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Let \(Y_{1}

Short Answer

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The required probability \(P\left(3 \leq Y_{4}\right)\) is given by \( 1 - [1 - e^{-3}]^4\)

Step by step solution

01

Identify known variables

In this problem, we know that the sample size is 4 (n = 4), and \(Y_{4}\) represents the maximum of all 4 samples. The random samples follow an Exponential Distribution with parameter \( \lambda = 1 \), hence pdf is given as \(f(x)=e^{-x}, 0<x<infty\). We are asked to find \(P\left(3 \leq Y_{4}\right)\).
02

Understanding the probability problem

The probability that the maximum of 4 random variables is greater than or equal to 3 is equal to one minus the probability that the maximum is less than 3. It's equivalent to say all four values are less than 3. So, \(P\left(Y_{4} \geq 3\right) = 1 - P\left(Y_{4} < 3\right) = 1 - P\left(Y_{1}<3, Y_{2}<3, Y_{3}<3, Y_{4}<3\right)\).
03

Using the cumulative distribution function (CDF)

The cumulative distribution function (CDF) of an exponential distribution can be expressed as \(1- e^{- \lambda x}\). Since all variables are independent and identically distributed, the joint probability is the product of individual probabilities. Hence, \(P\left(Y_{4} < 3\right) = [1 - e^{-3}]^4\).
04

Calculate the probability

Substitute the CDF into the probability equation found in Step 2 and calculate the result: \[P\left(Y_{4} \geq 3\right) = 1 - [1 - e^{-3}]^4\]

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Key Concepts

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

Exponential Distribution
To understand the problem at hand, we first need to get acquainted with the Exponential Distribution. This distribution is often used to model waiting times or lifespans—essentially scenarios where we're interested in the time until an event occurs. An Exponential Distribution is characterized by its constant rate parameter, often denoted by \(\textstyle \( \lambda \) \). The probability density function (pdf) for this distribution is given by \(f(x) = \lambda e^{-\lambda x}\) for \(x \geq 0\), and is zero elsewhere.

The key property of the Exponential Distribution is its memorylessness, which implies that the future probability of an event occurring is independent of when it last occurred. So, in the exercise problem where the pdf is \(f(x)=e^{-x}\), the rate parameter \(\lambda\) is 1, which means events occur at a rate of 1 per time unit. Understanding the Exponential Distribution is essential for solving problems related to order statistics, where 'ordering' refers to the ranking of the sample according to size or magnitude.
Cumulative Distribution Function (CDF)
Next, we delve into the Cumulative Distribution Function (CDF), which is a cornerstone concept in the field of probability. The CDF, denoted by \(F(x)\), is the probability that a random variable \(X\) takes a value less than or equal to \(x\). It effectively measures the probability that a random event will occur by a certain point, encompassing all possible values below \(x\).

For the Exponential Distribution, the CDF is particularly straightforward, expressed as \(F(x) = 1 - e^{-\lambda x}\) for \(x \geq 0\). To fully comprehend the exercise solution, it's vital to understand that the CDF encapsulates the probability of occurrences up to a certain threshold and that it is the integral of the pdf from minus infinity to \(x\). When working with maximum order statistics, the CDF is crucial as it represents the probability of all observations falling beneath a particular value, an insight used in calculating the joint probability of independent events.
Probability
Lastly, we focus on Probability, the measure of how likely it is for a certain event to occur. Probability ranges from 0 (impossible event) to 1 (certain event), and it involves various rules such as the addition and multiplication rules to compute the likelihood of compound events.

In the context of the exercise, we utilize the concept of probability in conjunction with order statistics. The problem inquires about the likelihood that the maximum value in a random sample (here represented by \(Y_{4}\)) is greater than or equal to a certain threshold. This is interpreted as 1 minus the probability that all sampled values are below that threshold. Mathematics offers us the tools to compute this probability through the CDF of the Exponential Distribution. Remember that in practice, to obtain this probability, one must meticulously parse the intersection of events, applying the rules of probability, like those for independent events, to arrive at the desired solution.

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Most popular questions from this chapter

Let \(p\) denote the probability that, for a particular tennis player, the first serve is good. Since \(p=0.40\), this player decided to take lessons in order to increase \(p\). When the lessons are completed, the hypothesis \(H_{0}: p=0.40\) will be tested against \(H_{1}: p>0.40\) based on \(n=25\) trials. Let \(y\) equal the number of first serves that are good, and let the critical region be defined by \(C=\\{y: y \geq 13\\}\). (a) Determine \(\alpha=P(Y \geq 13 ; ; p=0.40)\). (b) Find \(\beta=P(Y<13)\) when \(p=0.60\); that is, \(\beta=P(Y \leq 12 ; p=0.60)\) so that \(1-\beta\) is the power at \(p=0.60\).

. Let \(X_{1}, \ldots, X_{n}\) be a random sample from the \(\Gamma(2, \theta)\) distribution, where \(\theta\) is unknown. Let \(Y=\sum_{i=1}^{n} X_{i}\) (a) Find the distribution of \(Y\) and determine \(c\) so that \(c Y\) is an unbiased estimator of \(\theta\). (b) If \(n=5\), show that $$ P\left(9.59<\frac{2 Y}{\theta}<34.2\right)=0.95 $$ (c) Using Part (b), show that if \(y\) is the value of \(Y\) once the sample is drawn, then the interval $$ \left(\frac{2 y}{34.2}, \frac{2 y}{9.59}\right) $$ is a \(95 \%\) confidence interval for \(\theta\). (d) Suppose the sample results in the values, $$ \begin{array}{lllll} 44.8079 & 1.5215 & 12.1929 & 12.5734 & 43.2305 \end{array} $$

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(\Gamma(1, \beta)\) distribution. (a) Show that the confidence interval \(\left(2 n \bar{X} /\left(\chi_{2 n}^{2}\right)^{(1-(\alpha / 2))}, 2 n \bar{X} /\left(\chi_{2 n}^{2}\right)^{(\alpha / 2)}\right)\) is an exact \((1-\alpha) 100 \%\) confidence interval for \(\beta\). (b) Show that value of a \(90 \%\) confidence interval for the data of the example is \((64.99,136.69)\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(N\left(\mu, \sigma^{2}\right)\), where both parameters \(\mu\) and \(\sigma^{2}\) are unknown. A confidence interval for \(\sigma^{2}\) can be found as follows. We know that \((n-1) S^{2} / \sigma^{2}\) is a random variable with a \(\chi^{2}(n-1)\) distribution. Thus we can find constants \(a\) and \(b\) so that \(P\left((n-1) S^{2} / \sigma^{2}

. For \(\alpha>0\) and \(\beta>0\), consider the following accept/reject algorithm: (1) Generate \(U_{1}\) and \(U_{2}\) iid uniform \((0,1)\) random variables. Set \(V_{1}=U_{1}^{1 / \mathrm{a}}\) and \(V_{2}=U_{2}^{1 / \beta}\) (2) Set \(W=V_{1}+V_{2}\). If \(W \leq 1\), set \(X=V_{1} / W\), else go to Step (1). (3) Deliver \(X\).

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