Problem 64
Question
Suppose \(F_{Y}(y)=\frac{1}{12}\left(y^{2}+y^{3}\right), 0 \leq y \leq 2\). Find \(f_{Y}(y)\).
Step-by-Step Solution
Verified Answer
The probability density function \(f_{Y}(y)\) is given by \(f_{Y}(y) = \frac{1}{6}y + \frac{1}{4}y^2\), for \(0 \leq y \leq 2\).
1Step 1: Understand the problem
We are given \(F_{Y}(y)=\frac{1}{12}(y^{2}+y^{3})\), which is a cumulative distribution function (CDF). The task is to find the probability density function (PDF). Remember that the PDF is the derivative of the CDF.
2Step 2: Apply the power rule for differentiation
We apply the power rule to the function \(F_Y(y)\). The power rule states that if \(f(x) = x^n\), then \(f'(x) = nx^{(n-1)}\). So, we take the derivative of each term in the function. The derivative of \(y^2\) is \(2y\) and the derivative of \(y^3\) is \(3y^2\). Therefore, \(f_{Y}(y) = \frac{1}{12}(2y+3y^{2})\).
3Step 3: Simplify the resulting function
Simplify the function by distributing the \(\frac{1}{12}\) to each term: \(f_{Y}(y) = \frac{2}{12}y + \frac{3}{12}y^{2} = \frac{1}{6}y + \frac{1}{4}y^2\), for \(0 \leq y \leq 2\).
Key Concepts
Cumulative Distribution FunctionDifferentiationProbability Theory
Cumulative Distribution Function
A Cumulative Distribution Function (CDF) is a fundamental concept in probability theory. It is a function that describes the probability of a random variable taking on a value less than or equal to a specific point. For a random variable \( Y \), the CDF is denoted as \( F_Y(y) \). It provides a complete description of the probability distribution of a random variable. This function is non-decreasing and ranges from 0 to 1. In the given problem, the CDF is \( F_{Y}(y)=\frac{1}{12}(y^{2}+y^{3}) \). Here's what this means:
- It tells us the probability that the random variable \( Y \) is less than or equal to any specific value \( y \) within its domain, which is from 0 to 2.
Differentiation
Differentiation is an essential tool in both mathematics and statistics for finding rates of change. When dealing with functions, we use differentiation to find how a function is changing at any given point.In the context of probability, to convert a Cumulative Distribution Function (CDF) to a Probability Density Function (PDF), we take the derivative of the CDF.This is because the PDF is essentially a measure of the probability density; essentially, how 'tightly' packed the probabilities are at each point within a given range.
- Using the power rule, which states that if \( f(x) = x^n \), then \( f'(x) = nx^{(n-1)} \), we can differentiate our given CDF, \( F_Y(y)=\frac{1}{12}(y^{2}+y^{3}) \).
- The result of this differentiation process gives us \( f_{Y}(y)= \frac{1}{12}(2y+3y^{2}) \).
- Upon simplifying, we end up with \( f_{Y}(y)= \frac{1}{6}y + \frac{1}{4}y^2 \).
Probability Theory
Probability theory is the branch of mathematics concerned with the analysis of random phenomena. It is the foundation for statistics and involves concepts and methods for quantifying uncertainty, risk, and randomness.Key concepts in probability theory include: random variables, probability distributions, expected values, and events.
- A Cumulative Distribution Function (CDF) is a critical aspect of probability theory, providing a complete description of the probability distribution of a random variable.
- Transitioning from a CDF to a Probability Density Function (PDF) involves finding derivatives, showing the connection between calculus and probability theory.
- In our example, the PDF derived from the given CDF represents the likelihood of \( Y \) assuming any specific value over its range.
Other exercises in this chapter
Problem 62
A random variable \(Y\) has cdf $$ F_{Y}(y)= \begin{cases}0 & y
View solution Problem 63
The cdf for a random variable \(Y\) is defined by \(F_{Y}(y)=0\) for \(y1\). Find \(P\left(\frac{1}{4} \leq Y \leq \frac{3}{4}\right)\) by integrating \(f_{Y}(y
View solution Problem 65
In a certain country, the distribution of a family's disposable income, \(Y\), is described by the pdf \(f_{Y}(y)=\) \(y e^{-y}, y \geq 0\). Find \(F_{Y}(y)\).
View solution Problem 66
The logistic curve \(F(y)=\frac{1}{1+e^{-y}},-\infty
View solution