Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

The expected value of a function \(g(X)\) of a continuous random variable \(X\) having \(\operatorname{PDF} f(x)\) is defined to be \(E[g(X)]=\) \(\int_{A}^{B} g(x) f(x) d x .\) If the PDF of \(X\) is \(f(x)=\frac{15}{512} x^{2}(4-x)^{2}\), \(0 \leq x \leq 4\), find \(E(X)\) and \(E\left(X^{2}\right)\).

Step-by-Step Solution

Answer

E(X) = 2 and E(X^2) = \( \frac{16}{3} \).

Step 1: Identify the Probability Density Function (PDF)

The problem provides a probability density function (PDF) for the random variable \(X\): \( f(x) = \frac{15}{512} x^{2}(4-x)^{2} \) for \( 0 \leq x \leq 4 \). This function will be used as \( f(x) \) in our integral calculations.

Step 2: Define the Expected Value (E(X)) Integral

To find the expected value \(E(X)\), we need to set up the integral based on the formula \(E[g(X)]= \int_{A}^{B} g(x) f(x) \, dx\), where \(g(x) = x\). This gives us the expression \(E(X) = \int_{0}^{4} x \cdot \frac{15}{512} x^{2}(4-x)^{2} \, dx\).

Step 3: Simplify the Integral for E(X)

Simplify the integrand by calculating \( x \cdot \frac{15}{512} x^{2}(4-x)^{2} = \frac{15}{512} x^{3} (4-x)^{2} \). Expand \((4-x)^{2} = 16 - 8x + x^{2}\) and multiply by \(x^{3}\): \( \frac{15}{512} (16x^{3} - 8x^{4} + x^{5})\).

Step 4: Integrate to Find E(X)

Integrate \( \frac{15}{512} (16x^{3} - 8x^{4} + x^{5})\) from 0 to 4: 1. Integrate \(16x^{3}\) to get \(4x^{4}\).2. Integrate \(-8x^{4}\) to get \(-\frac{8}{5}x^{5}\).3. Integrate \(x^{5}\) to get \(\frac{1}{6}x^{6}\).4. Evaluate each term from 0 to 4 and sum the results to get \(E(X) = 2\).

Step 5: Define the Expected Value (E(X²)) Integral

Now, for \(E(X^2)\), set up the integral with \(g(x) = x^2\). So \(E(X^2) = \int_{0}^{4} x^2 \cdot \frac{15}{512} x^{2}(4-x)^{2} \, dx = \int_{0}^{4} \frac{15}{512} x^{4}(4-x)^{2} \, dx \).

Step 6: Simplify the Integral for E(X²)

Simplify the integrand: \( \frac{15}{512} x^{4}(4-x)^{2} = \frac{15}{512} (16x^4 - 8x^5 + x^6)\).

Step 7: Integrate to Find E(X²)

Integrate \( \frac{15}{512} (16x^4 - 8x^5 + x^6)\) from 0 to 4: 1. Integrate \(16x^4\) to get \(\frac{16}{5}x^{5}\).2. Integrate \(-8x^5\) to get \(-\frac{8}{6}x^{6}\).3. Integrate \(x^6\) to get \(\frac{1}{7}x^{7}\).4. Evaluate each term from 0 to 4 and sum the results to get \(E(X^2) = \frac{16}{3}\).

Key Concepts

Probability Density Function

A probability density function (PDF) is a crucial concept in probability theory when dealing with continuous random variables. Unlike discrete random variables, which have separate, distinct probabilities, continuous random variables can take any value within a certain range. Hence, the PDF is not about the probability of a single outcome but rather about how likely different outcomes are over an interval.

For a continuous random variable, the PDF must satisfy two primary conditions:

• The probability density function must be non-negative for all values. In mathematical terms, if a function \( f(x) \) is the PDF of a random variable \( X \), then \( f(x) \geq 0 \) for all \( x \).
• The integral of the PDF over the entire range of the variable must be equal to 1. That is, \( \int_{-\infty}^{\infty} f(x) \, dx = 1 \). This represents the total probability being equal to 1, as it should be for any random variable.

The given PDF in the exercise, \( f(x) = \frac{15}{512} x^2 (4-x)^2 \), is used to calculate the probabilities and expected values of outcomes over its defined interval \( 0 \leq x \leq 4 \). This PDF describes the likelihood of each outcome for the random variable \( X \), which is essential for determining expected values.

Continuous Random Variable

Continuous random variables are variables that can take an infinite number of values within a given range. This characteristic differentiates them from discrete random variables, which can only take specific, separate values. A classic example of a continuous random variable is the time taken to run a marathon, as it can assume any value over a range (for instance, from one hour to the record-breaking times).

In the exercise, the variable \( X \) represents a continuous random variable defined over the domain \( 0 \leq x \leq 4 \). This means \( X \) can take any value within the interval from 0 to 4. Thus, the calculations involve integrating over this interval because each infinitesimally small segment requires consideration to evaluate probabilities or expected values.

To analyze a continuous random variable, we employ techniques such as using PDFs and integrating over continuous intervals, allowing us to find probabilities of specific ranges and expected values, as in our exercise.

Integral Calculus

Integral calculus is a core mathematical tool when dealing with continuous random variables and their probability density functions. It allows us to find areas under curves, which, in the context of probability, represent the probability of a continuous variable falling within a particular interval.

In probability theory, we use definite integrals. A definite integral is written as \( \int_{A}^{B} f(x) \, dx \), and it calculates the area under the curve \( f(x) \) from \( A \) to \( B \). In the given exercise, the integral \( \int_{0}^{4} x \, f(x) \, dx \) helps us find the expected value of \( X \), and \( \int_{0}^{4} x^2 \, f(x) \, dx \) provides the expected value of \( X^2 \).

• When calculating \( E(X) \), integral calculus involves simplifying the expression by multiplying the PDF by \( x \) and integrating it over the defined range.
• Similarly, for \( E(X^2) \), the expression involves \( x^2 \), multiplied by the PDF, before integration.

Integral calculus thus plays a pivotal role in determining things like expected values and cumulative probabilities, underpinning a broad range of practical applications in statistics and probability.

Expected Value of a Function

The expected value of a function is a fundamental concept in probability and statistics, representing a kind of "average" outcome we anticipate for a random variable based on its probability distribution.

For a continuous random variable \( X \) with probability density function \( f(x) \), the expected value of a function \( g(X) \) is given as \( E[g(X)] = \int_{A}^{B} g(x) f(x) \, dx \). This formula calculates what you might "expect" the average value of \( g(X) \) to be, if you were to take countless samples from your random variable.

• In the exercise, the function \( g(x) = x \) gives us the expected value \( E(X) \), and \( g(x) = x^2 \) gives \( E(X^2) \). These represent different perspectives on the anticipated behavior of our variable \( X \).
• To find these expected values computationally involves setting up and evaluating the lateral integrals, as explained, which owe their groundwork to the mechanics of integral calculus.

Understanding expected values through this framework allows us to predict the long-term averages of different functions of our variable, making this a critical part of statistical analysis and decision-making.