/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Let \(X_{1}, X_{2}, \ldots, X_{n... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X_{1}, X_{2}, \ldots, X_{n}\) be random variables denoting \(n\) independent bids for an item that is for sale. Suppose each \(X_{i}\) is uniformly distributed on the interval \([100,200]\). If the seller sells to the highest bidder, how much can he expect to earn on the sale? [Hint: Let \(Y=\max \left(X_{1}, X_{2}, \ldots, X_{n}\right)\). First find \(F_{r}(y)\) by noting that \(Y \leq y\) iff each \(X_{i}\) is \(\leq y\). Then obtain the pdf and \(E(Y) .]\)

Short Answer

Expert verified
The expected amount the seller can earn is \(100 + \frac{100}{n+1}\).

Step by step solution

01

Define the Random Variable

Let the random variable \(Y\) represent the maximum bid. This means \(Y = \max(X_1, X_2, \ldots, X_n)\). Our goal is to find \(E(Y)\), the expected value of \(Y\).
02

Determine the Distribution Function

Since \(Y = \max(X_1, X_2, \ldots, X_n)\), \(Y \leq y\) if and only if each \(X_i \leq y\). The cumulative distribution function (CDF) of each \(X_i\) being less than or equal to \(y\) is \(F(x) = \frac{y-100}{100}\) for \(100 \leq y \leq 200\). Since the \(X_i\) are independent, \(F_Y(y) = (F(x))^n = \left(\frac{y-100}{100}\right)^n\).
03

Find the Probability Density Function

The probability density function (PDF) \(f_Y(y)\) is obtained by differentiating \(F_Y(y)\): \[f_Y(y) = \frac{d}{dy} \left( \frac{y-100}{100} \right)^n = n \left( \frac{y-100}{100} \right)^{n-1} \frac{1}{100}\] for \(100 \leq y \leq 200\).
04

Calculate Expected Value

To calculate \(E(Y)\), the expected value of the maximum bid, integrate \(y f_Y(y)\) over the interval from 100 to 200:\[E(Y) = \int_{100}^{200} y \cdot n \left( \frac{y-100}{100} \right)^{n-1} \frac{1}{100} \, dy\]Let \(u = \frac{y-100}{100}\), then \(dy = 100 \, du\). The bounds change to 0 and 1:\[E(Y) = 100n \int_{0}^{1} (100u+100) u^{n-1} \, du\]This can be simplified and solved using integration by parts, yielding:\[E(Y) = 100 + \frac{100}{n+1}\] for \(n\) bids.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Distribution
Imagine a fair playing field where every outcome within a certain range is equally likely. This situation is what we call a uniform distribution. In this exercise, each bid, represented by the random variable \(X_i\), is uniformly distributed. That means every bid is anywhere between \(100\) and \(200\) with equal probability.

The uniform distribution is simple yet powerful. It is defined by its minimum value, \(a\), and maximum value, \(b\). Here, \([a, b] = [100, 200]\).
  • Every number within this range has the same chance of being the bid amount.
  • There is no preference or bias toward any value between the given limits.
The key takeaway is that when every possible value within the given interval is equally likely, it is a uniform distribution. This assumption simplifies many computations and helps us to set a basis for further calculations like finding expected values.
Probability Density Function
The probability density function, or PDF, tells us about the likelihood of a random variable taking on a given value. It's like a guide through the sea of possibilities, providing direction on how likely each outcome is.

For the uniform distribution, the initial step is to establish the cumulative distribution function (CDF), then derive the PDF from it. The exercise explains this by first determining the CDF of the maximum bid \(Y\):
  • The CDF \(F(y)\) shows the probability that each \(X_i\), being a bid, is less than or equal to \(y\).
  • The PDF \(f_Y(y)\) is then derived by differentiating this CDF.
For our case, because all bids are independent and uniformly distributed, we find: \[f_Y(y) = n \left( \frac{y-100}{100} \right)^{n-1} \frac{1}{100}\] for \(100 \leq y \leq 200\). The PDF offers a picture of where the most likelihood is concentrated within the interval, acting as a backbone for determining expected values.
Cumulative Distribution Function
Think of the cumulative distribution function (CDF) as a step-by-step path leading to our final destination, the expected value. The CDF accumulates probabilities, giving us the probability that a random variable is less than or equal to a particular value.

In our problem, this concept is essential for handling the maximum bid, \(Y\), from a set of independent bids \(X_i\). Here's how it works:
  • If \(Y = \max(X_1, X_2, \ldots, X_n)\), then \(Y \leq y\) whenever each \(X_i \leq y\).
  • The CDF of each \(X_i\) being below \(y\) is \(F(x) = \frac{y-100}{100}\), based on uniform distribution.
  • Since all bids are independent, we can multiply the individual probabilities: \(F_Y(y) = \left( \frac{y-100}{100} \right)^n\).
The CDF connects what's known (individual bid behavior) to what's unknown (maximum bid behavior), allowing us to transition smoothly from probability to expected outcomes.
Independent Random Variables
Independence is freedom in probability, indicating that the outcome of one variable does not influence another. When dealing with independent random variables, like the bids \(X_1, X_2, \ldots, X_n\), each acts independently.

This independence is particularly significant in this problem because:
  • It means the maximum bid \(Y\) is influenced by all bids, but the exact value of each bid does not affect the others.
  • Therefore, when calculating probabilities for \(Y\), the probabilities of each bid can be multiplied together.
Consider this in practical terms: If you have a set of independent bids, no single bidder’s decision directly affects another's. This makes calculating the CDF and subsequently the expected value more straightforward, relying fully on the power of multiplication across the individual probabilities.

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Most popular questions from this chapter

A large but sparsely populated county has two small hospitals, one at the south end of the county and the other at the north end. The south hospital's emergency room has four beds, whereas the north hospital's emergency room has only three beds. Let \(X\) denote the number of south beds occupied at a particular time on a given day, and let \(Y\) denote the number of north beds occupied at the same time on the same day. Suppose that these two rv's are independent; that the pmf of \(X\) puts probability masses \(.1, .2, .3, .2\), and \(.2\) on the \(x\) values \(0,1,2,3\), and 4 , respectively; and that the pmf of \(Y\) distributes probabilities .1, .3, .4, and \(.2\) on the \(y\) values \(0,1,2\), and 3 , respectively. a. Display the joint pmf of \(X\) and \(Y\) in a joint probability table. b. Compute \(P(X \leq 1\) and \(Y \leq 1)\) by adding probabilities from the joint pmf, and verify that this equals the product of \(P(X \leq 1)\) and \(P(Y \leq 1)\). c. Express the event that the total number of beds occupied at the two hospitals combined is at most 1 in terms of \(X\) and \(Y\), and then calculate this probability. d. What is the probability that at least one of the two hospitals has no beds occupied?

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