Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Let \(X_{1}, \ldots, X_{n}\) be an exponential random sample with density \(\lambda \exp (-\lambda x), x>0, \lambda>0\) For simplicity suppose that \(n=m r\). Let \(Y_{1}\) be the total time at risk from time zero to the \(r\) th failure, \(Y_{2}\) be the total time at risk between the \(r\) th and the \(2 r\) th failure, \(Y_{3}\) the total time at risk between the \(2 r\) th and \(3 r\) th failures, and so forth. (a) Let \(X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}\) be the ordered values of the \(X_{j}\). Show that the joint density of the order statistics is $$ f_{X_{(1)}, \ldots, X_{(n)}}\left(x_{1}, \ldots, x_{n}\right)=n ! f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right), \quad x_{1}

Short Answer

Expert verified
The joint density is shown by \(n! f(x_1) \cdots f(x_n)\). Spacings \(Z_j\) are independent exponentials; hence, \(Y_j\) is gamma-distributed, with \(\log Y_j\) offering constant variance plots to estimate \(\lambda(t)\).

Step by step solution

01

Understanding the Joint Density of Order Statistics

The exponential random sample \(X_1, \dots, X_n\) has density \(f(x) = \lambda e^{-\lambda x}\) for \(x > 0\). The order statistics \(X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}\) represent the sorted values of \(X_j\). The joint density for order statistics is given by \(f_{X_{(1)}, \ldots, X_{(n)}}(x_1, \ldots, x_n) = n! f(x_1) f(x_2) \cdots f(x_n)\) with the constraint \(x_1 < x_2 < \cdots < x_n\). This accounts for all \(n!\) permutations of the sample.
02

Deriving Spacings as Independent Exponential Variables

Define \(Z_j = X_{(j)} - X_{(j-1)}\) with \(X_{(0)} = 0\), so \(X_{(1)} = Z_1, X_{(2)} = Z_1 + Z_2, \ldots, X_{(n)} = Z_1 + \cdots + Z_n\). Each \(Z_j\) represents the spacings between order statistics. Since each interval \(Z_j\) is a difference of exponential random variables, \(Z_j\sim \text{Exponential}((n+1-j)\lambda)\), making \(Z_j\) independent.
03

Showing Distribution of \(Y_j\) as Gamma

Consider \(Y_j = \sum_{k=(j-1)r+1}^{jr} Z_k\), which is the total time from \( (j-1)r \) to \(jr\) failures. Each \(Z_k\) contributing to \(Y_j\) follows an exponential distribution with different rates. The sum of \(r\) exponential variables with different rates results in a gamma distribution. Specifically, \(Y_j\sim \text{Gamma}(r, \lambda)\) with mean \(\mathbb{E}[Y_j] = r / \lambda\) and variance \(\operatorname{Var}(Y_j) = r / \lambda^2\).
04

Analyzing \(\log Y_j\) with Constant Hazard Rate

The logarithm of \(Y_j\), \(\log Y_j\), converts the gamma distribution to a different scale, maintaining independence among \(Y_j\). Since \(Y_j\) contains the sum of independent exponential variables, its log transforms to a normal-like distribution for large \(r\), with variance being consistent across \(j\), indicating constant variance distribution.
05

Estimating Non-Constant Hazard Rate Using \(\log Y_j\)

To estimate a non-constant hazard rate \(\lambda(t)\), plot values of \(\log Y_j\) against the midpoints of time intervals between \((r-1)j\)th and \(rj\)th failures. For a slowly-varying \(\lambda(t)\), these plots approximate \(\log \lambda(t)\), showing how the hazard rate changes over time.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Exponential Distribution
The exponential distribution is a fundamental probability distribution in statistics, especially useful for modeling the time between independent events that happen at a constant average rate. Its probability density function (PDF) is given by \( f(x) = \lambda e^{-\lambda x} \) for \( x > 0 \), where \( \lambda \) is the rate parameter. This rate parameter indicates how often events occur per unit time.
Some key characteristics of the exponential distribution include:
  • Memoryless property: The probability of an event occurring in the next instant is the same, regardless of how much time has already elapsed.
  • Mean: \( 1 / \lambda \)
  • Variance: \( 1 / \lambda^2 \)
Exponential distributions are commonly used to model wait times or lifetimes of certain processes, such as the time until machine failure or the duration of a phone call.
Order Statistics
Order statistics are statistics obtained from the ordered values of a random sample. If you have \( n \) observations in a dataset \( X_1, X_2, \ldots, X_n \), ordering these observations in increasing size gives you order statistics \( X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)} \).
In the context of an exponential sample, the joint density of these order statistics is strongly tied to the permutations of the sample, giving us the relation:
  • \( f_{X_{(1)}, \ldots, X_{(n)}}(x_1, \ldots, x_n) = n! f(x_1) f(x_2) \cdots f(x_n), \quad x_1 < x_2 < \cdots < x_n \)

