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Let \(F\) be a continuous distribution function. For some positive \(\alpha\), define the distribution function \(G\) by $$ \bar{G}(t)=(\bar{F}(t))^{\alpha} $$ Find the relationship between \(\lambda_{G}(t)\) and \(\lambda_{F}(t)\), the respective failure rate functions of \(G\) and \(F\).

Short Answer

Expert verified
The relationship between the failure rate functions \(\lambda_{G}(t)\) and \(\lambda_{F}(t)\) is given by: \[ \lambda_{G}(t) = \alpha \lambda_{F}(t) \frac{(1 - F(t))^{\alpha - 1}}{(\bar{F}(t))^{\alpha - 1}} \]

Step by step solution

01

Find the pdf of \(F\) and \(G\)

We know that the probability density function (pdf) of a distribution function can be calculated by finding its derivative with respect to time \(t\). Let's denote the pdf of \(F\) as \(f(t)\) and the pdf of \(G\) as \(g(t)\). From the exercise, the cumulative distribution function of \(G\) is given by \(\bar{G}(t)=(\bar{F}(t))^{\alpha}\). First, let's find the pdf of \(G\), \(g(t)\). To do that, we need to differentiate \(\bar{G}(t)\) with respect to \(t\): \[ g(t) = \frac{d}{dt} \bar{G}(t) = \frac{d}{dt} (\bar{F}(t))^{\alpha} \] Using the chain rule, we have: \[ g(t) = \alpha (\bar{F}(t))^{\alpha-1} \frac{d}{dt} \bar{F}(t) \] Since \(\bar{F}(t) = 1 - F(t)\) and \(f(t) = \frac{d}{dt} F(t)\), we can rewrite the pdf of G as: \[ g(t) = \alpha (1 - F(t))^{\alpha - 1}(-f(t)) \] Now we have found the pdfs of both \(F\) and \(G\).
02

Find the failure rate functions of \(F\) and \(G\)

The failure rate function for a distribution is defined as the ratio of the pdf to the survival function, which is given by \(1 - \text{cdf}\). Let's denote the failure rate function of \(F\) by \(\lambda_{F}(t)\) and the failure rate function of \(G\) by \(\lambda_{G}(t)\). We can compute the failure rate functions for both distributions: \[ \lambda_{F}(t) = \frac{f(t)}{1 - F(t)} \] \[ \lambda_{G}(t) = \frac{g(t)}{1 - G(t)} = \frac{g(t)}{\bar{G}(t)} \] Substitute the pdf of \(G\) and the survival function of \(G\) in the equation of \(\lambda_{G}(t)\): \[ \lambda_{G}(t) = \frac{\alpha (1 - F(t))^{\alpha - 1}(-f(t))}{(\bar{F}(t))^{\alpha}} \]
03

Find the relationship between \(\lambda_{G}(t)\) and \(\lambda_{F}(t)\)

Now we need to find the relationship between the failure rate functions of \(F\) and \(G\). Divide \(\lambda_{G}(t)\) by \(\lambda_{F}(t)\): \[ \frac{\lambda_{G}(t)}{\lambda_{F}(t)} = \frac{\alpha (1 - F(t))^{\alpha - 1}(-f(t))}{(\bar{F}(t))^{\alpha}} \cdot \frac{1 - F(t)}{f(t)} \] Simplify the equation: \[ \frac{\lambda_{G}(t)}{\lambda_{F}(t)} = \alpha \frac{(1 - F(t))^{\alpha - 1}}{(\bar{F}(t))^{\alpha - 1}} \] So, the relationship between the failure rate functions of \(G\) and \(F\) is given by: \[ \lambda_{G}(t) = \alpha \lambda_{F}(t) \frac{(1 - F(t))^{\alpha - 1}}{(\bar{F}(t))^{\alpha - 1}} \]

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Key Concepts

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

Continuous Distribution Function
In the realm of statistics, a Continuous Distribution Function, also known as a cumulative distribution function (CDF), is a fundamental concept that graphically represents the probability that a real-valued random variable X will have a value less or equal to x. Its main characteristic is that it's smooth, rather than jumping from one value to another as with discrete distributions.

To understand this more concretely, imagine plotting your probabilities on a graph. As you move along the x-axis, representing possible outcomes, the y-axis gives you the accumulated probability up to that point. This function will always start at 0, indicating that there is no probability of the random variable being less than the smallest value in its range, and as you move along the axis, this probability will increase up to 1, which represents certainty.

