/*! 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 105 A random variable \(Y\) has a be... [FREE SOLUTION] | 91Ó°ÊÓ

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A random variable \(Y\) has a beta distribution of the second kind, if, for \(\alpha>0\) and \(\beta>0\), its density is $$f_{y}(y)=\left\\{\begin{array}{ll} \frac{y^{\alpha-1}}{B(\alpha, \beta)(1+y)^{\alpha+\beta}}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ Derive the density function of \(U=1 /(1+Y)\)

Short Answer

Expert verified
The density function of \( U = \frac{1}{1+Y} \) is derived using transformation and PDF of \( Y \).

Step by step solution

01

Find the cumulative distribution function (CDF) of Y

To derive the density function of \( U = \frac{1}{1+Y} \), we start by finding the CDF of \( Y \), denoted as \( F_Y(y) \). This is given by the integral of the density function from 0 to \( y \). \[F_Y(y) = \int_0^y \frac{t^{\alpha-1}}{B(\alpha, \beta)(1+t)^{\alpha+\beta}} \, dt\] The CDF is needed to find the transformation of \( U \).
02

Transform Y to U

To find the distribution of \( U = \frac{1}{1+Y} \), we perform a transformation. The transformation \( U \) implies that \( Y = \frac{1-U}{U} \). We need the CDF of \( U \) through this transformation. Solve for \( Y \) to express it in terms of \( U \).
03

Use the transformation to find CDF of U

To find the CDF of \( U \), we use the relation:\[F_U(u) = P\left( \frac{1}{1+Y} \leq u \right) = P(Y \geq \frac{1-u}{u}) = 1 - F_Y\left( \frac{1-u}{u} \right)\]
04

Differentiate to find the PDF of U

To obtain the probability density function of \( U \), we differentiate the CDF found in Step 3 with respect to \( u \):\[f_U(u) = \frac{d}{du}(1 - F_Y\left( \frac{1-u}{u} \right)) = - \frac{d}{du} F_Y\left( \frac{1-u}{u} \right)\]Apply the chain rule in differentiation as:\[- \frac{d}{du} F_Y\left( \frac{1-u}{u} \right) = f_Y\left( \frac{1-u}{u} \right) \frac{d}{du}\left( \frac{1-u}{u} \right)\]Calculate the derivative of \( \frac{1-u}{u} \) and substitute back into the expression.
05

Simplify and express the PDF of U

The derivative \( \frac{d}{du}(\frac{1-u}{u}) \) is \( -\frac{1}{u^2} \). Substituting this in gives:\[f_U(u) = f_Y\left( \frac{1-u}{u} \right) \cdot \left(-\frac{1}{u^2}\right)\]Since \( f_Y(y) = \frac{y^{\alpha-1}}{B(\alpha, \beta)(1+y)^{\alpha+\beta}} \), replace \( y \) with \( \frac{1-u}{u} \) and simplify to find \( f_U(u) \). Compute and finalize the expression.

