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.
As I (and others) noted in the comments, the exercise seems to be really confusingly phrased (and not at all like any real auction or other transaction I've ever heard of), and the given solution doesn't look much better. So instead of trying to explain the book solution, let me rephrase the exercise in a (hopefully) slightly less confusing manner, and then show how I would solve it.
Solution:
Let $V \sim U(0,1)$ be a random variable denoting the nominal price of the item. The probability density function of $V$ is $$f(V) = \begin{cases} 1 & \text{if }0 < V < 1 \\ 0 & \text{otherwise.} \end{cases}$$
If you bid an amount $B$, your gain will be $$g(B,V) = \begin{cases} 2V -B & \text{if }V < B \\ 0 & \text{otherwise.} \end{cases}$$
Thus, your expected gain from bidding $B$ is $$\begin{aligned} \mathbb E_V[g(B,V)] &= \int_{-\infty}^\infty g(B,V)\: f(V)\: \mathrm dV \\ &= \int_0^1 g(B,V)\: \mathrm dV \\ &= \int_0^{\min(1,B)} (2V-B)\: \mathrm dV \\ &= \int_0^{\min(1,B)} 2V\: \mathrm dV - \int_0^{\min(1,B)} B\: \mathrm dV \\ &= V^2 \bigg|_{V=0}^{V=\min(1,B)} - BV \bigg|_{V=0}^{V=\min(1,B)} \\ &= (\min(1,B)^2 - 0^2) - (B\min(1,B) - B\cdot0) \\ &= \min(1,B)^2 - B\min(1,B) \\ &= \begin{cases} 1-B & \text{if }B > 1 \\ 0 & \text{if }B \le 1. \end{cases} \end{aligned}$$
(Since you said you had trouble following this part in the book solution, I included quite many intermediate steps. Let me know if there's still something you don't follow.)
As $1-B < 0$ whenever $B > 1$, you should never bid more than $1$. Instead, any bid of $1$ or less will result in an expected gain of $0$, so any such bid is as good as not bidding at all, which is the optimal strategy.
Actually, this came out looking more like that book solution after all, though hopefully a bit clearer. What I would probably do, if asked that in an exam or something, would be to start by noting that the chance of winning the item equals $1$ for any bid $B \ge 1$. Thus, any bid $B > 1$ is clearly suboptimal, as it increases the cost without changing the probability of winning. That out of the way, I'd the just calculate $$\mathbb E[g(B)] = \int_0^1 g(B,V)\: \mathrm dV = \int_0^B (2V-B)\: \mathrm dV = B^2 - B^2 = 0$$ for all $B \le 1$. (Or I might note that the integrand $2V-B$ has odd symmetry around the midpoint $V = B/2$ of the integration interval, so the integral has to be zero by symmetry considerations alone.)