Problem 65 - Calculus Single Variable

Question

In each of Exercises \(65-74\) calculate the expectation of a random variable whose probability density function is given. $$ \frac{1}{2} \sin (x) \quad 0 \leq x \leq \pi $$

Step-by-Step Solution

Answer

The expectation is \( \frac{\pi}{2} \).

Step 1: Understand the definition of Expectation

The expectation (or expected value) of a random variable with a continuous probability density function (pdf) \( f(x) \) is given by the formula \( E(X) = \int_{-f}^{f} x f(x) \, dx \). For this problem, we calculate \( E(X) \) by integrating over the range where the pdf is defined.

Step 2: Identify the bounds and the function

The pdf given is \( \frac{1}{2} \sin(x) \) and it is defined over the interval \( 0 \leq x \leq \pi \). Thus, the bounds for our integration will be from \( 0 \) to \( \pi \).

Step 3: Set up the integral for Expectation

The expectation is calculated as \( E(X) = \int_{0}^{\pi} x \cdot \frac{1}{2} \sin(x) \, dx \). This simplifies to \( \frac{1}{2} \int_{0}^{\pi} x \sin(x) \, dx \).

Step 4: Use Integration by Parts

Integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Choose \( u = x \) and \( dv = \sin(x) \, dx \). Then \( du = dx \) and \( v = -\cos(x) \).

Step 5: Apply Integration by Parts

Using the integration by parts formula: \[ \int x \sin(x) \, dx = -x \cos(x) \bigg|_0^\pi - \int_0^\pi -\cos(x) \, dx \].

Step 6: Calculate the Boundary Terms

Evaluate \(-x \cos(x) \bigg|_0^\pi\): at \( x = \pi \), it is \(-\pi \cdot (-1) = \pi \); at \( x = 0 \), it is \( 0 \cdot 1 = 0 \). Thus, it becomes \( \pi \).

Step 7: Solve the Remaining Integral

Solve \( \int_0^\pi \cos(x) \, dx = \sin(x) \bigg|_0^\pi \): at \( x = \pi \), it is \( 0 \); at \( x = 0 \), it is \( 0 \). So, this integral evaluates to 0.

Step 8: Combine the Results

The result of the integral \( \int_0^\pi x \sin(x) \, dx \) is \( \pi \). Multiply by \( \frac{1}{2} \) to match the original expectation integral: \( \frac{1}{2} \times \pi = \frac{\pi}{2} \).

Step 9: State the Final Expectation

The expected value \( E(X) \) is therefore \( \frac{\pi}{2} \).

Key Concepts

Probability Density Function

In probability theory, a Probability Density Function (pdf) is a fundamental concept used in the context of continuous random variables. A pdf is a function, denoted as \( f(x) \), that describes the likelihood of a continuous random variable taking on a particular value. However, because the probability of an exact value is essentially zero in a continuous framework, the pdf instead gives the probability **density**: essentially how likely it is for the variable to be near a particular value.
The key properties of a pdf are:

• Non-negativity: \( f(x) \geq 0 \) for all \( x \).
• Normalization: The integral of \( f(x) \) over its entire range must equal 1: \( \int_{-\infty}^{\infty} f(x) \, dx = 1 \).

For the exercise, the pdf is given as \( \frac{1}{2} \, \sin(x) \) over the interval \( 0 \leq x \leq \pi \). This means we are considering only those values within this interval for our probability calculations. Our first check would be to ensure that integrating this pdf over the given interval equals 1, confirming its suitability as a probability density function.

Integration by Parts

Integration by parts is a powerful technique in calculus used to integrate products of functions. Often in probability and statistics, especially when finding expectations, we encounter integrals that can be broken down utilizing this method.
The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]Where:

• \( u \) and \( dv \) are chosen parts of the original integral.
• \( du \) is the derivative of \( u \), and \( v \) the antiderivative of \( dv \).

In the context of the exercise, the function is \( x \sin(x) \). The typical choice for \( u \) is the polynomial part, so \( u = x \) and \( dv = \sin(x) \, dx \). Calculating further, we find \( du = dx \) and \( v = -\cos(x) \). By substituting these into the integration by parts formula, we simplify the integration process and move closer to our answer.

Expected Value Calculation

The expected value, or expectation, of a continuous random variable, is a measure of the central tendency of its probability distribution, similar to how the mean is used in a discrete setting. It provides a single value that summarizes the distribution of possible values that the random variable can take.
The expected value \( E(X) \) for a continuous random variable with a probability density function \( f(x) \) is calculated as:\[ E(X) = \int_{a}^{b} x f(x) \, dx \]Here, \( a \) and \( b \) define the interval over which the pdf is defined. For our pdf \( f(x) = \frac{1}{2} \sin(x) \), within \( 0 \leq x \leq \pi \), the expected value becomes:\[ E(X) = \int_{0}^{\pi} x \cdot \frac{1}{2} \sin(x) \, dx \]Evaluating this using the calculated integration by parts results allows us to find the expected value as \( \frac{\pi}{2} \), indicating the average value of \( X \) within the given constraints.

Continuous Random Variable

A continuous random variable is a type of variable that can take infinitely many values within a particular range, characterized by a probability density function. Unlike discrete random variables, which have countable outcomes, a continuous variable defines probabilities over intervals.
Key attributes include:

• Defined over an interval, e.g., \( [0, \pi] \).
• Values are not isolated points but rather on a continuum within the given bounds.
• Uses a probability density function to describe its distribution.

In our exercise, the continuous random variable \( X \) has values in the interval \( [0, \pi] \) with a pdf of \( \frac{1}{2} \sin(x) \). The continuous nature requires integral calculus to evaluate expectations or probabilities, contrary to using summation as with discrete variables.