/*! 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 40 There are two distinct methods f... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 40

There are two distinct methods for manufacturing certain goods, the quality of goods produced by method \(i\) being a continuous random variable having. distribution \(F_{i}, i=1,2\). Suppose that \(n\) goods are produced by method 1 and \(m\) by method 2. Rank the \(n+m\) goods according to quality and let $$ X_{j}= \begin{cases}1 & \text { if the } j \text { th best was produced from method } 1 \\ 2 & \text { otherwise }\end{cases} $$ For the vector \(X_{1}, X_{2}, \ldots, X_{n+m}\), which consists of \(n 1\) 's and \(m 2\) 's, let \(R\) denote the number of runs of 1 . For instance, if \(n=5, m=2\), and \(X=1\), \(2,1,1,1,1,2\), then \(R=2\). If \(F_{1}=F_{2}\) (that is, if the two methods produce identically distributed goods), what are the mean and variance of \(R\) ?

Short Answer

Expert verified
In conclusion, when both manufacturing methods produce identically distributed goods (i.e. \(F_1=F_2\)), the mean and variance of the number of runs (\(R\)) are as follows: Mean: \( E[R] = \frac{n+1}{2} \) Variance: \( Var[R] = E[R^2] - (E[R])^2 = \sum_{k=1}^{n-1} \frac{k^2 (n-k+1)}{2^{n+k}} - \left(\frac{n+1}{2}\right)^2 \)

Step by step solution

01

Understanding a Run and the Vector \(X\)

In the given context, a "run" refers to a consecutive set of goods produced from method 1. In other words, it is a sequence of consecutive 1's in the vector \(X\). The goal is to determine the mean and variance of the number of runs (\(R\)) when the quality of goods produced by both methods is identically distributed.
02

Deriving the Mean of R when \(F_1 = F_2\)

If \(F_1 = F_2\), the probabilities of each method producing a good of higher quality are equal, which is \(1/2\). Then, the probability of a particular good ending a run is the probability that the good coming after it is produced by method 2, which is \(1/2\). In the sequence with \(n+m\), there are \(n-1\) opportunities for a run to end in between items (excluding the last one produced by method 1), so the expected number of runs is: \( E[R] = (n - 1) \times \frac{1}{2} + 1 = \frac{n+1}{2} \) Here, we added 1 because we want to include the last run.
03

Deriving the Variance of R when \(F_1 = F_2\)

To calculate the variance, we need to find the expected value of the square of R (\(E[R^2]\)): \( E[R^2] = Var[R] + (E[R])^2 \) Let's first find \(E[R^2]\) by examining all possibilities for the last run to end. Here, \(k\) denotes the position where the last run ends (from the left): - When \(k = n -1\): There is one way for the last run to end at the \((n - 1)\)-th position. - When \(k = n - 2\): There are two ways for the last run to end at the \((n - 2)\)-th position. - ... - When \(k = 1\): There are \((n - 1)\) ways for the last run to end at the \(1\)-st position. In each case, the probability of this happening is \((1/2)^{n+k}\). Thus, we have: \( E[R^2] = \sum_{k=1}^{n-1} \frac{k^2 (n-k+1) }{2^{n+k}} \) Then, we find the variance: \( Var[R] = E[R^2] - (E[R])^2 \) Now we have both the expected value(mean) and variance for the number of runs, R: Mean: \( E[R] = \frac{n+1}{2} \) Variance: \( Var[R] = E[R^2] - (E[R])^2 \)

