Verifying Property 2 Of Probability Density Function For F(x) = (1/9)x^2

In the realm of probability and statistics, probability density functions (PDFs) play a crucial role in describing the likelihood of a continuous random variable falling within a specific range of values. To qualify as a legitimate PDF, a function must adhere to certain properties. One such property, often referred to as Property 2, states that the total area under the curve of the function over its entire domain must equal 1. This property ensures that the total probability of all possible outcomes is equal to 1, which is a fundamental requirement in probability theory.

In this article, we will delve into the verification of Property 2 for a given function, $f(x) = \frac{1}{9}x^2$, over a specified interval. We will explore the steps involved in calculating the area under the curve of this function within the given interval and demonstrate how this area relates to the property of a PDF. By the end of this discussion, you will gain a comprehensive understanding of how to verify Property 2 for a PDF and appreciate its significance in the context of probability theory.

Understanding Probability Density Functions (PDFs)

Before we dive into the specifics of verifying Property 2, let's take a moment to solidify our understanding of probability density functions (PDFs) themselves. In essence, a PDF is a function that describes the relative likelihood of a continuous random variable taking on a given value. Unlike discrete probability distributions, where we can directly assign probabilities to individual outcomes, continuous random variables can take on any value within a given range, making it necessary to use a function to represent the probability distribution.

The value of the PDF at a particular point does not directly represent the probability of the variable taking on that exact value. Instead, the probability of the variable falling within a specific interval is given by the area under the curve of the PDF over that interval. This is a crucial concept to grasp, as it forms the basis for understanding Property 2.

A valid PDF must satisfy two key properties:

1. Non-negativity: The function must be non-negative for all values within its domain. This makes intuitive sense, as probability cannot be negative.
2. Normalization: The total area under the curve of the function over its entire domain must equal 1. This property ensures that the total probability of all possible outcomes is equal to 1.

In this article, we are focusing on the second property, the normalization property, and how to verify it for a given function.

Verifying Property 2 for f(x) = (1/9)x^2

Now, let's turn our attention to the specific function we are interested in: $f(x) = \frac{1}{9}x^2$. Our goal is to verify Property 2, which states that the total area under the curve of this function over a given interval must equal 1 if it is a valid PDF within that interval. Let's assume the interval we are considering is $[a, b]$.

To calculate the area under the curve of $f(x)$ over the interval $[a, b]$, we need to perform a definite integral. The definite integral of a function between two limits represents the area bounded by the function's curve, the x-axis, and the vertical lines corresponding to the limits of integration.

In our case, the definite integral we need to evaluate is:

$$\int_{a}^{b} f(x) dx = \int_{a}^{b} \frac{1}{9}x^2 dx$$

Step 1: Evaluate the Definite Integral

To evaluate this definite integral, we first need to find the antiderivative of the function $\frac{1}{9}x^2$. Recall that the power rule of integration states that the antiderivative of $x^n$ is $\frac{x^{n+1}}{n+1}$, where $n$ is any constant except -1. Applying this rule to our function, we get:

$$\int \frac{1}{9}x^2 dx = \frac{1}{9} \int x^2 dx = \frac{1}{9} \cdot \frac{x^3}{3} + C = \frac{x^3}{27} + C$$

where $C$ is the constant of integration. However, when evaluating definite integrals, the constant of integration cancels out, so we can ignore it for our purposes.

Now, we can evaluate the definite integral using the fundamental theorem of calculus, which states that the definite integral of a function $f(x)$ from $a$ to $b$ is equal to the difference between the antiderivative evaluated at $b$ and the antiderivative evaluated at $a$:

$$\int_{a}^{b} \frac{1}{9}x^2 dx = \left[\frac{x^3}{27}\right]_{a}^{b} = \frac{b^3}{27} - \frac{a^3}{27}$$

Step 2: Set the Result Equal to 1 and Solve for the Interval

To verify Property 2, we need to determine the interval $[a, b]$ for which the area under the curve of $f(x)$ is equal to 1. In other words, we need to solve the following equation:

$$\frac{b^3}{27} - \frac{a^3}{27} = 1$$

Multiplying both sides by 27, we get:

$$b^3 - a^3 = 27$$

This equation tells us the relationship between the limits of integration, $a$ and $b$, that will result in the area under the curve being equal to 1. There are infinitely many possible intervals that satisfy this equation. For example, if we choose $a = 0$, then we have:

$$b^3 - 0^3 = 27$$

$$b^3 = 27$$

$$b = 3$$

So, the interval $[0, 3]$ is one possible interval over which the area under the curve of $f(x) = \frac{1}{9}x^2$ is equal to 1.

Step 3: Verify the Result

To verify our result, we can plug the values $a = 0$ and $b = 3$ back into the definite integral:

$$\int_{0}^{3} \frac{1}{9}x^2 dx = \frac{3^3}{27} - \frac{0^3}{27} = \frac{27}{27} - 0 = 1$$

This confirms that the area under the curve of $f(x) = \frac{1}{9}x^2$ over the interval $[0, 3]$ is indeed equal to 1, thus verifying Property 2 for this function over this interval.

Importance of Property 2

The normalization property, or Property 2, is not just a mathematical curiosity; it has profound implications in probability theory and practical applications. It ensures that the PDF represents a valid probability distribution, where the total probability of all possible outcomes is equal to 1. This is essential for making meaningful probabilistic statements and calculations.

For instance, consider using a PDF to model the distribution of heights in a population. If the total area under the PDF curve were not equal to 1, it would imply that the probabilities of all possible heights do not add up to 100%, which is nonsensical. Property 2 guarantees that the probabilities are properly normalized, allowing us to make accurate predictions and inferences about the distribution of heights.

In various fields, such as finance, engineering, and physics, PDFs are used extensively to model random phenomena. Verifying Property 2 ensures the reliability and validity of these models, enabling informed decision-making and accurate predictions.

Conclusion

In this article, we have explored the verification of Property 2 for the function $f(x) = \frac{1}{9}x^2$ over a given interval. We learned that Property 2, also known as the normalization property, is a fundamental requirement for a function to qualify as a probability density function (PDF). It states that the total area under the curve of the function over its entire domain must equal 1.

We demonstrated the steps involved in verifying Property 2, which include evaluating the definite integral of the function over the given interval, setting the result equal to 1, and solving for the interval. We found that the area under the curve of $f(x) = \frac{1}{9}x^2$ over the interval $[0, 3]$ is equal to 1, thus verifying Property 2 for this function over this interval.

Furthermore, we highlighted the importance of Property 2 in ensuring that a PDF represents a valid probability distribution and its significance in various practical applications. By understanding and verifying Property 2, we gain confidence in using PDFs to model random phenomena and make informed decisions based on probabilistic information.