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A member of the Pareto family of distributions (often used in economics to model income distributions) has a distribution function given by $$F(y)=\left\\{\begin{array}{ll} 0, & y<\beta \\ 1-\left(\frac{\beta}{y}\right)^{\alpha}, & y \geq \beta \end{array}\right.$$ where \(\alpha, \beta>0\) a. Find the density function. b. For fixed values of \(\beta\) and \(\alpha\), find a transformation \(G(U)\) so that \(G(U)\) has a distribution function of \(F\) when \(U\) has a uniform distribution on the interval (0,1) c. Given that a random sample of size 5 from a uniform distribution on the interval (0,1) yielded the values. 0058, .2048,7692,2475 and \(.6078,\) use the transformation derived in part \((\mathrm{b})\) to give values associated with a random variable with a Pareto distribution with \(\alpha=2, \beta=3\)

Short Answer

Expert verified
Distribution density is \(\frac{\alpha \beta^\alpha}{y^{\alpha+1}}\); transform \(G(U)=\frac{3}{(1-U)^{1/2}}\), yielding approx. 3.0087, 3.3767, 6.4723, 3.5740, 4.8015.

Step by step solution

01

Understanding the Distribution Function

The given distribution function is \(F(y) = \begin{cases} 0, & y < \beta \1 - \left(\frac{\beta}{y}\right)^\alpha, & y \geq \beta \end{cases}\) where \(\alpha\) and \(\beta\) are parameters greater than 0. This function describes the cumulative probability up to a point \(y\).
02

Deriving the Density Function

The density function, \(f(y)\), is the derivative of the distribution function \(F(y)\) with respect to \(y\).For \(y < \beta\), the density is 0. For \(y \geq \beta\), find \(f(y)\) by differentiating:\[f(y) = \frac{d}{dy}\left(1 - \left(\frac{\beta}{y}\right)^\alpha\right) = \alpha \beta^\alpha y^{-(\alpha + 1)}\] Thus, the density function is \(f(y) = \begin{cases} 0, & y < \beta \\frac{\alpha \beta^\alpha}{y^{\alpha+1}}, & y \geq \beta \end{cases}\).
03

Finding the Transformation Function

We need to find \(G(U)\) such that \(G(U)\) follows the distribution \(F(y)\). If \(U\) is uniformly distributed on (0,1), \(G(U) = F^{-1}(U)\).Solving for \(y\) in \(1 - \left(\frac{\beta}{y}\right)^\alpha = u\), we get:\[\left(\frac{\beta}{y}\right)^\alpha = 1-u\]\[\frac{\beta}{y} = (1-u)^{1/\alpha}\]\[y = \frac{\beta}{(1-u)^{1/\alpha}}\]So the transformation \(G(U) = \frac{\beta}{(1-U)^{1/\alpha}}\).
04

Applying the Transformation for Given Data

Using the derived transformation \(G(U) = \frac{3}{(1-U)^{1/2}}\) (with \(\beta = 3\) and \(\alpha = 2\)), apply it to the given sample data to find the corresponding Pareto random variables:- For \(U = 0.0058\): \[G(0.0058) = \frac{3}{(1-0.0058)^{1/2}} \approx 3.0087\]- For \(U = 0.2048\): \[G(0.2048) = \frac{3}{(1-0.2048)^{1/2}} \approx 3.3767\]- For \(U = 0.7692\): \[G(0.7692) = \frac{3}{(1-0.7692)^{1/2}} \approx 6.4723\]- For \(U = 0.2475\): \[G(0.2475) = \frac{3}{(1-0.2475)^{1/2}} \approx 3.5740\]- For \(U = 0.6078\): \[G(0.6078) = \frac{3}{(1-0.6078)^{1/2}} \approx 4.8015\]
05

Conclusion

The values associated with a random variable with a Pareto distribution for the given sample are approximately 3.0087, 3.3767, 6.4723, 3.5740, and 4.8015.

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Key Concepts

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

Density Function
The density function, also called the probability density function (pdf), is a fundamental concept in probability and statistics. It describes how the probability of a random variable is distributed over different values. In the case of the Pareto distribution, finding the density function involves taking the derivative of the given distribution function.
For the Pareto distribution, when dealing with the area where the value is greater than or equal to a threshold parameter \( \beta \), the density function is given by:- For \( y < \beta \), \( f(y) = 0 \)- For \( y \geq \beta \), the density function is \( f(y) = \frac{\alpha \beta^\alpha}{y^{\alpha+1}} \)
This result tells us how likely different values of \( y \) are in our distribution. The form \( \alpha \beta^\alpha y^{-(\alpha + 1)} \) shows how probability spreads as values increase, following a characteristic Pareto pattern typically skewed towards larger values in economical contexts like wealth distribution.
Inverse Transform Sampling
Inverse Transform Sampling is a powerful technique used for generating random numbers from any given distribution, and it relies on the cumulative distribution function (CDF). Here's how it works:
  • First, create a random number \( U \) that is uniformly distributed between 0 and 1.
  • Next, use the inverse of the CDF \( F^{-1}(U) \) to transform \( U \) into a random variable that follows the desired distribution.