One of the key aspects of continuous distributions is that the probability of observing any exact single value is essentially zero since there are infinitely many possibilities. Instead, probabilities are measured over intervals. For example, it would make sense to ask, 'What is the probability that a random variable will fall between two values?' This is directly related to the area under the curve of the probability density function (PDF) for the interval in question.
Probability Density Function
The Probability Density Function (PDF) plays a critical role in understanding continuous distributions. It's related to the CDF in that the PDF is the derivative of the CDF. In the simplest terms, it gives you the 'density' of our random variable at any point along the x-axis. This function helps us visualize where the values of our variable are most likely to occur.

Now, if you were to look at a graph of a PDF, it would show you how probabilities are distributed across different outcomes. The peak of the graph is where the random variable is most likely found. The total area under the graph of the PDF over the entire range of the variable is always 'normalized' to equal 1, representing the fact that the probability of some outcome occurring within the variable's range is certain.

Nevertheless, the PDF itself doesn't directly give us probabilities. Instead, to find the probability within a certain interval, you would calculate the area under the curve of the PDF between two points, essentially integrating the function within those points. This calculated area corresponds to the probability that the variable assumes a value within that interval.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF), which we initially noted as a continuous distribution function, can also be seen as a bridge between the PDF and the probability of intervals. It describes the accumulation of probability up to a certain point.

What's particularly important about the CDF is that it provides a direct way to calculate the probability that the random variable X is less than or equal to a certain value. By evaluating the CDF at any value x, we obtain the probability that the random variable is less than or equal to that value. Because of its nature as an accumulated function, you can also find the probability that X falls within an interval (a, b) by calculating the difference between the CDF evaluated at b and a.

The CDF is thus invaluable when we are interested in probabilities, and has a related function known as the 'survival function', which is simply one minus the CDF. In the context of failure rates, another term used is the 'reliability function', which again indicates the probability that a system or component continues to operate successfully up to a certain time t without failure. These concepts are key when we dive into understanding failure rate functions in reliability engineering and risk analysis.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote independent and identically distributed random variables and define the order statistics \(X_{(1)}, \ldots, X_{(n)}\) by $$ X_{(i)} \equiv i \text { th smallest of } X_{1}, \ldots, X_{n} $$ Show that if the distribution of \(X_{j}\) is IFR, then so is the distribution of \(X_{(i)}\).

Let \(t_{i}\) denote the time of failure of the \(i\) th component; let \(\tau_{\phi}(t)\) denote the time to failure of the system \(\phi\) as a function of the vector \(\mathrm{t}=\left(t_{1}, \ldots, t_{n}\right) .\) Show that $$ \max _{1 \leqslant j \leqslant s} \min _{i \in A_{j}} t_{i}=\tau_{\phi}(\mathbf{t})=\min _{1 \leqslant j \leqslant k} \max _{i \in C_{i}} t_{i} $$ where \(C_{1}, \ldots, C_{k}\) are the minimal cut sets, and \(A_{1}, \ldots, A_{s}\) the minimal path sets.

Prove that, for any structure function \(\phi\), $$ \phi(\mathbf{x})=x_{i} \phi\left(1_{i}, \mathbf{x}\right)+\left(1-x_{i}\right) \phi\left(0_{i}, \mathbf{x}\right) $$ where $$ \begin{aligned} &\left(1_{i}, \mathbf{x}\right)=\left(x_{1}, \ldots, x_{i-1}, 1, x_{i+1}, \ldots, x_{n}\right) \\ &\left(0_{i}, \mathbf{x}\right)=\left(x_{1}, \ldots, x_{i-1}, 0, x_{i+1}, \ldots, x_{n}\right) \end{aligned} $$

Prove the combinatorial identity $$ \left(\begin{array}{c} n-1 \\ i-1 \end{array}\right)=\left(\begin{array}{c} n \\ i \end{array}\right)-\left(\begin{array}{c} n \\ i+1 \end{array}\right)+\cdots \pm\left(\begin{array}{l} n \\ n \end{array}\right), \quad i \leqslant n $$ (a) by induction on \(i\) (b) by a backwards induction argument on \(i\) -that is, prove it first for \(i=n\), then assume it for \(i=k\) and show that this implies that it is true for \(i=k-1\).

Consider a structure in which the minimal path sets are \(\\{1,2,3\\}\) and \(\\{3,4,5\\}\). (a) What are the minimal cut sets? (b) If the component lifetimes are independent uniform \((0,1)\) random variables, determine the probability that the system life will be less than \(\frac{1}{2}\).

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