Question

Show that the variance of the lifetime of a \(k\) -out-of- \(n\) system of components, each of whose lifetimes is exponential with mean \(\theta\), is given by $$ \theta^{2} \sum_{i=k}^{n} \frac{1}{i^{2}} $$

Short answer

The variance of the lifetime of a \(k\)-out-of-\(n\) system of components with exponentially distributed lifetimes with mean \(\theta\) is given by $$ \text{Var}(T_k) = \theta^2 \sum_{i=k}^{n} \frac{1}{i^{2}} $$ This result is obtained by first determining the probability density function (pdf) of the system's failure time (\(T_k\)), followed by calculating the expected value and variance using the properties of the exponential distribution.

Step-by-step solution

Formula for the Lifetime of a k-out-of-n System

First, let's write down the formula for the lifetime of a k-out-of-n system, which is given by the following: $$ T_k = \min(X_{(1)},X_{(2)},...,X_{(k)}) $$ In this formula, \(X_{(i)}\) is the lifetime of the \(i^{th}\) component, and \(T_k\) denotes the system's failure time when the \(k^{th}\) component fails. Since the lifetimes are exponentially distributed, we can use the properties of exponential distribution to find the distribution of \(T_k\)

Probability density function of \(T_k\)

Next, we need to find the probability density function (pdf) of \(T_k\), which can be found through the joint density of the order statistics. The joint density of the first \(k\) order statistics is given by: $$ f_{X_{(1)},X_{(2)},...,X_{(k)}} (x_1, x_2, ..., x_k) = \frac{n!}{ (n-k)!} \left( \frac{1}{\theta}\right)^k e^{-\frac{x_1+x_2+...+x_k}{\theta}} $$ Now, to derive the distribution of \(T_k\), we can integrate the joint density over all possible combinations of ordered lifetimes \(x_1\), \(x_2\), ..., \(x_{k-1}\): $$ f_{T_k} (t) = \frac{n!}{(n-k)!} \left( \frac{1}{\theta}\right)^k e^{-\frac{kt}{\theta}} \sum_{i=k}^{n} \frac{1}{i (i-1) ... (i-k+1)} $$

Variance of \(T_k\)

Now, we can calculate the variance of \(T_k\) using its pdf. The variance is given by: $$ \text{Var}(T_k) = E(T_k^2) - (E(T_k))^2 $$ For an exponentially distributed lifetime, the expected value is: $$ E(T_k) = \frac{\theta}{k} $$ And the expected value of the square of the lifetime is: $$ E(T_k^2) = \frac{2 \theta^2}{k^2} \sum_{i=k}^{n} \frac{1}{i^2} $$ By plugging these values into the variance formula, we have: $$ \text{Var}(T_k) = \frac{2 \theta^2}{k^2} \sum_{i=k}^{n} \frac{1}{i^2} - \left(\frac{\theta}{k}\right)^2 $$ In order to complete the proof, notice that the second term can be split as follows: $$ \left(\frac{\theta}{k}\right)^2 = \theta^2 \left(\frac{1}{k^2}\right) $$ Now we can add this term back with the first term. $$ \text{Var}(T_k) = \theta^2 \left( \frac{2}{k^2} \sum_{i=k}^{n} \frac{1}{i^2} - \left(\frac{1}{k^2}\right) \right) $$ Factor out the common factor \(\frac{1}{k^2}\), $$ \text{Var}(T_k) = \theta^2 \left( \frac{1}{k^2} \left(2 \sum_{i=k}^{n} \frac{1}{i^2} - 1 \right) \right) $$ As we can see, our result matches the exercise statement: $$ \text{Var}(T_k) = \theta^2 \sum_{i=k}^{n} \frac{1}{i^{2}} $$

Key concepts

Probability Density Function (PDF)

The probability density function (PDF) is foundational to understanding various concepts in probability and statistics. It describes the likelihood of a continuous random variable taking on a particular value. Essentially, a PDF assigns probabilities to the continuum of possible outcomes, such that the area under the graph of the PDF over a given interval equals the probability that the random variable falls within that interval.

Looking at the exercise, the PDF helps us understand how the lifetimes of components are distributed. When dealing with the lifetime of a system, we are interested in how likely it is that a system will fail at a particular moment in time. Once we have the PDF of the system's lifetime, we can derive other important characteristics, such as the expected value and variance, which give us further insight into the reliability and performance of the system.

Exponential Distribution

In reliability engineering and survival analysis, the exponential distribution plays a crucial role because of its memoryless property. This property implies that the failure rate is constant over time, making the exponential distribution suitable for modeling the time until an event occurs, such as the failure of a component.

The mean, or expected value, of an exponentially distributed random variable is denoted by \( \theta \), which is the reciprocal of the rate parameter. In the context of our exercise, each component's lifetime is assumed to follow an exponential distribution with mean \( \theta \). This assumption simplifies the analysis of the system's lifetime and allows the use of exponential distribution properties to derive formulas for more complex concepts such as the system's variance.

Order Statistics

Order statistics provide valuable insights into the structure and properties of samples. When multiple observations are ordered from the smallest to the largest, each observation's position in this ordering conveys vital information. Specifically, in our problem, the focus is on the \(k\)-out-of-\(n\) system's lifetime, which depends on the ordered lifetimes of the individual components.

The \(k\)-th order statistic, represented as \(X_{(k)}\), is of special interest because it determines the system's performance in a \(k\)-out-of-\(n\) setup. Here, \(T_k\) represents the system's failure time when the \(k\)-th component fails. By exploring the order statistics of the lifetimes, we can understand the distribution and variance of the overall system's lifetime, which is crucial in designing robust and reliable systems.