For the Pareto distribution, we start with the given function, set it equal to \( U \), solve for \( y \), and hence derive the transformation \( G(U) = \frac{\beta}{(1-U)^{1/\alpha}} \). This transformation allows us to map uniform variables to variables adhering to the Pareto distribution, essentially "drawing" samples from it by performing this inverse transformation.
Uniform Distribution Transformation
A uniform distribution is the simplest form of continuous probability distribution, where all outcomes are equally likely over a given interval. When we talk about transforming a uniform distribution, we're referring to converting these equally likely values into another distribution with potentially different properties.
In the context of the Pareto distribution, the task is to transform a uniform random variable \( U \) on the interval (0,1) to fit the Pareto distribution. Using the inverse transformation function \( G(U) \), each value of \( U \) is transformed such that the result is a random variable reflecting the characteristics of a Pareto distribution. This is done using the formula:\[ y = \frac{\beta}{(1-U)^{1/\alpha}} \]This transformation ensures that the originally uniform samples now correctly reflect the long-tailed nature of Pareto distributions, making it crucial for simulations and modeling real-world phenomena, like income distributions.
Probability Density Function
The Probability Density Function (PDF) is a key concept that describes the likelihood of a continuous random variable to take on a specific value. Specifically, for continuous data, the PDF indicates how the density "builds" around different points in the range. In the Pareto distribution, this PDF is focused on how the density shifts as the variable moves away from the minimum threshold \( \beta \).
The Pareto PDF takes the form:\[ f(y) = \frac{\alpha \beta^\alpha}{y^{\alpha+1}} \quad \text{for} \quad y \geq \beta \]
This shows the density function's sensitivity to both the shape parameter \( \alpha \) and scale parameter \( \beta \). As \( y \) increases, the power \( -(\alpha + 1) \) ensures that the probability diminishes quickly, encapsulating the "heavy-tail" property that makes Pareto distributions particularly useful in fields like economics where extreme values have non-negligible probabilities.

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Most popular questions from this chapter

The Weibull density function is given by $$f(y)=\left\\{\begin{array}{ll} \frac{1}{\alpha} m y^{m-1} e^{-y^{m} / \alpha}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ where \(\alpha\) and \(m\) are positive constants. This density function is often used as a model for the lengths of life of physical systems. Suppose \(Y\) has the Weibull density just given. Find a. the density function of \(U=Y^{m}\) b. \(E\left(Y^{k}\right)\) for any positive integer \(k\)

Let \(Y_{1}\) and \(Y_{2}\) be independent Poisson random variables with means \(\lambda_{1}\) and \(\lambda_{2}\), respectively. Find the a. probability function of \(Y_{1}+Y_{2}\). b. conditional probability function of \(Y_{1}\), given that \(Y_{1}+Y_{2}=m\).

Suppose that \(n\) electronic components, each having an exponentially distributed length of life with mean \(\theta,\) are put into operation at the same time. The components operate independently and are observed until \(r\) have failed \((r \leq n) .\) Let \(W_{j}\) denote the length of time until the \(j\) th failure, with \(W_{1} \leq W_{2} \leq \cdots \leq W_{r} .\) Let \(T_{j}=W_{j}-W_{j-1}\) for \(j \geq 2\) and \(T_{1}=W_{1} .\) Notice that \(T_{j}\) measures the time elapsed between successive failures. a. Show that \(T_{j},\) for \(j=1,2, \ldots, r,\) has an exponential distribution with mean \(\theta /(n-j+1)\) b. Show that $$U_{r}=\sum_{j=1}^{r} W_{j}+(n-r) W_{r}=\sum_{j=1}^{r}(n-j+1) T_{j}$$ and hence that \(E\left(U_{r}\right)=r \theta .[U_{r} \text { is called the total observed life, and we can use } U_{r} / r\) as an . approximation to (or "estimator" of) \(\theta .]\)

In a missile-testing program, one random variable of interest is the distance between the point at which the missile lands and the center of the target at which the missile was aimed. If we think of the center of the target as the origin of a coordinate system, we can let \(Y_{1}\) denote the north-south distance between the landing point and the target center and let \(Y_{2}\) denote the corresponding eastwest distance. (Assume that north and east define positive directions.) The distance between the landing point and the target center is then \(U=\sqrt{Y_{1}^{2}+Y_{2}^{2}} \cdot\) If \(Y_{1}\) and \(Y_{2}\) are independent, standard normal random variables, find the probability density function for \(U\).

If \(Y_{1}\) and \(Y_{2}\) are independent exponential random variables, both with mean \(\beta\), find the density function for their sum. (In Exercise 5.7 , we considered two independent exponential random variables, both with mean 1 and determined \(P\left(Y_{1}+Y_{2} \leq 3\right) .\) )

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