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 ($v_h&gt;v_l$). 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$$\frac{3}{4}\bigg(\frac{S-v_l}{v_h-S}\bigg)$$ 
  
  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 $$\overline{S}=\frac{7v_h+3v_l}{10}$$
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

$$\begin{split}F(\overline{S}) = F\bigg(\frac{7v_h+3v_l}{10}\bigg) = k\bigg(\frac{\big(\frac{7v_h+3v_l}{10}\big)-v_l}{v_h-\big(\frac{7v_h+3v_l}{10}\big)}\bigg) &amp;= 1\\
k\bigg(\frac{7v_h+3v_l}{10}\bigg)-kv_l &amp;= v_h-\bigg(\frac{7v_h+3v_l}{10}\bigg)\\
\frac{7kv_h - 7kv_l}{10} &amp;= \frac{3v_h-3v_l}{10}\\
7kv_h - 7kv_l &amp;= 3v_h-3v_l\\
k &amp;= \frac{3}{7}\end{split}$$

Thus, I suspect that your mixed strategy for the high type is given by the cdf $$F(s) = \frac{3}{7}\bigg(\frac{S-v_l}{v_h-S}\bigg)$$