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Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent, exponentially distributed random variables with mean \(\beta\) a. Show that \(Y_{(1)}=\min \left(Y_{1}, Y_{2}, \ldots, Y_{\mathrm{n}}\right)\) has an exponential distribution, with mean \(\beta / n\) b. If \(n=5\) and \(\beta=2,\) find \(P\left(Y_{(1)} \leq 3.6\right)\)

Short Answer

Expert verified
a. Exponential distribution with mean \( \beta/n \). b. \( P(Y_{(1)} \leq 3.6) \approx 0.9999 \).

Step by step solution

01

Identify the Exponential Distribution Properties

The exponential distribution for each \( Y_i \) has the probability density function (pdf) given by \( f(y; \beta) = \frac{1}{\beta} e^{-y/\beta} \) for \( y \geq 0 \). The mean of \( Y_i \) is \( \beta \).
02

Find the Distribution of the Minimum

The cumulative distribution function (CDF) of \( Y_{(1)} = \min(Y_1, Y_2, \ldots, Y_n) \) is given by \( P(Y_{(1)} > y) = P(Y_1 > y) P(Y_2 > y) \ldots P(Y_n > y) \). For exponential rv's, \( P(Y_i > y) = e^{-y/\beta} \). Thus, \( P(Y_{(1)} > y) = (e^{-y/\beta})^n = e^{-ny/\beta} \).
03

Derive the CDF of the Minimum

The CDF of \( Y_{(1)} \) is \( P(Y_{(1)} \leq y) = 1 - P(Y_{(1)} > y) = 1 - e^{-ny/\beta} \). This shows that \( Y_{(1)} \) follows an exponential distribution with mean \( \beta/n \) since its CDF matches that of an exponential distribution with parameter \( \lambda = n/\beta \).
04

Apply the Exponential Distribution For \( n=5 \)

Given \( n = 5 \) and \( \beta = 2 \), then the mean of \( Y_{(1)} \) is \( \beta/n = 2/5 \). Thus, \( Y_{(1)} \) has an exponential distribution with rate \( \lambda = 5/2 \).
05

Calculate the Probability for Part b

We need to find \( P(Y_{(1)} \leq 3.6) \). Using the CDF \( 1 - e^{-ny/\beta} \), where \( y = 3.6 \), \( n = 5 \), and \( \beta = 2 \): \( P(Y_{(1)} \leq 3.6) = 1 - e^{-5 \times 3.6 / 2} = 1 - e^{-9} \).
06

Simplify the Expression for Probability

Calculate \( e^{-9} \) using a calculator or table. Thus, \( P(Y_{(1)} \leq 3.6) = 1 - e^{-9} = 1 - 0.0001234098 \approx 0.9998765902 \).

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Key Concepts

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

Probability Density Function
When discussing exponential distributions, a fundamental concept is the Probability Density Function (PDF). The PDF for an exponentially distributed random variable, such as each of the random variables in the set \( Y_1, Y_2, \ldots, Y_n \), is expressed mathematically by:
  • \( f(y; \beta) = \frac{1}{\beta} e^{-y/\beta} \) for \( y \geq 0 \).
This specific function captures how probability is distributed across different values in a continuous range. For the exponential distribution, the parameter \( \beta \) represents the mean of the distribution. Consequently, the shape of the PDF curve indicates how likely different outcomes are, with higher likelihoods signified by higher points on the curve.
Understanding the behavior of a PDF helps in calculating probabilities for different ranges of outcomes. In exponential distributions, it is common to look at the decay rate, which illustrates how quickly probabilities decline as you move away from the mean.
Cumulative Distribution Function
Another significant feature in probability theory is the Cumulative Distribution Function (CDF). The CDF of an exponential distribution helps determine the probability that the random variable is less than or equal to some value. For exponential distributions, this is formulated as:
  • \( F(y) = 1 - e^{-y/\beta} \).
The CDF accumulates probabilities up to a certain value of \( y \), providing insights into the probability of events falling within specified ranges. For minimum random variables like \( Y_{(1)} = \min(Y_1, Y_2, \ldots, Y_n) \), the CDF needs to account for all the independent variables.

Minimum Random Variable CDF

Utilizing the properties of independent random variables, the CDF of the minimum random variable \( Y_{(1)} \) is given by the formula:
  • \( P(Y_{(1)} \leq y) = 1 - e^{-ny/\beta} \)
This alteration reflects that the rate parameter is now adjusted by the number of variables, leading to an exponential distribution with mean \( \beta/n \). Understanding this CDF is key to solving problems like calculating probabilities for the minimum of exponential random variables.
Minimum of Random Variables
The behavior of the minimum of a set of random variables \( Y_{1}, Y_{2}, \ldots, Y_{n} \) is intriguing because it often reflects different statistical properties compared to individual variables. When these are independent and exponentially distributed with a common mean \( \beta \), the minimum, \( Y_{(1)} = \min(Y_{1}, Y_{2}, \ldots, Y_{n}) \), also yields an exponential distribution.
This distinctive property arises due to the nature of the exponential distribution—the only continuous distribution possessing the memoryless property. The CDF of \( Y_{(1)} \) becomes \( 1 - (e^{-y/\beta})^n = 1 - e^{-ny/\beta} \), pinpointing that \( Y_{(1)} \) has a new mean \( \beta/n \).
Grasping this concept is essential when investigating scenarios where identifying the first occurrence of an event among multiple trials is desired, a classic case of finding minima in statistical distributions.
Probability Calculation
Probability Calculation is the final aspect to explore, particularly as it applies to real-world scenarios. Let's solve a practical probability calculation using the fact that \( Y_{(1)} \) follows an exponential distribution with a mean \( \beta/n \), or a rate \( \lambda = n/\beta \). When \( n = 5 \) and \( \beta = 2 \), \( \lambda \) becomes \( 5/2 \).
For probability tasks like finding \( P(Y_{(1)} \leq 3.6) \), deploy the CDF of \( Y_{(1)} \):
  • \( 1 - e^{-n \times y/\beta} \)
Inserting the problem's particulars:
  • \( P(Y_{(1)} \leq 3.6) = 1 - e^{-5 \times 3.6 / 2} \)
Upon calculation, a negligible value for \( e^{-9} \) transforms the CDF to nearly \( 1 \), a strong indicator that the probability of \( Y_{(1)} \leq 3.6 \) is approximately \( 0.9999 \).
This approach elucidates how the exponential distribution elegantly simplifies solving such probability challenges.

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Most popular questions from this chapter

A parachutist wants to land at a target \(T\), but she finds that she is equally likely to land at any point on a straight line \((A, B),\) of which \(T\) is the midpoint. Find the probability density function of the distance between her landing point and the target.

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