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Let \(Y_{2}\) and \(Y_{n-1}\) denote the second and the \((n-1)\) st order statistics of a random sample of size \(n\) from a distribution of the continuous type having a distribution function \(F(x)\). Compute \(P\left[F\left(Y_{n-1}\right)-F\left(Y_{2}\right) \geq p\right]\), where \(0

Short Answer

Expert verified
The probability that \(P\left[F\left(Y_{n-1}\right)-F\left(Y_{2}\right) \geq p\right]\) is equal to \(1-(1-p)^{n-1}\).

Step by step solution

01

Understanding Order Statistics

Order statistics are the values of an observed sample that have been sorted in ascending or descending order. In this case, \(Y_{2}\) refers to the second smallest value and \(Y_{n-1}\) refers to the second largest value in the sample.
02

Calculate the Difference in CDFs

The problem asks to compute \(P\left[F\left(Y_{n-1}\right)-F\left(Y_{2}\right) \geq p\right]\). By definition of the cumulative distribution function (CDF), \(F\left(Y_{n-1}\right)\) is the probability that a random variable \(Y_{n-1}\) takes a value less than or equal to \(Y_{n-1}\) and \(F\left(Y_{2}\right)\) is the probability that the random variable \(Y_{2}\) takes a value less than or equal to \(Y_{2}\). The difference \(F\left(Y_{n-1}\right)-F\left(Y_{2}\right)\) represents the probability that \(Y_{2} \leq Y \leq Y_{n-1}\).
03

Understand the Probability Equation

The equation \(P\left[F\left(Y_{n-1}\right)-F\left(Y_{2}\right) \geq p\right]\) essentially asks, 'What is the probability that the difference between the probabilities that \(Y_{n-1}\) and \(Y_{2}\) take values less than or equal to themselves is greater than or equal to p?' Because 0<p<1, this is only possible when the values for \(Y_{n-1}\) and \(Y_{2}\) are spread farther apart. In other words, the larger the difference between the values of \(Y_{n-1}\) and \(Y_{2}\), the more likely that the inequality holds true.
04

Compute the Probability

The joint probability density function of the order statistics \(Y_{2}\) and \(Y_{n-1}\) is given by \(f\left(y_{2},y_{n-1}\right) = n(n-1)(F(y_{n-1})-F(y_{2}))^{n-3}f(y_{2})f(y_{n-1})\), where f(x) denotes the PDF. Take the integral of this function from \(p\) to \(1\) to obtain the desired probability. The function is symmetric in \(y_{2}\) and \(y_{n-1}\), hence the integral is equal to two times the integral from \(p\) to \(1-p\). The resulting expression simplifies to \(1-(1-p)^{n-1}\).

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Key Concepts

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

Cumulative Distribution Function (CDF)
The Cumulative Distribution Function, or CDF, is a crucial concept in statistics and probability theory. It helps us understand the distribution of random variables. A CDF is a function that gives the probability that a random variable is less than or equal to a certain value.

Imagine a continuous random variable like height, where values are measured on a continuum. The CDF will output the probability that a randomly selected measurement is less than or equal to a specific height. Mathematically, for a random variable X, the CDF is denoted as \( F(x) = P(X \leq x) \). It provides a view of the distribution of probabilities, accumulating from zero up to one as you move along the range of possible values.

The CDF is non-decreasing and right-continuous, meaning it never decreases in value and is smooth without abrupt jumps. Knowing the CDF helps in understanding the probabilities of an interval. For order statistics, like our earlier exercise, the difference \( F(Y_{n-1}) - F(Y_{2}) \) allows us to understand the spread between the second smallest and the second largest values within a sample.

In essence, the CDF serves as a tool that connects the concept of probability with the range of outcomes possible for the variable, giving a complete picture of the probability landscape.
Probability Density Function (PDF)
While the Cumulative Distribution Function (CDF) gives us cumulative probabilities, the Probability Density Function, or PDF, offers insight into the probability of a random variable taking a specific value. For continuous variables, the PDF is a function that shows the likelihood of outcomes within a particular range.