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Key Concepts

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

Random Variables
When we talk about random variables, we're referring to variables that represent possible outcomes of a random phenomenon. They can take on different values based on chance, making them an essential part of understanding probability in statistics.
In the exercise, the quality of goods produced by the manufacturing methods is seen as continuous random variables. This means that their quality can assume any value within a certain range.
Important points about random variables include:
  • They can be continuous or discrete. Continuous random variables, like the quality of goods, can take any value, whereas discrete ones take on specific values.
  • Random variables are associated with a probability distribution that describes the likelihood of different outcomes.
  • Understanding these variables is crucial for calculating expectations and variances, which provide insights into averages and variability.
In problems like this exercise, identifying random variables helps us structure our analysis and apply specific statistical methods to reach meaningful conclusions.
Probability Distribution
A probability distribution explains how the probabilities are distributed over the values of the random variable. In simpler words, it's a mathematical description of the likelihood of different outcomes.
In the exercise, the distribution of the quality of goods produced is captured by the functions \(F_1\) and \(F_2\). These functions specify how outcomes differ between two methods.
Key aspects to understand include:
  • If \(F_1 = F_2\), it means the quality across both methods is identically distributed, allowing us to treat them similarly in calculations.
  • The distribution can be portrayed as either discrete or continuous, which affects how we calculate probabilities.
  • Distributions like these form the basis for understanding how likely different events are, such as a product's placement in the ranking by quality.
This concept becomes essential when determining the expected number of runs as each distribution affects how often one method's goods outrank the other's.
Expectation
Expectation, or expected value, is the average value a random variable takes on over many trials. It's a prediction of the long-term average outcome of experiments.
In the exercise, our target is to understand the expected number of runs, \(E[R]\). With identically distributed quality across methods, the probability of ending a run becomes \( \frac{1}{2} \).
This leads to:
  • Calculating the expectation as \( E[R] = \frac{n+1}{2} \). This formula accounts for the opportunity each good has to end a run, plus an additional run including the last item.
The concept of expectation simplifies complex distributions to a single, average value, making it invaluable for decision making and predictions.
Variance
Variance is a measure of how spread out the numbers in a data set are around the mean value. It provides insight into the variability of the random variable.
For this exercise, variance tells us how much the number of runs, \(R\), varies as we would repeat the ranking of goods many times. It's calculated using the formula:
\[ Var[R] = E[R^2] - (E[R])^2 \]Understanding variance involves:
  • Calculating \(E[R^2]\), which requires evaluating the sum of squared probabilities for every possible scenario.
  • Identifying that higher variance means more fluctuation in the number of runs around the expected value.
  • Variance helps in understanding risk and variability in processes and results.
In conclusion, knowing both the expectation and variance of \(R\) provides a fuller picture of what to expect and how much that expectation could vary.

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Most popular questions from this chapter

A deck of \(n\) cards, numbered 1 through \(n\), is thoroughly shuffled so that all possible \(n !\) orderings can be assumed to be equally likely. Suppose you are to make \(n\) guesses sequentially, where the \(i\) th one is a guess of the card in position \(i\). Let \(N\) denote the number of correct guesses. (a) If you are not given any information about your earlier guesses show that, for any strategy, \(E[N]=1\). (b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that under this strategy $$ \begin{aligned} E[N] &=\frac{1}{n}+\frac{1}{n-1}+\cdots+1 \\ & \approx \int_{1}^{n} \frac{1}{x} d x=\log n \end{aligned} $$ (c) Suppose that you are told after each guess whether you are right or wrong. In this case it can be shown that the strategy that maximizes \(E[N]\) is one which keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy show that $$ \begin{aligned} E[N] &=1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !} \\ &=e-1 \end{aligned} $$

Each of \(m+2\) players pays 1 unit to a kitty in order to play the following game. A fair coin is to be flipped successively \(n\) times, where \(n\) is an odd number, and the successive outcomes noted. Each player writes down, before the flips, a prediction of the outcomes. For instance, if \(n=3\), then a player might write down \((H, H, T)\), which means that he or she predicts that the first flip will land heads, the second heads, and the third tails. After the coins are flipped, the players count their total number of correct predictions. Thus, if the actual outcomes are all heads, then the player who wrote \((H, H, T)\) would have 2 correct predictions. The total kitty of \(m+2\) is then evenly - split up among those players having the largest number of correct predictions. Since each of the coin flips is equally likely to land on either heads or tails, \(m\) of the players have decided to make their predictions in a totally random fashion. Specifically, they will each flip one of their own fair coins \(n\) times and then use the result as their prediction. However, the final 2 of the players have formed a syndicate and will use the following strategy. One of them will make predictions in the same random fashion as the other \(m\).

Suppose that balls are randomly removed from an um initially containing \(n\) white and \(m\) black balls. It was shown in Example \(2 \mathrm{~m}\) that \(E[X]=\) \(1+m /(n+1)\), when \(X\) is the number of draws needed to obtain a white ball. (a) Compute \(\operatorname{Var}(X)\). (b) Show that the expected number of balls that need be drawn to amass a total of \(k\) white balls is \(k[1+m /(n+1)]\). HINT: Let \(Y_{h}, i=1, \ldots, n+1\), denote the number of black balls withdrawn after the \((i-1)\) st white ball and before the \(i\) th white ball. Argue that the \(Y_{i}, i=1, \ldots, n+1\), are identically distributed.

Consider \(n\) independent trials each resulting in any one of \(r\) possible outcomes with probabilities \(P_{1}, P_{2}, \ldots, P_{r}\). Let \(X\) denote the number of outcomes that never occur in any of the trials. Find \(E[X]\) and show that among all probability vectors \(P_{1}, \ldots, P_{r}, E[X]\) is minimized when \(P_{i}=1 / r, i=1, \ldots, r\).

A coin, which lands on heads with probability \(p\), is continually flipped. Compute the expected number of flips that are made until a string of \(r\) heads in a row is obtained. HINT: Condition on the time of the first occurrence of tails, to obtain the equation $$ E[X]=(1-p) \sum_{i=1}^{r} p^{i-1}(i+E[X])+(1-p) \sum_{i=r+1}^{x} p^{i-1} r $$ Simplify and solve for \(E[X]\).
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