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Let \(Y_{1}

Short Answer

Expert verified
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

This exercise obtains a useful identity for the cdf of a Poisson cdf. (a) Use Exercise \(3.3 .5\) to show that this identity is true: $$ \frac{\lambda^{n}}{\Gamma(n)} \int_{1}^{\infty} x^{n-1} e^{-x \lambda} d x=\sum_{j=0}^{n-1} e^{-\lambda} \frac{\lambda^{j}}{j !} $$ for \(\lambda>0\) and \(n\) a positive integer. Hint: Just consider a Poisson process on the unit interval with mean \(\lambda\). Let \(W_{n}\) be the waiting time until the \(n\) th event. Then the left side is \(P\left(W_{n}>1\right)\). Why? (b) Obtain the identity used in Example \(4.3 .3\), by making the transformation \(z=\lambda x\) in the above integral.

Let \(Y_{1}0\), provided that \(x \geq 0\), and \(f(x)=0\) elsewhere. Show that the independence of \(Z_{1}=Y_{1}\) and \(Z_{2}=Y_{2}-Y_{1}\) characterizes the gamma pdf \(f(x)\), which has parameters \(\alpha=1\) and \(\beta>0 .\) That is, show that \(Y_{1}\) and \(Y_{2}\) are independent if and only if \(f(x)\) is the pdf of a \(\Gamma(1, \beta)\) distribution. Hint: Use the change-of-variable technique to find the joint pdf of \(Z_{1}\) and \(Z_{2}\) from that of \(Y_{1}\) and \(Y_{2}\). Accept the fact that the functional equation \(h(0) h(x+y) \equiv\) \(h(x) h(y)\) has the solution \(h(x)=c_{1} e^{c_{2} x}\), where \(c_{1}\) and \(c_{2}\) are constants.

Obtain the inverse function of the cdf of the Laplace pdf, given by \(f(x)=\) \((1 / 2) e^{-|x|}\), for \(-\infty

Suppose the pdf \(f(x)\) is symmetric about 0 with cdf \(F(x)\). Show that the probability of a potential outlier from this distribution is \(2 F\left(4 q_{1}\right)\), where \(F^{-1}(0.25)=\) \(q_{1}\) Use this to obtain the probability that an observation is a potential outlier for the following distributions. (a) The underlying distribution is normal. Use the \(N(0,1)\) distribution. (b) The underlying distribution is logistic; that is, the pdf is given by $$ f(x)=\frac{e^{-x}}{\left(1+e^{-x}\right)^{2}}, \quad-\infty

Recall For the baseball data (bb.rda), 15 out of 59 ballplayers are lefthanded. Let \(p\) be the probability that a professional baseball player is left-handed. Determine an exact \(90 \%\) confidence interval for \(p .\) Show first that the equations to be solved are: $$ \sum_{j=0}^{14}\left(\begin{array}{c} n \\ j \end{array}\right) \underline{\theta}^{j}(1-\underline{\theta})^{n-j}=0.95 \text { and } \sum_{j=0}^{15}\left(\begin{array}{c} n \\ j \end{array}\right) \bar{\theta}^{j}(1-\bar{\theta})^{n-j}=0.05 $$ Then do the following steps to obtain the confidence interval. (a) Show that \(0.10\) and \(0.17\) bracket the solution to the first equation. (b) Show that \(0.34\) and \(0.38\) bracket the solution to the second equation. (c) Then use the \(\mathrm{R}\) function binomci. \(\mathrm{r}\) to solve the equations.

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