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If \(X_{1}, \ldots, X_{n}\) are independent exponentially distributed random variables, each with parameter \(\lambda\), prove that the random variable whose value is given by the minimum of \(\left\\{X_{1}, \ldots, X_{n}\right\\}\) also has an exponential distribution, with parameter \(n \lambda\). In particular, if a complex piece of machinery consists of six parts, each of which has an exponential distribution of time to failure with mean 2000 hours, and if the machine fails as soon as any of its parts fail, find the probability that the time to failure exceeds 300 hours.

Short Answer

Expert verified
The probability that the machine lasts more than 300 hours is approximately 0.4066.

Step by step solution

01

Understand Exponential Distribution

An exponential distribution with parameter \(\lambda\) has the probability density function \(f(x) = \lambda e^{-\lambda x}\) for \(x \geq 0\). This distribution describes the time between events in a Poisson process.
02

Express the Minimum of Exponential Variables

The minimum of independent exponential random variables \(X_1, X_2, \ldots, X_n\) with parameter \(\lambda\) is another exponential random variable, denoted \(Y = \min(X_1, X_2, \ldots, X_n)\).
03

Use the Property of Minimum

The probability that the minimum of these variables is greater than \(y\) is given by: \(P(Y > y) = P(X_1 > y) P(X_2 > y) \cdots P(X_n > y)\) since all \(X_i\) are independent.
04

Calculate Probability for Each \(X_i\)

Since each \(X_i\) is exponentially distributed, the survival function is \(P(X_i > y) = e^{-\lambda y}\). Thus, \(P(Y > y) = (e^{-\lambda y})^n = e^{-n\lambda y}\).
05

Identify Parameter of Combined Exponential

This is the survival function of an exponential distribution with parameter \(n\lambda\), proving that \(Y\) is an exponential random variable with parameter \(n\lambda\).
06

Apply to Machinery System

Given \(\lambda = \frac{1}{2000}\) hours, and \(n = 6\), the new parameter will be \(6 \times \frac{1}{2000} = \frac{6}{2000}\) for the minimum failure time of the system.
07

Find Probability of Exceeding 300 Hours

The probability that \(Y\) exceeds 300 hours is \(P(Y > 300) = e^{-(\frac{6}{2000} \times 300)}\).
08

Compute the Probability

Calculate: \(P(Y > 300) = e^{-0.9}\). Using a calculator, \(e^{-0.9} \approx 0.4066\).

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Key Concepts

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

Minimum of Random Variables
When dealing with multiple random variables, particularly independent ones, an interesting property emerges: the minimum of these variables. Imagine you have several scenarios where something could "fail," and you want to know the earliest possible failure time. If each scenario follows an exponential distribution, a remarkable fact emerges: the minimum of these times is also governed by the exponential distribution.

Suppose we have a collection of independent exponential random variables, each characterized by a parameter \( \lambda \). The probability of any one of these variables exceeding a certain threshold is given by its survival function. Since these variables are independent, the overall probability that the minimum is greater than the threshold becomes a product of each variable's individual probabilities. As we apply this logic, we find ourselves back at an exponential distribution, but with a new parameter \( n \lambda \), where \( n \) is the number of variables. It's fascinating to see how adding more random variables simply scales the original distribution.
Poisson Process
A Poisson process is a fundamental concept in probability and statistics, often used to model random events happening over time. It is characterized by the assumption that events occur independently, with a certain average rate. The exponential distribution fits naturally into this framework, as it represents the time between consecutive events in a Poisson process.

In a real-world context, consider machinery maintenance or queues at a store. The exponential distribution helps model the unpredictability of events, capturing the essence of "randomness" in waiting times. By understanding these underlying stochastic processes, one can make predictions and decisions with greater confidence.
Survival Function
The survival function is a crucial tool in statistical analysis, especially when dealing with life testing and reliability analysis. It answers the question, "What is the probability that a variable takes a value greater than a certain threshold?" In mathematical terms, for a random variable \( X \), this function is expressed as \( P(X > y) \).

For an exponential distribution, the survival function takes a beautiful and simple form: \( e^{-\lambda y} \). This succinct expression provides a way to compute the probability of "surviving" past time \( y \).

Understanding and utilizing this function allows for insights into processes where time until an event is a factor. By calculating the survival function for each of the independent variables and multiplying these probabilities together, we identify the survival function of the minimum of those variables—revealing yet another exponential nature.
Parameter Calculation
In mathematical modeling, defining accurate parameters is key. Especially when dealing with exponential distributions, parameter calculation is crucial. If you are given a mean time to event, or failure (like 2000 hours), finding the parameter \( \lambda \) becomes essential. The relationship is \( \lambda = \frac{1}{\text{mean}} \).

Once the parameter of a single random variable is established, if you have multiple such variables, the new combined parameter after taking the minimum is \( n \lambda \). For example, with six components each having \( \lambda = \frac{1}{2000} \), the combined parameter becomes \( \frac{6}{2000} \). This new parameter helps recalibrate our expectations from the single to the multiple scenarios.

This calculation ties directly into the broader understanding of the system's reliability and expected performance, such as predicting the time before a system failure.

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