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Let \(Y_{1}

Short Answer

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After computing the above expression, we find that \(P\left(Y_{4} \geq 3\right)\approx 0.5665\).

Step by step solution

01

Define the Sample

Define \(Y_{4}\) as the biggest order statistic or the maximum of four identically distributed exponential random variables with rate 1, namely \(Y_{1}, Y_{2}, Y_{3}, Y_{4}\) .
02

Define the Joint PDF

In the case of exponential distribution with rate \(\lambda=1\), the joint PDF of \(n\) sample size distributed according to exponential distribution is given by \(f(y_{1}, y_{2}, ..., y_{n}) = n!e^{-ny_{n}}\) for \(0<y_{1}<y_{2}<...<y_{n}<\infty\).
03

Calculate the CDF

The Cumulative Distributive Function (CDF) of the maximum (or largest order statistic), \(Y_{4}\), of a random sample of size 4 from the exponential distribution is given by \(F_{Y_{4}}(y) =[1-e^{-y}]^{4}\) for \(0<y<\infty\), as it is the product of four identical exponential distributions.
04

Use the CDF to Calculate the Probability

To find \(P\left(Y_{4} \geq 3\right)\), we need to use the result from step 3. This is because \(P\left(Y_{4} \geq 3\right) = 1 - P\left(Y_{4} < 3\right) = 1 - F_{Y_{4}}(3)\). Substituting the values into the CDF, we get \(P\left(Y_{4} \geq 3\right) = 1 - [1 - e^{-3}]^{4}\).

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Key Concepts

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

Exponential Distribution
The exponential distribution is widely used to model the time until an event occurs, such as the lifespan of an electronic component or the time between phone calls at a call center. It is characterized by its constant hazard rate, which means that the likelihood of the event occurring in the next moment is independent of how much time has already elapsed.

Mathematically, the probability density function (pdf) for the exponential distribution is defined as:\[ f(x; \lambda) = \begin{cases} \lambda e^{-\lambda x}, & x \geq 0 \ 0, & x < 0 \end{cases} \]where \( \lambda > 0 \) is the rate parameter. The rate parameter defines how quickly events occur. For example, if \( \lambda \) is large, events occur more frequently, leading to a steeper decline in the pdf graph.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) represents the probability that a random variable takes on a value less than or equal to a specific value. For the exponential distribution, the CDF is derived by integrating the pdf from negative infinity to the value of interest. It's given by:\[ F(x; \lambda) = \begin{cases} 1 - e^{-\lambda x}, & x \geq 0 \ 0, & x < 0 \end{cases} \]With the CDF, you can find probabilities for intervals. For instance, to find the probability that the random variable \( X \) is between two values \( a \) and \( b \), one would calculate \( F(b) - F(a) \). The CDF can also be used to find the median or percentile of a given distribution.
Probability Density Function (pdf)
The probability density function (pdf) is a fundamental concept in statistics that specifies the likelihood of a random variable taking on a particular value. For continuous distributions, like the exponential distribution, the pdf gives the height of the probability distribution at any point.

The area under the curve of the pdf over a particular interval represents the probability of the random variable falling within that interval. However, it's important to note that the value of the pdf itself is not a probability but rather a density. This is why looking at the area under the curve (which can be found using the CDF) is essential for calculating probabilities.
Random Sample
In statistics, a random sample refers to a set of observations that is drawn from a larger population, where each member of the population has an equal chance of being selected. This ensures that the sample is representative of the population, making the conclusions drawn from the sample more likely to be valid for the entire population.

A random sample is used to estimate population parameters (like the mean or variance) and to test hypotheses about the population. Random sampling helps to eliminate bias and allows for the use of probability theory to make inferences about the population, which is the basis for many statistical methods and analyses.

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Most popular questions from this chapter

Let \(X\) have a Poisson distribution with mean \(\theta\). Consider the simple hypothesis \(H_{0}: \theta=\frac{1}{2}\) and the alternative composite hypothesis \(H_{1}: \theta<\frac{1}{2} .\) Thus \(\Omega=\left\\{\theta: 0<\theta \leq \frac{1}{2}\right\\}\). Let \(X_{1}, \ldots, X_{12}\) denote a random sample of size 12 from this distribution. We reject \(H_{0}\) if and only if the observed value of \(Y=X_{1}+\cdots+X_{12} \leq 2\). Show that the following \(\mathrm{R}\) code graphs the power function of this test: theta=seq \((.1, .5, .05) ;\) gam=ppois \((2\), theta*12 \()\) plot (gam "theta, pch=" ", xlab=expression(theta), ylab=expression(gamma)) lines (gam "theta) Run the code. Determine the significance level from the plot.

Two numbers are selected at random from the interval \((0,1) .\) If these values are uniformly and independently distributed, by cutting the interval at these numbers, compute the probability that the three resulting line segments can form a triangle.

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a \(N(0,1)\) distribution. Then the probability that the random interval \(\bar{X} \pm t_{\alpha / 2, n-1}(s / \sqrt{n})\) traps \(\mu=0\) is \((1-\alpha)\). To verify this empirically, in this exercise, we simulate \(m\) such intervals and calculate the proportion that trap 0, which should be "close" to \((1-\alpha)\). (a) Set \(n=10\) and \(m=50\). Run the \(\mathrm{R}\) code mat=matrix (rnorm \((\mathrm{m} * \mathrm{n}), \mathrm{n} \overline{\mathrm{col}=\mathrm{n}})\) which generates \(m\) samples of size \(n\) from the \(N(0,1)\) distribution. Each row of the matrix mat contains a sample. For this matrix of samples, the function below computes the \((1-\alpha) 100 \%\) confidence intervals, returning them in a \(m \times 2\) matrix. Run this function on your generated matrix mat. What is the proportion of successful confidence intervals? (b) Run the following code which plots the intervals. Label the successful intervals. Comment on the variability of the lengths of the confidence intervals.

Let \(x_{1}, x_{2}, \ldots, x_{n}\) be the values of a random sample. A bootstrap sample, \(\mathbf{x}^{* \prime}=\left(x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}\right)\), is a random sample of \(x_{1}, x_{2}, \ldots, x_{n}\) drawn with replacement. (a) Show that \(x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}\) are iid with common cdf \(\widehat{F}_{n}\), the empirical cdf of \(x_{1}, x_{2}, \ldots, x_{n}\) (b) Show that \(E\left(x_{i}^{*}\right)=\bar{x}\) (c) If \(n\) is odd, show that median \(\left\\{x_{i}^{*}\right\\}=x_{((n+1) / 2)}\). (d) Show that \(V\left(x_{i}^{*}\right)=n^{-1} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\).

Consider the sample of data (data are in the file ex4.4.3data.rda): \(\begin{array}{rrrrrrrrrrr}13 & 5 & 202 & 15 & 99 & 4 & 67 & 83 & 36 & 11 & 301 \\ 23 & 213 & 40 & 66 & 106 & 78 & 69 & 166 & 84 & 64 & \end{array}\) (a) Obtain the five-number summary of these data. (b) Determine if there are any outliers. (c) Boxplot the data. Comment on the plot.

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