Problem 24 - Calculus

Question

Use \(\int(x-\mu)^{2} p(x) d x\) and also \(\int x^{2} p(x) d x-\mu^{2}\) to find \(\sigma^{2}\) for the uniform distribution: \(p(x)=1\) for \(0 \leqslant x \leqslant 1\).

Step-by-Step Solution

Answer

The variance \(\sigma^2\) is \(\frac{1}{12}\).

Step 1: Define the Uniform Distribution

The uniform distribution provided is defined on the interval \(0 \leq x \leq 1\) with a probability density function (PDF) of \(p(x) = 1\). This means the distribution is constant over this interval.

Step 2: Calculate the Mean \(\mu\)

The mean \(\mu\) of a uniform distribution on the interval \([a, b]\) is given by \(\mu = \frac{a + b}{2}\). For this problem, \(a = 0\) and \(b = 1\), so:\[\mu = \frac{0 + 1}{2} = \frac{1}{2}\]

Step 3: Use the First Formula for Variance

We start by using the formula for variance: \(\int (x - \mu)^2 p(x) \, dx\). Substitute \(x \in [0,1]\), \(p(x) = 1\), and \(\mu = \frac{1}{2}\):\[\int_{0}^{1} \left(x - \frac{1}{2}\right)^2 \, dx\]

Step 4: Integrate to Find Variance

Solve the integral:\[\int_{0}^{1} \left(x - \frac{1}{2}\right)^2 \, dx = \int_{0}^{1} \left(x^2 - x + \frac{1}{4}\right) \, dx = \left[\frac{x^3}{3} - \frac{x^2}{2} + \frac{x}{4}\right]_{0}^{1}\]Evaluate at bounds:\[\left(\frac{1}{3} - \frac{1}{2} + \frac{1}{4}\right) = \frac{1}{3} - \frac{2}{4} + \frac{1}{4} = \frac{1}{12}\]

Step 5: Use the Second Formula for Variance

The second formula for variance is: \(\int x^2 p(x) \, dx - \mu^2\). Substitute \(p(x) = 1\) and \(\mu = \frac{1}{2}\):\[\int_{0}^{1} x^2 \, dx - \left(\frac{1}{2}\right)^2\]

Step 6: Integrate to Find Second Variance

Evaluate \(\int_{0}^{1} x^2 \, dx\):\[\left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1}{3}\]Subtract \(\mu^2 = \frac{1}{4}\):\[\frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}\]

Step 7: Confirm the Variance \(\sigma^2\)

Both methods have indicated the variance \(\sigma^2\) for the given distribution is \(\frac{1}{12}\). This confirms the consistency and correctness of the solution.

Key Concepts

Uniform Distribution

A uniform distribution is a type of probability distribution where every outcome in a specific range is equally likely. The graph of this distribution is flat or constant, resembling a rectangle. This type of distribution is common when there is no additional information to favor one outcome over another within the specified boundaries. In this problem, we see a continuous uniform distribution within the interval \(0 \leq x \leq 1\). This means any value of \(x\) in this range is equally probable, which is why the probability density function is \(p(x) = 1\). The uniform distribution is often described as being simple and fair, as each element within the interval has equal chance of occurrence.

Probability Density Function

The Probability Density Function (PDF) is a function that describes the likelihood of a random variable to take a particular value. For continuous distributions like the uniform distribution, the PDF is used to find the probability of the random variable falling within a particular range. It is crucial to understand that the probability of a continuous random variable taking an exact value is zero. Instead, the PDF helps us understand the distribution over an interval. For the uniform distribution in this problem, the PDF is given by \(p(x) = 1\) for the interval \(0 \leq x \leq 1\).

This means the height of the PDF is constant, spanning the entire interval. The total area under the PDF curve should equal 1, which confirms that the sum of probabilities over the interval is complete and certain.

Mean Calculation

Calculating the mean or expected value of a distribution is an essential statistical measure that provides the central tendency of the distribution. For a uniform distribution defined on the interval \([a, b]\), the mean is calculated using the formula:

• \(\mu = \frac{a + b}{2}\)

In our example, the interval is \([0, 1]\), so the mean is

• \(\mu = \frac{0 + 1}{2} = \frac{1}{2}\)

This indicates that the middle point of the interval 0 to 1 is 0.5, which is expected as every point within the range is equally likely.

Integration of Functions

Integration is a fundamental concept in calculus that is used to calculate areas under curves, among other things. When dealing with probability distributions, integration aids in determining probabilities and expected values, such as the mean and variance. In this exercise, we deal with integrating functions over the interval \(0 \leq x \leq 1\).

To find the variance of the uniform distribution, we use two integral setups:

• The first integral setup: \(\int_{0}^{1} (x - \mu)^2 \cdot p(x) \, dx\) computes how far each point squared deviates from the mean.
• The second integral setup: \(\int_{0}^{1} x^2 \cdot p(x) \, dx - \mu^2\) calculates the expected value of \(x^2\) and subtracts the square of the mean.

Both methods rely heavily on the skill of performing integration correctly, evaluating it to find values that represent variance. Appropriately integrating these expressions provides confidence in the calculation results, ensuring that both methods confirm each other mathematically. Understanding these integral calculations is crucial for determining distribution characteristics accurately.