# Bidding Item Problem

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.

$$P(S) = \begin{cases} 2S - B & B \geq S\\ 0 & \text{otherwise} \end{cases}$$

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.

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Why would you bid more than $1$ if the item's value is distributed between $0$ and $1$? –  Cocopuffs Jul 4 '12 at 19:20
Why is the bidder's profit $2V-B$? Shouldn't it be $V-B$? If I get a \$100 item for \$90, my profit is \$10, not \$110. –  Ross Millikan Jul 4 '12 at 20:19
@Ross: Because the question is silly. They define $V$ as the "true value" of the item, which then somehow magically doubles if you get it. So the value of the item to you is $2V$. Blame the author of the exercise. –  Ilmari Karonen Jul 4 '12 at 20:22
@Cocopuffs The value doubles just before you collect. So since when you can win more than a dollar you might decide to bid mote than a dollar to increase your chances to win. –  Michael Chernick Jul 4 '12 at 20:28
@Michael: ...except that bidding more than a dollar cannot increase your chances of winning, under the assumptions as stated. But it's really a badly phrased question. –  Ilmari Karonen Jul 4 '12 at 20:30

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.

Exercise: You are bidding on an item that has an unknown nominal value $V$ uniformly distributed between $0$ and $1$. You do not know the nominal value of the item, but you do know that the value of the item to you is twice its nominal value. You know that, if you bid more than the nominal value of the item, you will win the item and have to pay your bid; otherwise you don't get the item and don't have to pay anything. How much should you bid to maximize your expected gain?

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.)

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Why do your bounds go from $[0,1]$ to $[0, \min (1,B) ]$. What is the intuition behind this? –  asdfasdf Mar 19 at 5:14
Look at the definition of $g(B,V)$. (Ps. I had a small typo in it; maybe it's a bit clearer now?) It's defined piecewise, with the piece boundary at $B=V$, so to integrate over it, we need to integrate over each piece separately: $\int_0^1 g(B,V)\,dV = \int_0^{\min(1,B)} g(B,V)\,dV + \int_{\min(1,B)}^1 g(B,V)\,dV = \int_0^{\min(1,B)} (2V - B)\,dV + \int_{\min(1,B)}^1 0\,dV$. But the second integral is identically zero, so it falls out, and we're left with $\int_0^{\min(1,B)} (2V - B)\,dV$. –  Ilmari Karonen Mar 19 at 11:31

Edit: This replaces an earlier solution that had a major error.

We explore the consequences of offering $b$, where $0 \le b\le 1$. Let random variable $W$ denote the value of the item, given that the offer was accepted. For $0 \le w \le b$ we have $$Pr(W \le w)=Pr(V\le w|V\le b)=\frac{\Pr(V \le w)}{\Pr(V \le b)}=\frac{w}{b}.$$ Thus $W$ has uniform distribution on $[0,b]$, and therefore has expectation $b/2$. The profit is $2W-b$. This has expectation $2E(W)-b$, which is $0$ whatever $b$ in $[0,1]$ is chosen for the offer.

Offering $\lt 0$ trivially also gives expectation of profit equal to $0$. If we offer $b\gt 1$, the profit is $2V-b$, which has mean $1-b \lt 0$.

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Can you expound a little bit on how you calculated $E(2V-b)$? Thanks for the answer. –  Matt Jul 4 '12 at 18:34
It is general properties of expectation. Expectation of a linear combination is the linear combination of the expectations. So $E(2V-b)=2E(V)-b$. But the mean of a uniform on $[0,1]$ is $1/2$, the midway point. One could calculate $\int_0^1 (v)(1)\,dv$ instead. Or without appeal to linearity calculate $\int_0^1 (2v-b)(1)\,dv$. –  André Nicolas Jul 4 '12 at 18:38
Ah ok, yes that simplifies things. Thanks for the explanation. Then by maximize do you mean maximizing $b(1-b)$, so I should get $1-2b=0, b=.5$ as my final answer? –  Matt Jul 4 '12 at 18:40
@Matt: Yes. The problem asks one to maximize the expected profit, and $b(1-b)$ is the expected profit. The maximum is reached at $b=1/2$. –  André Nicolas Jul 4 '12 at 18:42
Thanks once again. Before I accept your answer, can you help me with one last thing? I posted the solution of this problem from the book I'm working off of in the OP, and it seems their final answer is different than what you have explained to me. Can you clarify why this is? –  Matt Jul 4 '12 at 19:02