Problem 79 - Calculus Single Variable

Question

Let \(p\) be a positive constant. Let \(X\) be a random variable with range [0,1] and probability density function \(f(x)=\) \((p+1) x^{p}, 0 \leq x \leq 1\). Show that \(\bar{X} \neq f_{\text {avg. }}\) (In general, the average of a random variable is not equal to the average of its probability density function.)

Step-by-Step Solution

Answer

The expected value is \(\frac{p+1}{p+2}\), which differs from \(f_{\text{avg}} = 1\). Thus, \(\bar{X} \neq f_{\text{avg}}\).

Step 1: Calculate the Expected Value of X

The expected value (or mean) of a random variable \(X\) with probability density function \(f(x)\) is given by \(E(X) = \int_{0}^{1} x f(x) \, dx\). Given \(f(x) = (p+1) x^{p}\), calculate the expected value as follows:\[ E(X) = \int_{0}^{1} x (p+1) x^p \, dx = (p+1) \int_{0}^{1} x^{p+1} \, dx. \]To integrate, apply the power rule for integration, resulting in:\[ E(X) = (p+1) \left[ \frac{x^{p+2}}{p+2} \right]_{0}^{1} = \frac{p+1}{p+2}. \]

Step 2: Calculate the Average of the Probability Density Function (f_avg)

The average value of a function \(f(x)\) over an interval \([a, b]\) is given by \(f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\). For the probability density function \(f(x) = (p+1)x^p\) over \([0, 1]\), calculate \(f_{\text{avg}}\) as follows:\[ f_{\text{avg}} = \int_{0}^{1} (p+1) x^p \, dx = (p+1) \int_{0}^{1} x^p \, dx. \]Integrate using the power rule for integration:\[ f_{\text{avg}} = (p+1) \left[ \frac{x^{p+1}}{p+1} \right]_{0}^{1} = 1. \]

Step 3: Compare E(X) and f_avg

From previous steps, we have \(E(X) = \frac{p+1}{p+2}\) and \(f_{\text{avg}} = 1\). To show that \(\bar{X} eq f_{\text{avg}}\), compare these values:- \(E(X) = \frac{p+1}{p+2}\) which is always less than 1 for all positive \(p\).- \(f_{\text{avg}} = 1\), a constant value.Clearly, \(E(X)\) is not equal to \(f_{\text{avg}}\), confirming that the average of a random variable is not necessarily equal to the average of its probability density function.

Key Concepts

Probability Density Function

A probability density function (PDF) is a critical concept in statistics and probability theory used to specify the probability of a random variable to have a particular value. In simpler terms, you can think of the PDF as a function that tells you how likely different outcomes are. PDFs are specific to continuous random variables and have some unique properties:

• The probability that a random variable equals a specific value, denoted by its PDF, is always non-negative.
• The integral of a PDF over its entire range is equal to 1, which ensures that the total probability over all possible outcomes is 1.
• For a given interval, the area under the PDF graph represents the probability that the random variable falls within that interval.

In the problem, you are given the PDF for a random variable within the interval [0,1]. The function is defined as \( f(x) = (p+1)x^{p} \), where \( p \) is a positive constant. Here, the PDF offers a way to understand how the values of the random variable are distributed across the interval.

Power Rule for Integration

The power rule for integration is a fundamental technique in calculus used to integrate functions in the form of \( x^n \). It's especially handy because it simplifies the process of finding the antiderivative or integrating a polynomial expression.The formula for using the power rule is as follows:

• For a function \( x^n \), the integral is given by \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] where \( C \) is the constant of integration.
• This rule applies to all real numbers \( n \) except \( n = -1 \), where the integration leads to a logarithmic term.

In the exercise, the power rule is applied to integrate terms like \( x^{p} \) and \( x^{p+1} \) when calculating the expected value and the average of the PDF. You adjust the exponent by 1 and divide by the new exponent, offering a straightforward method to solve the integration tasks.

Random Variable

A random variable is a foundational concept in probability and statistics that represents a numerical outcome of a random phenomenon. Random variables are categorized into two main types: discrete and continuous. In this specific problem, we deal with a continuous random variable denoted by \( X \). Here are some key aspects:

• Random variables have a range or domain that describes all possible values they can take. For this exercise, the range is between 0 and 1.
• The behavior and distribution of a random variable are described by its probability function, which in this case is a PDF.
• One of the primary measures associated with a random variable is the expected value, which represents its mean or average outcome over repetitions of the random phenomenon.

Understanding how random variables are defined and how they function with their probability distributions is crucial for solving complex probability and statistics problems as demonstrated in the exercise.