Problem 52 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose \(f_{Y}(y)=4 y^{3}, 0 \leq y \leq 1\). Find \(P(0 \leq\) \(\left.Y \leq \frac{1}{2}\right)\).

Step-by-Step Solution

Answer

The probability is \(1/16\).

Step 1: Setting Up the Integral

Since we are given the probability density function (pdf) and the range of \(y\), we find the probability that \(y\) is between 0 and 1/2 by integrating the pdf from 0 to 1/2. This gives us \(\int_{0}^{1/2} 4y^3 dy\).

Step 2: Calculating the Integral

Next is to evaluate the integral. The antiderivative of \(4y^3\) with respect to \(y\) is \(y^4\). So, the integral becomes \([y^4]_0^{1/2}\).

Step 3: Applying the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus tells us that we can evaluate this definite integral by subtracting the value of the antiderivative at 0 from the value of the antiderivative at 1/2. Therefore, we get \((1/2)^4 - (0)^4 = 1/16 - 0 = 1/16\). This is the value of the integral, which represents the probability that \(Y\) is between 0 and 1/2.

Key Concepts

Probability Density Function

A probability density function (PDF) is a key concept in probability theory, used to specify the probability of a continuous random variable falling within a particular range of values. In simpler terms, a PDF helps us understand the likelihood of different outcomes for continuous data. For example, with a random variable like the one represented by the function \(f_{Y}(y) = 4y^{3}\), it's crucial to know how to find the probability of \(Y\) being in a certain range.

PDFs have certain properties:

• The PDF is non-negative: \(f(y) \geq 0\) for all \(y\).
• The total area under the PDF curve over its entire range is equal to 1. This represents 100% probability or certainty. For our function, the area from \(0\) to \(1\) must sum to 1.

When working with PDFs, especially for exercises like finding \(P(0 \leq Y \leq \frac{1}{2})\), we're often interested in calculating the area under the curve within specific limits. This is done using integration.

Integration

In calculus, integration helps us find the area under a curve. This is particularly useful for determining probabilities from a PDF. When we're given a function like \(f_{Y}(y) = 4y^{3}\), integration allows us to calculate the chance that \(Y\) falls between specified limits, such as \(0\) and \(\frac{1}{2}\).

Here's the process:

• Set up the integral based on the PDF and the interval: \(\int_{0}^{1/2} 4y^3 \, dy\).
• Find the antiderivative to represent the accumulated area under the curve. For \(4y^3\), the antiderivative is \(y^4\).

By evaluating this definite integral from \(0\) to \(1/2\), you actually determine the probability of \(Y\) being in this range.

Why Integration Matters

Integration is not just about calculating areas in geometry; it ties directly into real-world probabilities and helps quantify uncertainty, which is widespread in fields like statistics, physics, and engineering.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a powerful tool that bridges the gap between differentiation and integration. It tells us that the process of finding the derivative (rate of change) is essentially the reverse of finding the integral (accumulated area).

This theorem is crucial to solve problems like finding \(P(0 \leq Y \leq \frac{1}{2})\). It allows us to evaluate the definite integral of a function by taking the difference between the values of its antiderivative at two boundary points, such as \(1/2\) and \(0\).

Here's how it works with our problem:

• Find the antiderivative \(y^4\).
• Compute the difference: \([y^4]_{0}^{1/2} = (1/2)^4 - 0^4\).
• This gives \(1/16\), which is the probability of \(Y\) between 0 and \(1/2\).

Understanding this theorem provides not only a method for solving integrals but also a deeper insight into the intrinsic relationship between accumulation and change.