Question

Computing a mean and variance. Consider the probability distribution \(p(x)=a x^{n}, 0 \leq x \leq 1\), for a positive integer \(n\). (a) Derive an expression for the constant \(a\), to normalize \(p(x)\). (b) Compute the average \(\langle x\rangle\) as a function of \(n\). (c) Compute \(\sigma^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2}\) as a function of \(n\).

Short answer

The constant \a \ is \+1\. The average \langle x \rangle is \frac{n+1}{n+2}\. The variance \sigma^2\ is \frac{n+1}{n+3} - \frac{(n+1)^2}{(n+2)^2}\.

Step-by-step solution

Title - Derive an Expression for the Constant a

To normalize the probability density function (pdf) \(p(x) = a x^n\), we need the integral of \(p(x)\) over the interval \[0, 1\] to equal 1. This gives us: \[ \int_{0}^{1} p(x) \, dx = 1 \]Substituting \(p(x) = a x^n \), we get: \[ \int_{0}^{1} a x^n \, dx = 1 \]To solve this integral, we have: \[ a \int_{0}^{1} x^n \, dx = 1 \]The integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\), so:\[ a \left[ \frac{x^{n+1}}{n+1} \right]_{0}^{1} = 1\]\[ a \left( \frac{1}{n+1} - 0 \right) = 1\]\[ a \frac{1}{n+1} = 1\] Solving for \(a\): \[ a = n+1 \]

Title - Compute the Average \langle x\rangle

To find the average \langle x \rangle\, we calculate: \[ \langle x \rangle = \int_{0}^{1} x p(x) \, dx \]Substituting \(p(x) = (n+1) x^n \), we get: \[ \langle x \rangle = \int_{0}^{1} x (n+1) x^n \, dx \]This simplifies to: \[ \langle x \rangle = (n+1) \int_{0}^{1} x^{n+1} \, dx \]The integral of \(x^{n+1}\) is \(\frac{x^{n+2}}{n+2}\), so:\[ \langle x \rangle = (n+1) \left[ \frac{x^{n+2}}{n+2} \right]_{0}^{1} \]\[ \langle x \rangle = (n+1) \frac{1}{n+2} \]\[ \langle x \rangle = \frac{n+1}{n+2} \]

Title - Compute the Variance \sigma^2\

To find the variance \sigma^2\, use the formula: \[ \sigma^2 = \langle x^2 \rangle - \langle x \rangle^2 \]First, we need \langle x^2 \rangle:\[ \langle x^2 \rangle = \int_{0}^{1} x^2 p(x) \, dx \]Substituting \(p(x) = (n+1) x^n\), we get: \[ \langle x^2 \rangle = (n+1) \int_{0}^{1} x^{n+2} \, dx \]The integral of \(x^{n+2}\) is \(\frac{x^{n+3}}{n+3}\), so:\[ \langle x^2 \rangle = (n+1) \left[ \frac{x^{n+3}}{n+3} \right]_{0}^{1} \]\[ \langle x^2 \rangle = (n+1) \frac{1}{n+3} \]\[ \langle x^2 \rangle = \frac{n+1}{n+3} \]

Title - Compute \sigma^2\

Finally, use the results of the previous steps: \[ \sigma^2 = \frac{n+1}{n+3} - \left( \frac{n+1}{n+2} \right)^2 \]Expanding the square:\[ \sigma^2 = \frac{n+1}{n+3} - \frac{(n+1)^2}{(n+2)^2} \]

Key concepts

Probability Distribution Normalization

In probability theory, normalization is a key step that ensures the total probability of a random variable sums up to 1. For a probability density function (pdf) like the one given in the exercise, we follow the rule:
\(\[\begin{equation} \int_{0}^{1} p(x) \, dx = 1. \end{equation}\]\)
By solving this integral, we essentially find the constant factor (let's call it \(a\)) that makes sure our pdf, \( p(x) = a x^n\), is a valid probability distribution. The procedure typically involves integrating over the entire range of possible values for our random variable (here, from 0 to 1), and solving for \(a\) to set this integral to 1.
This normalization ensures that when you apply the pdf to any real-world problem, you correctly reflect the inherent probabilities of the different outcome values.

Statistical Thermodynamics

Statistical thermodynamics bridges the macroscopic properties of materials and their microscopic behaviors. One key component is understanding distributions and probability densities. The normalization step in your probability distribution relates to ensuring that the sum of all possible microscopic states (microstates) probabilities equals 1.
In this context, the mean (average) value of a distribution corresponds to observable macroscopic thermodynamic properties, like temperature or pressure. Similarly, the variance indicates the spread or fluctuation of these thermodynamic quantities around their average values.
By integrating fundamental statistical principles, we can better comprehend and predict thermodynamic behaviors, creating a powerful connection between microscopic statistical probabilities and macroscopic thermodynamic quantities.

Mean and Variance Calculation

The concepts of mean (or expected value, \( \langle x \rangle \)) and variance (\( \sigma^2 \)) are fundamental in statistics and probability theory. They provide key insights into the distribution of a random variable.

• Mean: The mean of a probability distribution gives you the central value, essentially capturing what you would expect as an average outcome. For our function, we calculated:
  \( \langle x \rangle = \int_{0}^{1} x p(x) \, dx = \frac{n+1}{n+2}. \)
• Variance: The variance measures the spread of the distribution, giving us a sense of how much the values differ from the mean. To compute this, we first find \( \langle x^2 \rangle \), then use: \(\[\begin{equation} \sigma^2 = \frac{n+1}{n+3} - \left( \frac{n+1}{n+2} \right)^{-2}. \end{equation}\]\)

These calculations help you understand both the central tendency and the dispersions of your distribution, which are critical for various applications in statistics and science.

Probability Density Function

A probability density function (pdf) is a function that describes the likelihood of a continuous random variable to take on a particular value. For a pdf, the probability that the random variable falls within a particular range is given by the integral of this function over that range. In other words, if \(p(x) = ax^n\) represents our pdf, then the area under the curve of this function over \(0 \leq x \leq 1\) must sum to 1 to maintain proper probability distribution characteristics.
PDFs are indispensable in statistics as they allow analyzing and interpreting continuous data. They help to deduce properties such as the mean and variance, which are vital statistics that summarize critical aspects of the data distribution.
Understanding the shape and behavior of pdfs assists not only in visualizing data but also in applying strategical approaches to solve real-life problems efficiently.