Problem 69 - Calculus Single Variable

Question

In each of Exercises \(65-74\) calculate the expectation of a random variable whose probability density function is given. $$ \frac{3}{16} \sqrt{4-x} \quad 0 \leq x \leq 4 $$

Step-by-Step Solution

Answer

The expectation of the random variable is \( \frac{32}{15}. \)

Step 1: Identify the Formula for Expectation

The expectation of a continuous random variable with a given probability density function (pdf) is calculated using the formula \( E[X] = \int_{a}^{b} x f(x) \, dx \), where \( f(x) \) is the probability density function over the interval \( [a, b] \).

Step 2: Set Up the Integral

Given the probability density function \( f(x) = \frac{3}{16} \sqrt{4-x} \) and the interval \( 0 \leq x \leq 4 \), the integral becomes: \[ E[X] = \int_{0}^{4} x \left( \frac{3}{16} \sqrt{4-x} \right) \, dx. \]

Step 3: Simplify the Integral

Simplify the integral by distributing \( x \) into the function: \[ E[X] = \frac{3}{16} \int_{0}^{4} x \sqrt{4-x} \, dx. \] This integral needs to be solved to find the expectation.

Step 4: Use Substitution Method

Use substitution with \( u = 4 - x \), so \( du = -dx \) and when \( x = 0 \), \( u = 4 \), and when \( x = 4 \), \( u = 0 \). Rewrite the integral: \[ E[X] = \frac{3}{16} \int_{4}^{0} (4-u) \sqrt{u} \, (-du). \]

Step 5: Correct the Integral Limits and Simplify

Flipping the limits changes the sign: \[ E[X] = \frac{3}{16} \int_{0}^{4} (4-u) \sqrt{u} \, du. \] Now distribute and separate the integrals: \[ E[X] = \frac{3}{16} \left( \int_{0}^{4} 4\sqrt{u} \, du - \int_{0}^{4} u\sqrt{u} \, du \right). \]

Step 6: Solve the First Integral

For the first integral \( \int_{0}^{4} 4\sqrt{u} \, du = 4\left[ \frac{2}{3}u^{3/2} \right]_{0}^{4} \), compute it to get \( 4 \times \frac{2}{3} \times 8 = \frac{64}{3}. \)

Step 7: Solve the Second Integral

For the second integral \( \int_{0}^{4} u\sqrt{u} \, du = \int_{0}^{4} u^{3/2} \, du = \left[ \frac{2}{5}u^{5/2} \right]_{0}^{4} \), calculate it to get \( \frac{2}{5} \times 32 = \frac{64}{5}. \)

Step 8: Compute the Final Expectation Value

Combine the results: \[ E[X] = \frac{3}{16} \left( \frac{64}{3} - \frac{64}{5} \right). \] Simplify the expression \( \frac{64}{3} - \frac{64}{5} = \frac{64 \times 5 - 64 \times 3}{15} = \frac{128}{15}. \) Thus, \( E[X] = \frac{3}{16} \times \frac{128}{15} = \frac{32}{15}. \)

Key Concepts

Probability Density Function

A probability density function (pdf) is a vital concept in the study of continuous random variables. It describes the likelihood of a random variable taking on a particular value. For a continuous random variable, the probability of exactly one value occurring is zero. Hence, we focus on probability over an interval.
In our exercise, the pdf is given by \( f(x) = \frac{3}{16} \sqrt{4-x} \) for the interval \( 0 \leq x \leq 4 \). This function tells us how the probabilities are distributed across this range of \( x \). The function must satisfy two key properties:

• The function is non-negative for all values of \( x \) within the interval.
• The integral of the pdf over the entire range of \( x \) equals 1, which ensures that the total probability over the interval is 100%.

Understanding the pdf helps us in calculating important measures like the expectation of the random variable.

Continuous Random Variable

A continuous random variable can assume an infinite number of values within a given range. It is different from a discrete random variable, which can only take specific values. Think of measuring something like time, weight, or in this case, the variable \( x \) between 0 and 4.
Since we're dealing with a continuous variable, we employ concepts like the probability density function to analyze it. The pdf allows us to work with probabilities over intervals rather than individual points.
For the expectation of the continuous random variable discussed in this exercise, we calculate it using an integral. This calculation is essential because it gives us a "weighted average" of all possible outcomes, taking into account the likelihood of each value according to the pdf.

Integral Calculation

The calculation of integrals is a central tool in finding the expectation of continuous random variables. Integrals help us accumulate a continuous set of values to find totals or averages, like expectation.
In our problem, we aim to calculate \( E[X] = \int_{0}^{4} x f(x) \, dx \). This integral represents the sum of all values that \( x \) takes, weighted by their probabilities.
The steps involve:

• Setting up the integral with the given pdf: \( E[X] = \frac{3}{16} \int_{0}^{4} x \sqrt{4-x} \, dx \).
• Using methods like substitution, which simplifies the calculation by changing the variable of integration.
• Breaking down the integral into manageable parts if needed.

Solving these integrals requires some practice, but it's an essential skill for working with continuous variables.

Substitution Method

The substitution method is a technique used to simplify complex integrals. It involves changing variables to transform the integral into a more straightforward form. This method is particularly useful when direct integration is difficult.
In our example, substitution was used to replace \( x \) with \( u = 4 - x \). This transformation makes it easier to handle the \( \sqrt{4-x} \) term. Moreover, it requires changing the limits of integration; initially from \( 0 \) to \( 4 \) in terms of \( x \) to \( 4 \) to \( 0 \) in terms of \( u \).
Key points when using substitution:

• Choose a substitution that simplifies the integrand.
• Express all parts of the integral, including limits, in terms of the new variable.
• Solve the new, simpler integral and then substitute back if needed.

This method turns a complex problem into a more manageable one, aiding us in finding the expectation value efficiently.