{v_h-S}\bigg)$$