You are bidding on an item that has an unknown value uniformly distributed between 0 and 1. You do not know the true value of the item, but you know that if you end up winning the bid for the item, the item will increase its value to 2x its original value. Your bid can only go through if its at least as large as the original value of the item. How do you bid to maximize expected payoff.
Here's what I have:
Let V be the true value of the item
Let B be the bid you make
Let f(V) represent the profit you make given V as the true original value
$f(V) = \begin{cases} 2V - B & B \geq V\\ 0 & B< V \end{cases}$
Where I get confused is when I need to start applying integrals to calculate how to maximize the expected value.
Thanks for any help.
EDIT: Here is the solution from the book I'm working off of. I do not understand how they are doing the calculus.
Let B be your bid. Let S be the true value of the item. The density function of S equals unity for $0 \leq S \leq 1$, and 0 otherwise.
Your payoff P is
$P(S) = \begin{cases} 2S - B & B \geq S\\ 0 & \text{otherwise} \end{cases}$
The maximum post bid item value is 2, so you should be no more than 2. You want to maximize $E[P(S)]$ with respect to choice of B in the interval [0, 2]. Your expected payoff is:
$\begin{aligned} E[P(S)] &= \int_{S=0}^{S=1} P(S)*1*\,\mathrm{d}S \\ &= \int_{S=0}^{S=\min(B,1)} (2S-B)\,\mathrm{d} \\ &= \left.(S^2-BS)\right|_{S=0}^{S=\min(B,1)} \\ &= \begin{cases} 0 & B\leq1\\ 1 - B & B>1 \end{cases} \end{aligned}$
so you should bid less than or equal to 1 and expect to break even.