Optimal bid strategy

Question

You have a box filled with cash. Cash value is uniformly randomly distributed from $0$ to $1000$. You are trying to win the box in an auction: you win the box if you bid at least the value of the cash in the box; you win nothing if you bid less (but you lose nothing). If you win the box, you can resell it for $150\%$ of its value.

How much should you bid to maximize the expected value of your profit (resale of box minus bid)?

My attempt:

Let $p$ be the cash value of the box and let's find the expected profit given a bid $b$:

$E=\frac{1000-b}{1000}\cdot0 + \frac{b}{1000} \cdot (1.5\cdot E[p]-b)$

Now my problem is: is $E[p]=500$, or, do I have to compute $E[p]$ given that $p<b$, which would yield $E[p]=\frac{b}{2}$?

How about this alternate approach: We find $b$ that maximizes $\int_{b}^{1000} (1.5p-b) $. Is this logically sound?

Answer 1

To avoid confusion, I will use capital letters for random variables and lower case letters for deterministic quantities. Your bidding strategy is controlled by a single real number, $b$. The random variable $U$ is uniformly distributed on $[0,1000]$ and represents the cash value of the prize.

The value of your profit is also a random variable, call it $T$. Then, $$ T=\begin{cases}0,&U>b\\ \tfrac{3}{2}U-b,& U<b\end{cases} $$

The expected value of $T$ is then equal to the probability that $U<b$ multiplied by the conditional expectation $\mathbb E[\tfrac{3}{2}U-b\mid U<b]$. Conditioning a uniform random variable results in another uniform random variable, but distributed on a smaller interval. In this case, if we consider a new random variable $V$ which is uniformly distributed on $[0,b]$ then we have $$ \mathbb E[\tfrac{3}{2}U-b\mid U<b]=\mathbb E[\tfrac{3}{2}V-b]=\frac{3}{2}\cdot \frac{b}{2}-b=-\frac{b}{4}. $$

Returning to $T$, it follows that $$ \mathbb ET = \mathbb P(U<b)\cdot \mathbb E[\tfrac{3}{2}U-b\mid U<b] = \frac{b}{1000}\cdot \frac{-b}{4}=\frac{-b^2}{4000}. $$

So we find that the optimum expectation occurs when $b=0$, i.e. it is a game where the only winning move is not to play.

Epilogue. The intuitive explanation of why we lose money on average is that no matter what we bet, we expect the true value to be about half that on average. So to recoup our losses from over valuing the item, we would need to be able to double its value in resale. But 150 percent is less than 200 percent, so we would lose money.

Now things would change significantly if we had a more educated guess on the value, for instance if we had a better model for the true value - something other than the uniform distribution.

Moral of the story: don't bid on things when you have no idea of their true value.

Answer 2

"You win the box if you bid at lesat the value of the cash in the box; you win nothing if you bid less (but you lose nothing). If you win the box you can resell it for 150% of its value

How much should you bid to maximize the expected value of your profit (resale of box minus bid)"

Please forgive me switch of notation, but it is conventional (at least in most probability theory texts) to use upper-case letters to denote random variables.

Let $U$ be $\mathcal{U}(0,1000)$ and $b>0$. The payout function is $1.5 \cdot U \mathbb{1}_{U \leq b}$, where $\mathbb{1}_A$ is the indicator function of the set $A$, that is $\mathbb{1}_A(\omega)=1$ if $\omega \in A$ and $\mathbb{1}_A(\omega)=0$ if $\omega \notin A$. That is, if you bid a value $b$ greater than the (unknown and random) value of the cash in the box, you receive the box, and can sell it at $1.5$ its value $U$, whatever that is. Otherwise, you earn no payout.

Thus the profit function is
$$h(U,b)=1.5 \cdot U \cdot \mathbb{1}_{U\leq b}-b$$ and the expected profit function is $g(b)=\mathbb{E}(h(U, b))$, which we compute as $$1.5 \mathbb{E}(U \cdot \mathbb{1}_{U \leq b})-b=1.5\int_0^b u f_U(u) du-b,$$ which results in $g(b)=\frac{1.5}{1000 \cdot 2} b^2-b$ which then one can compute critical points $g'(b)=0$ to find that $b^*=2000/3$ is the only critical point. Unfortunately, by checking via derivative tests or your favorite method, we actually conclude this is a global minimum and that before this point, i.e. on the interval $[0, b^*]$, the profit function is decreasing therefore the maximum value occurs at the boundary: $b=0$.