/*! 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 14 Let \(X\) be a random variable w... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) be a random variable with distribution function \(m_{X}(x)\) defined by $$ m_{X}(-1)=1 / 5, \quad m_{X}(0)=1 / 5, \quad m_{X}(1)=2 / 5, \quad m_{X}(2)=1 / 5 $$ (a) Let \(Y\) be the random variable defined by the equation \(Y=X+3\). Find the distribution function \(m_{Y}(y)\) of \(Y\). (b) Let \(Z\) be the random variable defined by the equation \(Z=X^{2}\). Find the distribution function \(m_{Z}(z)\) of \(Z\).

Short Answer

Expert verified
(a) \(m_{Y}(2) = \frac{1}{5}, m_{Y}(3) = \frac{1}{5}, m_{Y}(4) = \frac{2}{5}, m_{Y}(5) = \frac{1}{5}\); (b) \(m_{Z}(0) = \frac{1}{5}, m_{Z}(1) = \frac{3}{5}, m_{Z}(4) = \frac{1}{5}\).

Step by step solution

01

Understand Variable Transformations

We have two transformations to work with: for part (a), we will transform variable \(X\) into \(Y\) using the equation \(Y = X + 3\); for part (b), \(Z = X^2\). With each transformation, we need to adjust the given probabilities to find the distribution of the resultant variables \(Y\) and \(Z\).
02

Find Distribution of Y

Given that \(Y = X + 3\), we substitute each possibility for \(X\) into this equation:- If \(X = -1\), then \(Y = -1 + 3 = 2\) with \(P(Y = 2) = P(X = -1) = \frac{1}{5}\).- If \(X = 0\), then \(Y = 0 + 3 = 3\) with \(P(Y = 3) = P(X = 0) = \frac{1}{5}\).- If \(X = 1\), then \(Y = 1 + 3 = 4\) with \(P(Y = 4) = P(X = 1) = \frac{2}{5}\).- If \(X = 2\), then \(Y = 2 + 3 = 5\) with \(P(Y = 5) = P(X = 2) = \frac{1}{5}\).Thus, the distribution of \(Y\) is:\[ m_{Y}(2) = \frac{1}{5}, \quad m_{Y}(3) = \frac{1}{5}, \quad m_{Y}(4) = \frac{2}{5}, \quad m_{Y}(5) = \frac{1}{5}. \]
03

Find Distribution of Z

Given that \(Z = X^2\), we substitute each possibility for \(X\) into this equation:- If \(X = -1\), then \(Z = (-1)^2 = 1\) with \(P(Z = 1) = P(X = -1) = \frac{1}{5}\).- If \(X = 0\), then \(Z = 0^2 = 0\) with \(P(Z = 0) = P(X = 0) = \frac{1}{5}\).- If \(X = 1\), then \(Z = 1^2 = 1\) with \(P(Z = 1) = P(X = 1) = \frac{2}{5}\).- If \(X = 2\), then \(Z = 2^2 = 4\) with \(P(Z = 4) = P(X = 2) = \frac{1}{5}\).We combine probabilities for \(Z = 1\) as it appears in two different cases:\[ P(Z = 1) = \frac{1}{5} + \frac{2}{5} = \frac{3}{5}. \]Thus, the distribution of \(Z\) is:\[ m_{Z}(0) = \frac{1}{5}, \quad m_{Z}(1) = \frac{3}{5}, \quad m_{Z}(4) = \frac{1}{5}. \]

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Key Concepts

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

Random Variable Transformations
When dealing with random variables, transformations are an important tool. They allow us to obtain new variables from existing ones, based on specific rules. In this exercise, we explore two types of transformations: linear and non-linear.

For example, transforming a random variable using a linear transformation like \( Y = X + 3 \) involves simply adding a constant to the variable \(X\). This operation shifts the entire distribution of \(X\) by the constant amount, producing outcomes for \(Y\) that are each shifted accordingly.

On the other hand, a non-linear transformation such as \( Z = X^2 \) involves squaring the values of \(X\). This can significantly change the shape of the distribution. Instead of a shift, we are modifying the values themselves, potentially creating entirely new outcomes and altering the probability distribution accordingly. Transformations are key in adapting probability models to new situations and analyzing different scenarios.
Probability Distribution
A probability distribution describes how the probabilities are distributed over the values of a random variable. The distribution function, usually denoted by \( m_X(x) \), is a function that defines this spread. It provides the probability of each value that the random variable can assume, reflecting the likelihood of each outcome occurring.

In discrete random variables, this function assigns a probability to each possible value. For example, if a discrete random variable can take on values \(-1, 0, 1,\) and \(2\), the probability distribution might look like \( m_{X}(x) = \frac{1}{5}, \frac{1}{5}, \frac{2}{5}, \frac{1}{5} \) respectively.

The key takeaway is that the probabilities must add up to 1, representing the certainty that one of the possible values will occur. Understanding the probability distribution is crucial, as it represents the entire behavior of a random variable and provides insight into the outcomes you can expect.
Discrete Random Variables
Discrete random variables are those that have a finite or countable number of possible values. Unlike continuous variables, which can take on any value within a range, discrete variables are more straightforward in terms of defining and calculating probabilities.

Each possible value of a discrete random variable is associated with a particular probability, defined by the probability distribution function. For instance, in our exercise, the random variable \(X\) can only take the values \(-1, 0, 1, 2\). The probability associated with each of these values is provided in the distribution function \( m_{X}(x) \).

This simplicity allows for easier calculations and transformations, as seen when \(X\) is transformed into \(Y\) and \(Z\). Nevertheless, despite their simplicity, discrete random variables can represent a wide variety of scenarios and are a fundamental element of probability theory.

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Most popular questions from this chapter

A die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. (For example, a six is three times as probable as a two.) What is the probability of getting an even number in one throw?

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