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


Then we have something like this


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

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

  • $\begingroup$ "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. $\endgroup$
    – Giskard
    Mar 29, 2018 at 20:35
  • $\begingroup$ @denesp I've updated the question. $\endgroup$
    – user526463
    Mar 29, 2018 at 22:27
  • $\begingroup$ 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)? $\endgroup$
    – Herr K.
    Mar 29, 2018 at 22:43

1 Answer 1


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 $$F(s) = \frac{3}{7}\bigg(\frac{S-v_l}{v_h-S}\bigg)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.