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Chapter 5: Problem 86

The random variable \(X\) has the probability distribution $$ f_{X}(x)=\frac{x}{8}, \quad 0 \leq x \leq 4 $$ Determine the probability distribution of \(Y=(X-2)^{2}\).

Short Answer

Expert verified
The distribution of Y is given by \(f_{Y}(y) = \frac{2 + \sqrt{y}}{16 \sqrt{y}}\) for \(0 \leq y \leq 4\).

Step by step solution

01

Understand the Functions

We are given the probability distribution function (PDF) of random variable \(X\) as \(f_{X}(x)=\frac{x}{8}\) for the interval \(0 \leq x \leq 4\). We need to find the distribution of \(Y = (X-2)^2\). This requires finding how the transformation \(Y = g(X)\) affects the distribution.
02

Find the Inverse Function

The transformation \(Y = (X-2)^{2}\) can be solved for \(X\) to express \(X\) in terms of \(Y\). Solving for \(X\), we get two branches: \(X = 2 + \sqrt{Y}\) and \(X = 2 - \sqrt{Y}\). Since \(0 \leq x \leq 4\), \(X = 2 + \sqrt{Y}\) is the valid transformation in this context.
03

Determine the Range of Y

To determine the range for \(Y\), we consider the range of \(X\). With \(0 \leq X \leq 4\), substituting these into \((X-2)^2\), we have \(0 \leq Y \leq 4\). However, by examining smaller segments, we see \(X\) ranges from 0 to 2 results in \(0 \leq Y \leq 4\), and from 2 to 4 results in a different interpretation under the transformation. Hence, \(0 \leq Y \leq 4\) but notice that as \(X = 2 + \sqrt{Y}\), the effective range is limited to \([0, 4]\).
04

Change of Variables Formula

We utilize the change of variables formula for transformation: \(f_{Y}(y) = f_{X}(g^{-1}(y)) \cdot \left| \frac{d}{dy} g^{-1}(y) \right|\). Here, \(g^{-1}(y) = 2 + \sqrt{y}\). Compute the derivative \( \frac{d}{dy}g^{-1}(y) = \frac{1}{2 \sqrt{y}}\).
05

Compute the Probability Distribution of Y

Compute the distribution function \(f_{Y}(y)\) by substituting \(g^{-1}(y)\) into \(f_{X}(x)\): \(f_{Y}(y) = \frac{2 + \sqrt{y}}{8} \times \frac{1}{2 \sqrt{y}}\) for \(0 \leq y \leq 4\). Simplifying, we get: \(f_{Y}(y) = \frac{2 + \sqrt{y}}{16 \sqrt{y}}\).

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Key Concepts

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

Random Variables
A random variable is a numerical value that results from a random event. In essence, it's a way of quantifying randomness. For example, think about flipping a coin. If we let heads be 1 and tails be 0, the result is a random variable.

Random variables can be continuous or discrete:
  • **Discrete random variables** take on a finite or countable number of values, like the number of heads in a series of coin flips.
  • **Continuous random variables** can take any value within an interval, such as the height of a randomly chosen person.

In the given exercise, we deal with a continuous random variable, denoted by \(X\). Its values range between 0 and 4. By using the probability density function, we understand the likelihood of \(X\) taking on a given value within this range. This concept is crucial when predicting outcomes based on random events.
Probability Density Function
The probability density function (PDF) is a key component in probability theory for continuous random variables. It describes how the values of a random variable are distributed.

The given PDF is \(f_{X}(x) = \frac{x}{8}\) for \(0 \le x \le 4\). The PDF provides the relative likelihood that a random variable \(X\) will take on a value \(x\). Here鈥檚 how it works:
  • The PDF is not the probability; rather, it indicates where the values are most densely packed.
  • The area under the PDF curve gives the probability. For continuous random variables, this requires integration.
  • For the PDF to be valid, the total area under the curve (from \(-\infty\) to \(+\infty\)) must equal 1.

