Math Problem Statement

You are currently bidding for a painting. You know that the value of the painting is between $0 and $100,000 uniformly. If your bid is greater than the value of the painting, you win and sell it to an art museum at a price of 1.5 times the value. What’s your bid to max your profit? If you can not profit, bid $0.

Solution

To solve this problem, we need to set up a strategy for bidding that maximizes your expected profit.

Problem Breakdown:

  1. Value of the painting VV: The value is uniformly distributed between 00 and 100,000100,000, so the probability density function (pdf) of VV is fV(v)=1100,000f_V(v) = \frac{1}{100,000} for v[0,100,000]v \in [0, 100,000].

  2. Bidding: Let your bid be BB. If you bid BB and the value of the painting is less than or equal to your bid (VBV \leq B), you win the painting. In this case, you sell it to the art museum at 1.5×V1.5 \times V.

  3. Profit Function: If you win, your profit is: Profit=1.5×VB\text{Profit} = 1.5 \times V - B If you don't win, your profit is 00.

  4. Expected Profit: The expected profit, π(B)\pi(B), is the integral of the profit over all possible values of VV, weighted by the probability of VV being in that range.

Let's calculate the expected profit as a function of your bid BB.

For BB, your expected profit is the integral over VV from 00 to BB (since if VBV \leq B, you win). The probability density function for VV is uniform, so the probability that VBV \leq B is B100,000\frac{B}{100,000}. Therefore, the expected profit is:

π(B)=0B(1.5×vB)1100,000dv\pi(B) = \int_0^B \left( 1.5 \times v - B \right) \frac{1}{100,000} \, dv

Now, let's compute this integral:

π(B)=1100,0000B(1.5vB)dv\pi(B) = \frac{1}{100,000} \int_0^B \left( 1.5v - B \right) \, dv

First, split the integral:

π(B)=1100,000[0B1.5vdv0BBdv]\pi(B) = \frac{1}{100,000} \left[ \int_0^B 1.5v \, dv - \int_0^B B \, dv \right]

Now compute each integral:

0B1.5vdv=1.5B22=0.75B2\int_0^B 1.5v \, dv = \frac{1.5B^2}{2} = 0.75B^2 0BBdv=B2\int_0^B B \, dv = B^2

So, the expected profit becomes:

π(B)=1100,000[0.75B2B2]\pi(B) = \frac{1}{100,000} \left[ 0.75B^2 - B^2 \right] π(B)=1100,000(0.25B2)\pi(B) = \frac{1}{100,000} \left( -0.25B^2 \right) π(B)=0.25B2100,000\pi(B) = -\frac{0.25B^2}{100,000}

Maximizing the Profit:

The expected profit function π(B)\pi(B) is negative for all values of B>0B > 0, which suggests that you cannot profit from bidding any positive amount. Hence, the optimal strategy is to bid $0 to avoid any losses, because your expected profit is always negative.

Conclusion:

The optimal bid to maximize your profit is $0.


Would you like any further details or clarifications? Here are some related questions you may find useful:

  1. How does the value distribution (uniform vs. normal) affect the optimal bidding strategy?
  2. What would happen if the selling price to the art museum was a fixed amount rather than 1.5 times the value?
  3. How would this scenario change if the probability distribution of the painting's value was skewed or non-uniform?
  4. What other factors (like transaction costs) should be considered in real-world bidding strategies?
  5. How can you model the scenario if the value of the painting is known with certainty?

Tip: For optimal bidding in situations involving uncertainty, consider risk aversion or using expected utility theory.

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Math Problem Analysis

Mathematical Concepts

Expected Value
Probability
Integration
Uniform Distribution

Formulas

Expected Profit: \pi(B) = \int_0^B (1.5v - B) \frac{1}{100,000} dv

Theorems

Uniform Distribution
Integration of Linear Functions

Suitable Grade Level

Grades 11-12