The PDF is denoted as \( f(x) \) for a random variable X, with the integral of \( f(x) \) over a range providing the probability that X falls within that range. In simpler terms, while the PDF gives us the density of the probability at different points, the area under the PDF curve across an interval gives the actual probability of the event.

It's important to note that the PDF itself doesn't give probabilities; probabilities are found using integrals of the PDF over intervals. This is because continuous random variables have infinite outcomes in any specific point; hence, the probability at a point is zero, and we must look for ranges instead.

In the context of order statistics and probability, we often need to use PDFs to determine the joint probability of several variables. For example, in finding \( P[F(Y_{n-1}) - F(Y_2) \geq p] \), we use the joint PDF of our order statistics \( Y_2 \) and \( Y_{n-1} \). Understanding how PDFs work is fundamental in calculating different probabilities and making sense of data in statistical analysis.
Random Sample
In statistics, a random sample is a subset of individuals chosen from a larger set, or population, in an unbiased manner. The idea is that every possible sample of the same size has an equal chance of being selected. This randomness is what ensures that the sample is representative of the whole population.

Random sampling is crucial because it helps prevent bias and enables us to make inferences about the population from which the sample is drawn. When we say that a sample was obtained randomly, this means the selection was based on probability theory, ensuring that each member of the population had a known and non-zero chance of selection.

In exercises like those involving order statistics, the concept of a random sample is pivotal. It ensures that the values we are analyzing are not skewed or systematically biased in any way. For example, when analyzing order statistics \( Y_2 \) and \( Y_{n-1} \), it becomes useful to assume the sample's randomness to correctly apply probability distributions and calculate probabilities, like \( P[F(Y_{n-1}) - F(Y_2) \geq p] \).

Thus, understanding how and why we obtain random samples is essential in statistical analysis and ensures the reliability and validity of the statistical conclusions we draw.

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Most popular questions from this chapter

For the proof of Theorem 5.8.1, we assumed that the cdf was strictly increasing over its support. Consider a random variable \(X\) with cdf \(F(x)\) which is not strictly increasing. Define as the inverse of \(F(x)\) the function $$ F^{-1}(u)=\inf \\{x: F(x) \geq u\\}, \quad 0

Let \(\bar{X}\) and \(\bar{Y}\) be the means of two independent random samples, each of size \(n\), from the respective distributions \(N\left(\mu_{1}, \sigma^{2}\right)\) and \(N\left(\mu_{2}, \sigma_{2}\right)\), where the common variance is known. Find \(n\) such that $$ P\left(\bar{X}-\bar{Y}-\sigma / 5<\mu_{1}-\mu_{2}<\bar{X}-\bar{Y}+\sigma / 5\right)=0.90 $$

Let \(\bar{X}\) denote the mean of a random sample of size 25 from a gamma-type distribution with \(\alpha=4\) and \(\beta>0 .\) Use the Central Limit Theorem to find an approximate \(0.954\) confidence interval for \(\mu\), the mean of the gamma distribution. Hint: \(\quad\) Use the random variable \((\bar{X}-4 \beta) /\left(4 \beta^{2} / 25\right)^{1 / 2}=5 \bar{X} / 2 \beta-10\).

Let a random sample of size 17 from the normal distribution \(N\left(\mu, \sigma^{2}\right)\) yield \(\bar{x}=4.7\) and \(s^{2}=5.76 .\) Determine a 90 percent confidence interval for \(\mu .\)

Suppose the number of customers \(X\) that enter a store between the hours 9:00 AM and 10:00 AM follows a Poisson distribution with parameter \(\theta\). Suppose a random sample of the number of customers for 10 days results in the values, $$ \begin{array}{llllllllll} 9 & 7 & 9 & 15 & 10 & 13 & 11 & 7 & 2 & 12 \end{array} $$ Based on these data, obtain an unbiased point estimate of \(\theta .\) Explain the meaning of this estimate in terms of the number of customers.

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