/*! 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 33 The proportion of impurities in ... [FREE SOLUTION] | 91Ó°ÊÓ

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The proportion of impurities in certain ore samples is a random variable \(Y\) with density function given by $$f(y)=\left\\{\begin{array}{ll} (3 / 2) y^{2}+y, & 0 \leq y \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ The dollar value of such samples is \(U=5-(Y / 2) .\) Find the probability density function for \(U\).

Short Answer

Expert verified
The density function for \( U \) is \( f_U(u) = 12u^2 - 124u + 320 \) for \( 4.5 \leq u \leq 5. \)

Step by step solution

01

Define the Relationship Between Y and U

We are given that the relationship between the variables is \[ U = 5 - \frac{Y}{2}. \]This can be rearranged to express \( Y \) in terms of \( U \):\[ Y = 2(5 - U) = 10 - 2U. \] This will help us in changing the variable in the probability density function.
02

Determine the Range for U

Given that \(0 \leq Y \leq 1\), we can plug these bounds into the relation \(Y = 10 - 2U\) to find the corresponding range for \(U\).- For \(Y = 0\), \(0 = 10 - 2U\) implies \(U = 5.\)- For \(Y = 1\), \(1 = 10 - 2U\) gives \(2U = 9\), or \(U = 4.5.\)Thus, \(4.5 \leq U \leq 5.\)
03

Find the Derivative of Y in terms of U

We need to find \( \frac{dY}{dU} \) to use the change of variable formula for density functions. We have:\[ Y = 10 - 2U. \]Taking the derivative with respect to \( U \):\[ \frac{dY}{dU} = -2. \]
04

Use the Change of Variables Formula

The change of variables formula for probability density functions is\[ f_U(u) = f_Y(g(u)) \left| \frac{dg(u)}{du} \right| \]where \( g(u) \) is the inverse of \( U = g(Y) \) which we found as \( Y = g(U) = 10 - 2U \).Using this, the density function for \( U \) becomes:\[ f_U(u) = f_Y(10 - 2u) \cdot | -2 |, \]where\[ f_Y(y) = \frac{3}{2}y^2 + y. \]Thus,\[ f_U(u) = \left( \frac{3}{2}(10 - 2u)^2 + (10 - 2u) \right) \times 2. \]
05

Substitute and Simplify

Substitute \( y = 10 - 2u \) into \( f_Y(y) \):\[ f_Y(10 - 2u) = \frac{3}{2}(10 - 2u)^2 + (10 - 2u) \]Now plug into the expression for \( f_U(u) \):\[ f_U(u) = 2 \left( \frac{3}{2}(10 - 2u)^2 + (10 - 2u) \right). \]Simplifying,\[ f_U(u) = 3(10 - 2u)^2 + 2(10 - 2u). \]
06

Final Step: Simplify to Find the PDF for U

Expand and simplify:\( (10 - 2u)^2 = 100 - 40u + 4u^2 \).\[ f_U(u) = 3(100 - 40u + 4u^2) + 20 - 4u \]\[ = 300 - 120u + 12u^2 + 20 - 4u \]\[ = 320 - 124u + 12u^2. \]Thus, the probability density function for \( U \) is:\[ f_U(u) = 12u^2 - 124u + 320 \quad \text{for} \quad 4.5 \leq u \leq 5, \]and \( f_U(u) = 0 \) elsewhere.

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Key Concepts

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

Random Variable
A random variable is a measurable function from a set of possible outcomes to a measurable space. In simple terms, it's a variable that can take on different values, each with a certain probability. For instance, in the context of this exercise, the random variable is the proportion of impurities \( Y \) in ore samples. This \( Y \) is characterized by a probability distribution, defined by a density function \( f(y) \).The idea is that every outcome of the process of picking an ore sample has a certain likelihood, described by \( f(y) \), which provides the probability of the impurity level being within a particular interval. For \( Y \) in the given exercise, the density function is quadratic, which means \( Y \) might have several values, weighted by the function's formula. In mathematical form, \( f(y) \) tells us about the "density" of probabilities accumulated at any given point \( y \).Understanding random variables helps in making predictions and in performing statistical analysis, which can be very powerful in decision-making processes based on observed data.
Change of Variables Formula
The change of variables formula is a mathematical approach to transform the variable of a probability density function from one form to another. In our exercise, we want to understand how the random variable \( Y \) (the impurities) relates to another random variable \( U \) (the value in dollars). We begin by expressing \( Y \) (the initial variable) in terms of \( U \) (the new variable), which is derived as \( Y = 10 - 2U \). By using this relationship, we aim to find the density function of \( U \). This transformation process is essential when the desired quantity or insight involves a new variable or measurement, especially when performing integration over a probability density function. Properly changing variables allows us to understand how changing \( Y \) directly affects \( U \), and adjust our approach to probability accordingly.
Derivative in Probability
The derivative plays a crucial role in probability theory when changing variables. It helps us understand how the value of one variable affects another variable, particularly when converting between different probability distributions. In our exercise, we compute \( \frac{dY}{dU} = -2 \) to determine how probabilities related to \( Y \) should be adjusted to find those related to \( U \). Essentially, this derivative acts as a "scaling factor," capturing how the probability density stretches or contracts as we switch from \( Y \) to \( U \). This changing of scales through differentiation is especially important when applying the change of variables formula for probability density functions. It helps maintain the integrity of the probability distribution during transformations, ensuring that it remains accurate and meaningful.
Density Function Transformation
Transforming a density function is about modifying the function to describe a different variable, in this case, from \( Y \) to \( U \). This involves determining the function form of \( U \) from the known function of \( Y \). We utilize the formula:\[ f_U(u) = f_Y(g(u)) \cdot \left| \frac{dg(u)}{du} \right| \]This formula ensures that the properties of distributions are preserved and allows us to compute \( f_U(u) \), the density function for the dollar value \( U \). By substituting \( Y = 10 - 2U \) into the original density equation for \( Y \), we recalculate and simplify to determine the resulting density function:\[ f_U(u) = 12u^2 - 124u + 320 \quad \text{for} \quad 4.5 \leq u \leq 5 \]Transformations like this are common in probability when dealing with real-world data that need to be interpreted in different contexts. Understanding these transformations helps in crafting suitable models that reflect changes in defined variables.

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Most popular questions from this chapter

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 \(\alpha=2, \beta=3\)

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 \(U=1 /(1+Y)\)

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent, exponentially distributed random variables with mean \(\beta\) a. Show that \(Y_{(1)}=\min \left(Y_{1}, Y_{2}, \ldots, Y_{\mathrm{n}}\right)\) has an exponential distribution, with mean \(\beta / n\) b. If \(n=5\) and \(\beta=2,\) find \(P\left(Y_{(1)} \leq 3.6\right)\)

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 $$ f_{U_{1}}\left(u_{1}\right)=\int_{-\infty}^{\infty} f_{Y_{1}}\left(u_{1} u_{2}\right)\left|u_{2}\right| d u_{2} $$

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent random variables such that each \(Y_{i}\) has a gamma distribution with parameters \(\alpha_{i}\) and \(\beta\). That is, the distributions of the \(Y\) 's might have different \(\alpha\) 's, but all have the same value for \(\beta\). Prove that \(U=Y_{1}+Y_{2}+\cdots+Y_{n}\) has a gamma distribution with parameters \(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}\) and \(\beta\).

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