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Key Concepts

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

Cumulative Distribution Function
To understand how the cumulative distribution function (CDF) helps in transforming a variable, we need to explore its role in probability distributions. The CDF, denoted as \( F_Y(y) \) for a random variable \( Y \), represents the probability that \( Y \) will take a value less than or equal to \( y \). Mathematically, it is an integral of the probability density function (PDF) \( f_Y(y) \):
  • \( F_Y(y) = \int_0^y f_Y(t) \, dt \)
This process of integrating the PDF over a range allows us to obtain the cumulative probability up to a specific point \( y \).
In the context of the Beta distribution of the second kind, where \( f_Y(y) = \frac{y^{\alpha-1}}{B(\alpha, \beta)(1+y)^{\alpha+\beta}} \), finding the CDF involves integrating this expression. This integral not only provides insight into the distribution of \( Y \), but also serves as a crucial step when transforming \( Y \) into another random variable, such as \( U = 1/(1+Y) \).
Understanding and computing the CDF are essential for solving complex problems where the nature of a distribution changes, making it a critical concept in statistical transformations.
Probability Density Function
The probability density function (PDF) is a fundamental aspect of continuous probability distributions, like the Beta distribution. It describes the relative likelihood for a random variable to take on a particular value.
For example, the PDF of a Beta distribution of the second kind is given by:
  • \( f_Y(y) = \frac{y^{\alpha-1}}{B(\alpha, \beta)(1+y)^{\alpha+\beta}} \)
This function provides a way to compute probabilities for different intervals of the random variable's values by integrating over those intervals.
The PDF is also crucial when performing transformations of variables. When transforming \( Y \) into a new variable \( U = 1/(1+Y) \), finding the PDF of \( U \) involves not only the CDF but also differentiation of the CDF concerning \( u \). This leads to the expression:
  • \( f_U(u) = -\frac{d}{du} F_Y\left( \frac{1-u}{u} \right) \)
By substituting the expression derived from \( Y \), the PDF \( f_U(u) \) allows us to capture the behavior of \( U \) and understand how its probabilities are distributed over potential outcomes.
Therefore, PDFs provide the foundational basis for exploring statistical behaviors and serve as a bridge in transforming and understanding different distributions.
Transformation of Variables
Transforming variables is a method used to derive new probability distributions from existing ones. This is particularly important when we deal with non-standard distributions, like the transformation of \( Y \) depicted in this exercise.
The transformation \( U = 1/(1+Y) \) changes the perspective from \( Y \) to \( U \), effectively "flipping" the distribution while changing its scale. This requires finding the distribution of \( U \), which involves several mathematical operations:
  • Express \( Y \) in terms of \( U \): \( Y = \frac{1-U}{U} \).
  • Use this relationship in the cumulative distribution function to find \( F_U(u) \).
  • Differentiation to unveil the probability density function of \( U \).
These steps unlock how the patterns of \( Y \) are mapped onto \( U \). The PDF transformation in this example shows how such methods cater to calculating \( f_U(u) \) based on variable relationships:
  • \( f_U(u) = f_Y\left( \frac{1-u}{u} \right) \cdot \left(-\frac{1}{u^2}\right) \).
By understanding transformations, we gain insights into the flexibility of statistical distributions, enabling us to apply probability theory across various disciplines, and solve complex problems by changing the random variable in question.

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Most popular questions from this chapter

Let \(Y_{1}\) and \(Y_{2}\) be independent random variables, both uniformly distributed on \((0,1) .\) Find the probability density function for \(U=Y_{1} Y_{2}\).

Let \(\left(Y_{1}, Y_{2}\right)\) have joint density function \(f_{Y_{1}, Y_{2}}\left(y_{1}, y_{2}\right)\) and let \(U_{1}=Y_{1} / Y_{2}\) and \(U_{2}=Y_{2}\) a. Show that the joint density of \(\left(U_{1}, U_{2}\right)\) is \(=f_{Y_{1}, Y_{2}}\left(u_{1} u_{2}\right.\) b. Show that the marginal density function for \(U_{1}\) is $$ f_{U_{1}}\left(u_{1}\right)=\int_{-\infty}^{\infty} f_{Y_{1}, Y_{2}}\left(u_{1} u_{2}, u_{2}\right)\left|u_{2}\right| d u_{2} $$ c. If \(Y_{1}\) and \(Y_{2}\) are independent, show that the marginal density function for \(U_{1}\) is $$ f_{U_{1}}\left(u_{1}\right)=\int_{-\infty}^{\infty} f_{Y_{1}}\left(u_{1} u_{2}\right)\left|u_{2}\right| d u_{2} $$

The time until failure of an electronic device has an exponential distribution with mean 15 months. If a random sample of five such devices are tested, what is the probability that the first failure among the five devices occurs a. after 9 months? b. before 12 months?

The weight (in pounds) of "medium-size" watermelons is normally distributed with mean 15 and variance 4. A packing container for several melons has a nominal capacity of 140 pounds. What is the maximum number of melons that should be placed in a single packing container if the nominal weight limit is to be exceeded only \(5 \%\) of the time? Give reasons for your answer.

Suppose that \(Y_{1}\) and \(Y_{2}\) are independent and that both are uniformly distributed on the interval \((0,1),\) and let \(U_{1}=Y_{1}+Y_{2}\) and \(U_{2}=Y_{1}-Y_{2}.\) a. Show that the joint density of \(U_{1}\) and \(U_{2}\) is given by $$ f_{U_{1}, U_{2}}\left(u_{1}, u_{2}\right)=\left\\{\begin{array}{ll} 1 / 2, & -u_{1}0\) c. Show that the marginal density of \(U_{1}\) is $$ f_{U_{1}}\left(u_{1}\right)=\left\\{\begin{array}{ll} u_{1}, & 0

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