Problem 2
Question
We generate a bootstrap dataset \(x_{1}^{*}, x_{2}^{*}, x_{3}^{*}, x_{4}^{*}\) from the empirical distribution function of the dataset \(\begin{array}{llll}1 & 3 & 4 & 6 .\end{array}\) a. Compute the probability that the bootstrap sample mean is equal to \(1 .\) b. Compute the probability that the maximum of the bootstrap dataset is equal to 6 . c. Compute the probability that exactly two elements in the bootstrap sample are less than \(2 .\)
Step-by-Step Solution
Verified Answer
a. \( \frac{1}{256} \); b. \( \frac{175}{256} \); c. \( \frac{27}{128} \).
1Step 1: Understand the Empirical Distribution Function
The empirical distribution function of a dataset is a probability distribution that assigns equal probability to each element in the dataset. For our dataset \( \{1, 3, 4, 6\} \), each element has a probability of \( \frac{1}{4} \) of being selected in the bootstrap sample.
2Step 2: Calculate Probability of Bootstrap Sample Mean Equal to 1
Since the mean of a sample cannot be less than the minimum value in the sample, for the mean to be 1, all selected bootstrap values must be 1. The probability of selecting 1 for each sample point is \( \frac{1}{4} \). Therefore, the probability that \( x_1^* = x_2^* = x_3^* = x_4^* = 1 \) is \( \left( \frac{1}{4} \right)^4 = \frac{1}{256} \).
3Step 3: Calculate Probability of Bootstrap Maximum Equal to 6
For the maximum to be 6, at least one of the sampled elements must be 6. The probability of not choosing 6 in any one trial is \( \frac{3}{4} \), so for four trials, it is \( \left( \frac{3}{4} \right)^4 \). Therefore, the probability of at least one element being 6 (i.e., the maximum being 6) is \( 1 - \left( \frac{3}{4} \right)^4 = \frac{175}{256} \).
4Step 4: Calculate Probability of Exactly Two Elements Less than 2
Elements less than 2 in the dataset are \{1\}. Find the probability of choosing 1 exactly twice from 4 positions. The probability of choosing 1 in a position is \( \frac{1}{4} \), and choosing not 1 (3, 4, or 6) is \( \frac{3}{4} \). The required probability is given by binomial probability: \( \binom{4}{2} \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^2 = 6 \times \frac{1}{16} \times \frac{9}{16} = \frac{27}{128} \).
Key Concepts
Empirical Distribution FunctionBootstrap Sample MeanBootstrap MaximumBinomial Probability
Empirical Distribution Function
The empirical distribution function (EDF) is a crucial concept in statistics for understanding the nature of your data. Imagine you have a dataset with elements, as in our case: \( \{1, 3, 4, 6\} \). The EDF is simply a step function that increases by \( \frac{1}{n} \) at each of the \( n \) data points.
- Each data point in the distribution, therefore, has an equal probability of being chosen. In our dataset, each number has a \( \frac{1}{4} \) chance of selection.
- This function helps in generating bootstrap samples, which replicate the sampling distribution by repeatedly drawing samples from the data.
Bootstrap Sample Mean
The bootstrap sample mean is a way to estimate the average of a bootstrapped dataset. In bootstrap sampling, we repeatedly resample data with replacement.
- For instance, to find the probability that the bootstrap sample mean equals 1, all elements in the sample must be 1.
- Because each element in the dataset has a \( \frac{1}{4} \) chance of being picked, the probability that every selected value is 1 is \( \left( \frac{1}{4} \right)^4 \) or \( \frac{1}{256} \).
Bootstrap Maximum
The term bootstrap maximum refers to the largest value in a bootstrapped dataset derived from an empirical distribution. To determine the probability that the maximum of the sample is 6, you must consider that at least one of the draws is a 6.
- The probability that a particular draw does not result in a 6 is \( \frac{3}{4} \), because there are three other numbers (1, 3, 4).
- The probability that none of the four draws results in a 6 is \( \left( \frac{3}{4} \right)^4 \).
- Thus, to find the probability that at least one is 6, we calculate \( 1 - \left( \frac{3}{4} \right)^4 \), which equals \( \frac{175}{256} \).
Binomial Probability
Binomial probability comes into play when you want to determine the chances of a particular number of successes in a sequence of independent experiments. In our example, we look at the probability of having exactly two elements in a bootstrap sample that are less than 2.
- The only element less than 2 in the dataset is 1, and its probability is \( \frac{1}{4} \).
- To compute the probability of exactly two ones being selected, we use the binomial coefficient: \( \binom{4}{2} \), which counts how many ways we can choose 2 successes out of 4 trials.
- The full probability calculation is \( \binom{4}{2} \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^2 \), giving us \( \frac{27}{128} \).
Other exercises in this chapter
Problem 1
\(\square\) We generate a bootstrap dataset \(x_{1}^{*}, x_{2}^{*}, \ldots, x_{6}^{*}\) from the empirical distribution function of the dataset \(\begin{array}{
View solution Problem 3
boxplus\( We generate a bootstrap dataset \)x_{1}^{*}, x_{2}^{*}, \ldots, x_{10}^{*}\( from the empirical distribution function of the dataset \)\begin{array}{l
View solution Problem 5
\(\square\) Suppose we have a dataset \(\begin{array}{lll}0 & 3 & 6\end{array}\) which is the realization of a random sample from a distribution function \(F\).
View solution Problem 6
Suppose that the dataset \(x_{1}, x_{2}, \ldots, x_{n}\) is a realization of a random sample from an \(\operatorname{Exp}(\lambda)\) distribution with distribut
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