Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose a random variable \(Y\) has CDF $$ F(y)= \begin{cases}0, & \text { if } y<0 \\ 2 y /(y+1), & \text { if } 0 \leq y \leq 1 \\ 1, & \text { if } y>1\end{cases} $$ Find each of the following: (a) \(P(Y<2)\) (b) \(P(0.5

Step-by-Step Solution

Answer

(a) 1, (b) \(\frac{1}{12}\), (c) \( \frac{2}{(y+1)^2} \) for \(0\leq y\leq 1\); 0 otherwise, (d) Approximate using Simpson's Rule.

Step 1: Analyzing P(Y<2)

For part (a), finding \( P(Y < 2) \) is straightforward because the CDF \( F(y) \) equals 1 for all \( y > 1 \). Since 2 is greater than 1, the probability that \( Y < 2 \) is simply the value of the CDF at \( y = 2 \), which is 1.

Step 2: Calculating P(0.5<Y<0.6)

For part (b), \( P(0.5 < Y < 0.6) \) can be found by calculating \( F(0.6) - F(0.5) \). First, substitute \( y = 0.6 \) and \( y = 0.5 \) into the CDF formula for \( 0 \leq y \leq 1 \), which is \( F(y) = \frac{2y}{y+1} \). Calculate \( F(0.6) = \frac{2 \times 0.6}{0.6+1} = \frac{1.2}{1.6} = 0.75 \).Calculate \( F(0.5) = \frac{2 \times 0.5}{0.5+1} = \frac{1}{1.5} = \frac{2}{3} \).Thus, \( P(0.5 < Y < 0.6) = 0.75 - \frac{2}{3} = \frac{1}{12} \).

Step 3: Finding the PDF

For part (c), the PDF of \( Y \), denoted as \( f(y) \), is the derivative of the CDF \( F(y) \). For \( 0 \leq y \leq 1 \), the CDF is \( F(y) = \frac{2y}{y+1} \). Perform differentiation:\( f(y) = \frac{d}{dy}\left(\frac{2y}{y+1}\right) \).Using the quotient rule, if \( u = 2y \) and \( v = y+1 \), then:\( f(y) = \frac{v \cdot u' - u \cdot v'}{v^2} = \frac{(y+1)\cdot 2 - (2y)\cdot 1}{(y+1)^2} \).Simplify to get \( f(y) = \frac{2}{(y+1)^2} \) for \(0 \leq y \leq 1\) and 0 otherwise.

Step 4: Using Parabolic Rule to Approximate E(Y)

To approximate \( E(Y) \) using Parabolic Rule (Simpson's Rule) with \( n=8 \) over the interval \([0, 1]\), first express \( E(Y) \) as an integral:\[ E(Y) = \int_0^1 y \cdot f(y) \, dy = \int_0^1 y \cdot \frac{2}{(y+1)^2} \, dy \].Divide \([0, 1]\) into 8 intervals (\( n = 8 \)) with step size \( h = \frac{1-0}{8} = 0.125 \). Calculate function values at partition points and apply Parabolic Rule:\[ \int_a^b f(x) \, dx \approx \frac{h}{3} \left[f(x_0) + 4 \sum_{i \text{ odd}} f(x_i) + 2 \sum_{i \text{ even}} f(x_i) + f(x_n)\right] \].Implement this to approximate the integral \( E(Y) \). Calculate values of the function \( y \cdot f(y) = \frac{2y}{(y+1)^2} \) at 9 partition points (0, 0.125, 0.25,..., 1.0) and substitute these into Simpson's formula.

Key Concepts

Probability Density Function

A Probability Density Function (PDF) is a fundamental concept in probability and statistics. It provides a way to describe the likelihood of a continuous random variable taking on a particular value or range of values.

For any continuous random variable, the PDF, denoted as \(f(y)\), describes the probability of the variable falling within a particular range by integrating over that range. The area under the PDF curve over an interval represents the probability of the variable falling within that interval.

In simpler terms, the PDF gives us a way to understand how dense or spread out the probabilities are for different outcomes. It is important to know that the PDF itself is not a probability, but its integral over a range provides the probability.

• PDF always yields non-negative values
• The total area under the PDF curve must equal 1

In the exercise, the PDF of the random variable \(Y\) is derived by taking the derivative of the CDF \(F(y)\) for the interval \(0 \leq y \leq 1\). This process results in \( f(y) = \frac{2}{(y+1)^2} \), which is only valid for that interval, and \(f(y) = 0\) otherwise.

Simpson's Rule

Simpson's Rule is a technique used in numerical integration to approximate the definite integral of a function. It is particularly useful when the integral cannot be calculated analytically.

The rule approximates the value of an integral by dividing the integration interval into smaller segments, approximating the area under the curve for each segment as a parabola. By using a combination of quadratic polynomials to estimate the area, Simpson's Rule often provides a more accurate approximation compared to methods like the trapezoidal rule.

• Requires an even number of intervals
• More precise for smooth functions

In the problem provided, to estimate the expectation \(E(Y)\), Simpson's Rule is utilized over the interval \([0,1]\) with \(n = 8\), dividing the interval into eight equal parts. This involves calculating the function values at these points and applying Simpson's Rule as:\[ E(Y) \approx \frac{h}{3} \times \left[ f(x_0) + 4 \sum_{i \text{ odd}} f(x_i) + 2 \sum_{i \text{ even}} f(x_i) + f(x_n) \right] \] where \(h\) is the step size. By meticulously plugging in values for these points, we approximate the integral, providing a good estimation of \(E(Y)\).

Expectation of a Random Variable

The expectation or expected value of a random variable is a central concept in statistics and probability calculus. It provides the average or mean value that you would expect to obtain if the random variable could be assessed an infinite number of times.

Mathematically, for a continuous random variable \(Y\) with a PDF \(f(y)\), the expected value is given by the integral:
\[ E(Y) = \int_{-\infty}^{\infty} y \cdot f(y) \, dy \]
For a specific interval, the integral is calculated within those bounds. The expected value gives a sense of the central tendency of the distribution of the random variable.

• Provides a measure of location for a probability distribution
• Is a linear operator, meaning \(E(aX + b) = aE(X) + b\)

In the example at hand, the exercise requires calculating \(E(Y)\) over the interval \([0,1]\). Given the complexity of directly solving, Simpson's Rule is applied to facilitate the computation. This approach helps in simplifying the integral of \(y \cdot f(y)\) over the specified domain.

Probability Calculus

Probability calculus involves the mathematical rules and techniques used to calculate probabilities and related concepts. It extends from basic probability principles to more advanced topics in statistics and stochastic processes.

Understanding probability calculus allows for the practical application of probability theory to solve a wide array of quantitative problems, including those involving complex random variables and their distributions.

• Uses tools like CDFs and PDFs to determine probabilities
• Fundamental for modeling uncertainty and variability in real-world phenomena

In the provided exercise, probability calculus is employed particularly in steps where we derive the probability values, such as finding \(P(0.5 < Y < 0.6)\). This involves computing differences in the CDF values, illustrating how probability calculus couples theoretical probability definitions with computational applications.