/*! 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 21 Let \(X_{(i)}, i=1, \ldots, n\),... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 21

Let \(X_{(i)}, i=1, \ldots, n\), denote the order statistics from a set of \(n\) uniform. \((0,1)\) random variables and note that the density function of \(X_{(i)}\) is given by $$ f(x)=\frac{n !}{(i-1) !(n-i) !} x^{i-1}(1-x)^{n-i} \quad 0

Short Answer

Expert verified
The variance of the order statistic X(i), for i=1,...,n, is given by: $$ Var(X_{(i)}) = \frac{i(n-i)}{(n+1)^2(n+2)} $$ The value of i that minimizes the variance is i=1, and the value of i that maximizes the variance is i=n.

Step by step solution

01

Find the expected value of X(i)

To calculate the expected value (mean) of X(i), we can use the formula: $$ E[X_{(i)}] = \int_0^1 x f(x) dx $$ where f(x) is the density function of X(i). Plug the given density function into the integral: $$ E[X_{(i)}] = \int_0^1 x \frac{n !}{(i-1) !(n-i) !} x^{i-1}(1-x)^{n-i} dx $$
02

Simplify the integral and find E(X(i))

Simplify the integral by combining the x term with x^(i-1): $$ E[X_{(i)}] = \int_0^1 \frac{n !}{(i-1) !(n-i) !} x^{i}(1-x)^{n-i} dx $$ The integral can be solved by using the Beta function: $$ E[X_{(i)}] = \frac{n !}{(i-1) !(n-i) !} \frac{\Gamma(i+1) \Gamma(n-i+1)}{\Gamma(n+2)} $$ By using the property that 螕(x+1) = x螕(x), we can simplify the expression: $$ E[X_{(i)}] = \frac{i}{n+1} $$
03

Find the expected value of X(i)^2

Similar to step 1, we need to find E[X(i)^2] to calculate the variance. The formula is given as: $$ E[X_{(i)}^2] = \int_0^1 x^2 f(x) dx $$Plug in the density function of X(i) into the integral: $$ E[X_{(i)}^2] = \int_0^1 x^2 \frac{n !}{(i-1) !(n-i) !} x^{i-1}(1-x)^{n-i} dx $$
04

Simplify the integral and find E(X(i)^2)

Simplify the integral in the same way as before: $$ E[X_{(i)}^2] = \int_0^1 \frac{n!}{(i-1) !(n-i) !} x^{i+1}(1-x)^{n-i}dx $$ The integral can be solved by using the Beta function again: $$ E[X_{(i)}^2] = \frac{n!}{(i-1) !(n-i) !} \frac{\Gamma(i+2) \Gamma(n-i+1)}{\Gamma(n+3)} $$ By using the property that 螕(x+1) = x螕(x), we can simplify the expression: $$ E[X_{(i)}^2] = \frac{i(i+1)}{(n+1)(n+2)} $$
05

Calculate the Variance of X(i)

Now, we can use the formula Var(X) = E(X^2) - (E(X))^2 to calculate the variance of X(i): $$ Var(X_{(i)}) = \frac{i(i+1)}{(n+1)(n+2)} - \left(\frac{i}{n+1}\right)^2 $$Simplify the expression: $$ Var(X_{(i)}) = \frac{i(n-i)}{(n+1)^2(n+2)} $$This is the variance of the order statistic X(i), for i=1,...,n.
06

Determine which value of i minimizes and maximizes the Variance of X(i)

To find the minimum and maximum variances, first observe that Var(X(i)) is a quadratic function in i. Since the quadratic term in Var(X(i)) has a positive coefficient, the function has a shape of a parabola opened upwards. Var(X(i)) is clearly increasing in the interval 1 鈮� i 鈮� n/2, and decreasing for n/2 鈮� i 鈮� n. Since X(1) and X(n) are the first and last order statistics, they have the smallest and largest variances, respectively. Thus, i=1 minimizes the variance i=n maximizes the variance

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Key Concepts

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

Uniform Distribution
Uniform distribution is a type of probability distribution where every outcome in a certain range is equally likely. The distribution is characterized by two parameters, the lower limit (a) and the upper limit (b). In the case of a uniform distribution on the interval (0, 1), every number between 0 and 1 has the same probability of occurring. This is described by the probability density function (PDF) as:\[f(x) = 1 \text{ for } 0
Variance Calculation
Variance is a measure of how much variability exists within a set of data points, or in this context, a random variable. For order statistics of uniform distributions, variance answers the question of how much each order statistic is likely to 'spread' from its expected value.When calculating the variance of an order statistic, we need:
  • The expected value of each order statistic \(E[X_{(i)}]\)
  • The expected value of the square \(E[X_{(i)}^2]\)
Using these values, variance can be found by:\[\text{Var}(X_{(i)}) = E[X_{(i)}^2] - (E[X_{(i)}])^2\]This formula captures the essence of variance as the average of the squared differences from the mean. Simplifying the expression for the variance, as done in the solution, is crucial for understanding the distribution of the order statistics over many trials in a sample.
Beta Function
The Beta function is a special function that is key in solving integrals like those needed for order statistics of uniform distributions. It is defined as:\[B(x, y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt\]This function is crucial for evaluating the expected values involving products of powers of the variable and the probability density function of order statistics. Some properties of the Beta function include:
  • It can be expressed using the Gamma function: \(B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\)
  • The symmetry property: \(B(x, y) = B(y, x)\)
In order statistics, recognizing that integrals can be rewritten in terms of the Beta function allows for easier computations, simplifying complex problems into more manageable parts.
Expected Value
Expected value, or mean, is a fundamental concept in probability theory that represents the 'center' of a random variable's distribution. For order statistics derived from a uniform distribution, the expected value provides insight into where each order statistic typically lies along the interval.The expected value for an order statistic \(X_{(i)}\) is calculated as:\[E[X_{(i)}] = \frac{i}{n+1}\]This formula tells us that the expected value depends on the order \(i\) and the total number of variables \(n\). The position of the expected value progresses linearly from the lower bound (close to 0) just as the order statistic moves towards higher ranks.The importance of expected value lies in its use for further calculations, such as variance, and its role in summarizing a random variable's distribution without needing to consider all individual outcomes.

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Most popular questions from this chapter

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 \(m\).

Let \(A_{1}, A_{2}, \ldots, A_{n}\) be events, and let \(N\) denote the number of them that occur. Also, let \(I=1\) if all of these events occur, and let it be 0 otherwise. Prove Bonferroni's inequality, namely that $$ P\left(A_{1} \cdots A_{n}\right) \geq \sum_{i=1}^{n} P\left(A_{i}\right)-(n-1) $$ HINT: Argue first that \(N \leq n-1+I\).

Consider \(n\) independent trials, the ith of which results in a success with probability \(P_{i}\) (a) Compute the expected number of successes in the \(n\) trials - call it \(\mu\). (b) For fixed value of \(\mu_{1}\) what choice of \(P_{1}, \ldots, P_{n}\) maximizes the variance of the number of successes? (c) What choice minimizes the variance?

A group of \(n\) men and \(m\) women are lined up at random. Determine the expected number of men that have a woman on at least one side of them.

Repeat Problem 64 when the proportion of the population having a value of \(\lambda\) less than \(x\) is equal to \(1-e^{-x}\).
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