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

<blockquote>
  <p>Consider a two-bidder auction with two types of players, high type and
  low type ($v_h&gt;v_l$). 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$$\frac{3}{4}\bigg(\frac{S-v_l}{v_h-S}\bigg)$$ </p>
  
  <p>What is the value of the upper bound of the support $s$?</p>
</blockquote>

<p>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</p>

<p>$$\frac{S-v_l}{v_h-S},$$</p>

<p>Then we have something like this</p>

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

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

<hr>

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

CC BY-SA 3.0
2018-03-29T20:35:27.130Z
"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
CC BY-SA 3.0
2018-03-29T22:27:38.677Z
@denesp I've updated the question.
user526463
CC BY-SA 3.0
2018-03-29T22:43:43.327Z
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.
<mathematical-economics><game-theory><auctions>
Deriving support in bidding strategy
user526463
# Deriving support in bidding strategy

(Asked by: username_0 on 2018-03-29)

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 $$\overline{S}=\frac{7v\_h+3v\_l}{10}$$

**Tags:** mathematical-economics, game-theory, auctions

> **Comments:**
> **username_1**: "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.  
> **username_0**: @denesp I've updated the question.  
> **username_2**: 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)?  

---

## 0 answers