In this example, the PDF indicates that values of \(X\) closer to 4 are more likely than values closer to 0. This PDF is used to determine how \(X\) behaves and, subsequently, how \(Y\), a function of \(X\), distributes its values.
Transformation of Variables
Transformation of variables is a technique used to find the probability distribution of a new random variable, which is a function of another random variable. Consider \(Y = (X-2)^2\), a transformation of \(X\). This method allows us to explore how changes in \(X\) affect \(Y\).

Here鈥檚 the process:
  • **Expressing the Transformation:** To find \(Y\)鈥檚 distribution, express \(X\) in terms of \(Y\). Solving \(Y = (X-2)^2\), we obtain \(X = 2 + \sqrt{Y}\). Given the original domain, we ensure we use the correct branch of the function.
  • **Validation of Range:** We check the range of \(Y\) using \(X\)'s bounds, resulting in \(0 \leq Y \leq 4\).
  • **Applying Change of Variables:** Calculate the PDF of \(Y\) using the formula \(f_{Y}(y) = f_{X}(g^{-1}(y)) \cdot \left| \frac{d}{dy} g^{-1}(y) \right|\). This involves simplifying the expression to find whether the transformation maintains a valid PDF for \(Y\).

The transformation helps in shifting our perspective from \(X\) to analyzing outcomes of scenarios dependent on \(Y\), illustrating the probability landscape of the transformed variable.

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Most popular questions from this chapter

The computational time of a statistical analysis applied to a data set can sometimes increase with the square of \(N,\) the number of rows of data. Suppose that for a particular algorithm, the computation time is approximately \(T=0.004 N^{2}\) seconds. Although the number of rows is a discrete measurement, assume that the distribution of \(N\) over a number of data sets can be approximated with an exponential distribution with a mean of 10,000 rows. Determine the probability density function and the mean of \(T\).

Determine the value of \(c\) such that the function \(f(x, y)=c x y\) for \(01.8,1

The rate of return of an asset is the change in price divided by the initial price (denoted as \(r\) ). Suppose that \(\$ 10,000\) is used to purchase shares in three stocks with rates of returns \(X_{1}, X_{2}, X_{3}\). Initially, \(\$ 2500, \$ 3000,\) and \(\$ 4500\) are allocated to each one, respectively. After one year, the distribution of the rate of return for each is normally distributed with the following parameters: \(\mu_{1}=0.12, \sigma_{1}=0.14, \mu_{2}=0.04, \sigma_{2}=0.02, \mu_{3}=0.07, \sigma_{3}=0.08\) (a) Assume that these rates of return are independent. Determine the mean and variance of the rate of return after one year for the entire investment of \(\$ 10,000\). (b) Assume that \(X_{1}\) is independent of \(X_{2}\) and \(X_{3}\) but that the covariance between \(X_{2}\) and \(X_{3}\) is \(-0.005 .\) Repeat part (a). (c) Compare the means and variances obtained in parts (a) and (b) and comment on any benefits from negative covariances between the assets.

The systolic and diastolic blood pressure values (mm Hg) are the pressures when the heart muscle contracts and relaxes (denoted as \(Y\) and \(X,\) respectively). Over a collection of individuals, the distribution of diastolic pressure is normal with mean 73 and standard deviation \(8 .\) The systolic pressure is conditionally normally distributed with mean \(1.6 x\) when \(X=x\) and standard deviation of \(10 .\) Determine the following: (a) Conditional probability density function \(f_{Y \mid 73}(y)\) of \(Y\) given \(X=73\) (b) \(P(Y<115 \mid X=73)\) (c) \(E(Y \mid X=73)\) (d) Recognize the distribution \(f_{X Y}(x, y)\) and identify the mean and variance of \(Y\) and the correlation between \(X\) and \(Y\)

Assume that the weights of individuals are independent and normally distributed with a mean of 160 pounds and a standard deviation of 30 pounds. Suppose that 25 people squeeze into an elevator that is designed to hold 4300 pounds. (a) What is the probability that the load (total weight) exceeds the design limit? (b) What design limit is exceeded by 25 occupants with probability \(0.0001 ?\)
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