Order statistics are useful for estimating parameters and quantifying reliability, and they play a significant role in non-parametric statistics.
Gamma Distribution
The gamma distribution is a continuous probability distribution defined for \( x > 0 \), and it is often used to model the sum of exponential random variables. Particularly, it plays a critical role in survival analysis and queuing theory.
The gamma distribution is parameterized by two quantities: shape \( k \) and rate \( \lambda \). Its probability density function is:
  • \( f(x; k, \lambda) = \frac{\lambda^k}{\Gamma(k)} x^{k-1} e^{-\lambda x} \)
The shape parameter \( k \) dictates the distribution’s form, while the rate \( \lambda \) scales the distribution.
Key properties of the gamma distribution include:
  • Mean: \( k / \lambda \)
  • Variance: \( k / \lambda^2 \)
In our context, \( Y_j \) is the sum of exponential variables, leading it to be gamma distributed. This relationship helps determine various statistical properties like means and variances, simplifying the analysis of complex systems.
Hazard Rate Estimation
The hazard rate, also known as the failure rate, is a function that describes the instantaneous potential for failure occurrence at a particular time, given that survival has occurred up to that time. In exponential distributions, the hazard rate is constant and simply equals the rate parameter \( \lambda \).
For situations where the hazard rate varies with time, it is often expressed as a smooth function, \( \lambda(t) \). By analyzing the function, particularly by plotting \( \log Y_j \) against midpoints of failure time intervals, one can explore how the hazard rate may change over time.
Key considerations in hazard rate estimation include:
  • The relationship between time and increasing risk of failure.
  • Using logs of gamma distributed backup quantities to stabilize variance and normalize the data, aiding in proper hazard rate estimation.
Estimation of a non-constant hazard rate can reveal trends crucial for reliability engineering and risk management.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Suppose that \(Y_{1}\) and \(Y_{2}\) have gamma densities (2.7) with parameters \(\lambda, \kappa_{1}\) and \(\lambda, \kappa_{2}\). Show that the conditional density of \(Y_{1}\) given \(Y_{1}+Y_{2}=s\) is $$ \frac{\Gamma\left(\kappa_{1}+\kappa_{2}\right)}{s^{\kappa_{1}+\kappa_{2}-1} \Gamma\left(\kappa_{1}\right) \Gamma\left(\kappa_{2}\right)} u^{\kappa_{1}-1}(s-u)^{\kappa_{2}-1}, \quad 00 $$ and establish that this is an exponential family. Give its mean and variance. (b) Show that \(Y_{1} /\left(Y_{1}+Y_{2}\right)\) has the beta density. (c) Discuss how you would use samples of form \(y_{1} /\left(y_{1}+y_{2}\right)\) to check the fit of this model with known \(v_{1}\) and \(v_{2}\).

The mean excess life function is defined as \(e(y)=\mathrm{E}(Y-y \mid Y>y)\). Show that $$ e(y)=\mathcal{F}(y)^{-1} \int_{y}^{\infty} \mathcal{F}(u) d u $$ and deduce that \(e(y)\) satisfies the equation \(e(y) Q^{\prime}(y)+Q(y)=0\) for a suitable \(Q(y)\). Hence show that provided the underlying density is continuous, $$ \mathcal{F}(y)=\frac{e(0)}{e(y)} \exp \left\\{-\int_{0}^{y} \frac{1}{e(u)} d u\right\\} $$ As a check on this, find \(e(y)\) and hence \(\mathcal{F}(y)\) for the exponential density. One approach to modelling survival is in terms of \(e(y)\). For human lifetime data, let \(e(y)=\gamma(1-y / \theta)^{\beta}\), where \(\theta\) is an upper endpoint and \(\beta, \gamma>0\). Find the corresponding survivor and hazard functions, and comment.

Show that the geometric density $$ f(y ; \pi)=\pi(1-\pi)^{y}, \quad y=0,1, \ldots, 0<\pi<1 $$ is an exponential family, and give its cumulant-generating function. Show that \(S=Y_{1}+\cdots+Y_{n}\) has negative binomial density $$ \left(\begin{array}{c} n+s-1 \\ n-1 \end{array}\right) \pi^{n}(1-\pi)^{s}, \quad s=0,1, \ldots $$ and that this is also an exponential family.

Suppose that \(\varepsilon\) has known density \(f\) with support on the unit circle in the complex plane, and that \(Y=e^{i \theta} \varepsilon\) for \(\theta \in \mathbb{R}\). Show that this is a group transformation model. Is it transitive? Is the action free?

Suppose \(Y=\tau \varepsilon\), where \(\tau \in \mathbb{R}_{+}\)and \(\varepsilon\) is a random variable with known density \(f\). Show that this scale model is a group transformation model with free action \(g_{\tau}(y)=\tau y\). Show that \(s_{1}(Y)=\bar{Y}\) and \(s_{2}(Y)=\left(\sum Y_{j}^{2}\right)^{1 / 2}\) are equivariant and find the corresponding maximal invariants. Sketch the orbits when \(n=2\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free