/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Compute the hazard rate function... [FREE SOLUTION] | 91Ó°ÊÓ

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Compute the hazard rate function of \(X\) when \(X\) is uniformly distributed over \((0, a)\).

Short Answer

Expert verified
The hazard rate function of \(X\) when \(X\) is uniformly distributed over \((0, a)\) is \(h(x) = \frac{1/a}{1 - x/a}\) for \(0 \le x \le a\).

Step by step solution

01

Find the pdf of the uniform distribution

The probability density function (pdf) of a continuous random variable with a uniform distribution over an interval \((0,a)\) is given by: \[ f(x) = \frac{1}{a}, \quad 0 < x < a \] and \[f(x) = 0, \quad \text{otherwise}.\]
02

Find the CDF of the uniform distribution

The cumulative distribution function (CDF) is the integral of the pdf. For the given uniform distribution, compute the CDF as follows: \[ F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} \frac{1}{a} dt = \frac{1}{a}(t)\Big\rvert_0^x = \frac{x}{a}, \quad 0 \le x \le a \] and \[F(x) = 0, \quad x < 0, \quad F(x) = 1, \quad x > a.\]
03

Find the survival function of the uniform distribution

The survival function, \(S(x)\), is the complement of the CDF, and it is given by: \[ S(x) = 1 - F(x) \] For the uniform distribution, the survival function is: \[ S(x) = 1 - \frac{x}{a}, \quad 0 \le x \le a \] and \[S(x) = 1, \quad x < 0, \quad S(x) = 0 , \quad x > a.\]
04

Compute the hazard rate function

The hazard rate function, \(h(x)\), is defined as the ratio of the pdf and the survival function: \[ h(x) = \frac{f(x)}{S(x)} \] For the uniform distribution, the hazard rate function is: \[ h(x) = \frac{\frac{1}{a}}{1 - \frac{x}{a}}, \quad 0 \le x \le a \] and \[h(x) = 0, \quad \text{otherwise}.\] So the hazard rate function of \(X\) when \(X\) is uniformly distributed over \((0, a)\) is \(h(x) = \frac{1/a}{1 - x/a}\) for \(0 \le x \le a\).

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Key Concepts

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

Uniform Distribution
The concept of a uniform distribution is a staple in the study of probability. A uniform distribution is one where every outcome in a given interval is equally likely. Think of it like a perfectly fair game where every option is just as probable as the others. This is visualized as a flat, horizontal line when we graph the probability over the range.
For the uniform distribution over an interval \((0,a)\), the probability of landing on any specific value within this interval is \( rac{1}{a}\). However, it's important to note that outside this interval, the probability density function (pdf) drops to zero. This means that any value less than zero or greater than \(a\) is impossible in this distribution.
In practical scenarios, uniform distributions are rare but serve as a simplified model for basic probability problems, helping to elucidate more complex concepts.
Probability Density Function
The Probability Density Function (pdf) is a crucial part of understanding distributions. It shows how the density of the probability is spread over different values. For a uniform distribution across the interval \(0, a\), this function is constant, highlighted by:
\[ f(x) = \frac{1}{a}, \quad 0 < x < a \]
This equation tells us that the likelihood is evenly spread between 0 and \(a\). Picture it like spreading butter evenly across a slice of bread: every part inside the bounds gets the same amount.
If you're outside the interval, meaning \(x < 0\) or \(x > a\), the pdf value is zero. This is a literal representation of the probability: outside our slice of bread, there is no butter. In other words, values outside the interval are not possible.
While it can be a bit abstract, the pdf is foundational, helping determine other functions like the cumulative distribution function (CDF).
Cumulative Distribution Function
The Cumulative Distribution Function, or CDF, is your probability roadmap. Instead of telling us the precise density at any given value, the CDF gives cumulative probabilities, essentially summing up the probabilities up to a point \(x\).
For our uniform distribution over \(0, a\), we compute it as:
\[ F(x) = \frac{x}{a}, \quad 0 \le x \le a \]
This formula might appear different, but it's simply adding up probabilities. It tells us the likelihood that our variable is less than or equal to \(x\). If \(x < 0\), the CDF reads \(0\), because no probability exists before we hit the starting point. If \(x > a\), every possible outcome has been covered, making the probability one, thus a flat line at its peak.
The CDF is invaluable, telling us not just if, but how likely we are to fall within a specific range in the data set.
Survival Function
The Survival Function offers a different perspective compared to the CDF. It reflects the probability that a variable exceeds a particular value \(x\). It's essentially the opposite story of the CDF. For a uniform distribution on \(0, a\), the survival function is given by:
\[ S(x) = 1 - \frac{x}{a}, \quad 0 \le x \le a \]
This mirrors real-world scenarios like the chance of a machine still operating beyond a certain time. It says "Here's what's left," after accounting for all probabilities up to \(x\). Before touching 0, there's nothing to "survive," so \(S(x)=1\). Past \(a\), everything's been accounted for so the survival rate is zero.
Understanding the survival function aids greatly in evaluating failure rates, longevity, and the remaining potential across various fields, from reliability engineering to actuarial sciences.

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Most popular questions from this chapter

The median of a continuous random variable having distribution function \(F\) is that value \(m\) such that \(F(m)=\frac{1}{2}\). That is, a random variable is just as likely to be larger than its median as it is to be smaller. Find the median of \(X\). if \(X\) is (a) uniformly distributed over \((a, b)\); (b) normal with parameters \(\mu, \sigma^{2}\); (c) exponential with rate \(\lambda\).

Jones figures that the total number of thousands of miles that an auto can be driven before it would need to be junked is an exponential random variable with parameter \(\frac{1}{20}\). Smith has a used car that he claims has been driven only 10,000 miles. If Jones purchases the car, what is the probability that she would get at least 20,000 additional miles out of it? Repeat under the assumption that the lifetime mileage of the car is not exponentially distributed but rather is (in thousands of miles) uniformly distributed over \((0,40)\).

The speed of a molecule in a uniform gas at equilibrium is a random variable whose probability density function is given by $$ f(x)= \begin{cases}a x^{2} e^{-b x^{2}} & x \geq 0 \\ 0 & x<0\end{cases} $$ where \(b=m / 2 k T\) and \(k, T\), and \(m\) denote, respectively, Boltzmann's constant, the absolute temperature, and the mass of the molecule. Evaluate \(a\) in terms of \(b\).

(a) A fire station is to be located along a road of length \(A, A<\infty\). If fires will occur at points uniformly chosen on \((0, A)\), where should the station be located so as to minimize the expected distance from the fire? That is, choose \(a\) so as to minimize \(E[|X-a| \mathrm{I}\) when \(X\) is uniformly distributed over \((0, A)\). (b) Now suppose that the road is of infinite length-stretching from point 0 outward to \(\infty\). If the distance of a fire from point 0 is exponentially distributed with rate \(\lambda\), where should the fire station now be located? That is, we want to minimize \(E[|X-a|]\) where \(X\) is now exponential with rate \(\lambda\).

Compute the hazard rate function of a gamma random variable with parameters \((t, \lambda)\) and show it is increasing when \(t \geq 1\) and decreasing when \(t \leq 1\).

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