Question

Find the moments of the "square distribution" defined by $$ P(x)=0 \quad \text { for }|x|>a, \quad P(x)=(2 a)^{-1} \quad \text { for }|x|

Short answer

For even n, the n-th moment of the square distribution is \( \mu_n = \frac{a^n}{n+1} \) and for odd n, \( \mu_n = 0 \) due to symmetry.

Step-by-step solution

Understanding the Square Distribution

The square distribution given has a constant probability density of \( (2a)^{-1} \) for the interval \( -a < x < a \) and a probability density of 0 outside that interval. This distribution is symmetrical about the y-axis. The moments of a distribution are expected values of powers of the variable.

Calculating the n-th Moment

The n-th moment of a distribution is given by \( \mu_n = \int_{-\infty}^{\infty} x^n P(x) \, dx \) where \( P(x) \) is the probability density function.

Simplifying the Integral

Since \( P(x) \) is 0 outside \( -a < x < a \) and constant within this range, the integral simplifies to \( \mu_n = \int_{-a}^{a} x^n (2a)^{-1} \, dx \) for the n-th moment.

Evaluating the Integral for Even Powers

For even powers of n, the integral \( \mu_n = \int_{-a}^{a} x^n (2a)^{-1} \, dx \) is symmetric about the y-axis and hence \( \mu_n = 2 \int_{0}^{a} x^n (2a)^{-1} \, dx \) which evaluates to \( \mu_n = \frac{1}{(n+1)(2a)} x^{n+1} \biggr|_0^a = \frac{a^n}{n+1} \) since \( x^n \) is even.

Evaluating the Integral for Odd Powers

For odd powers of n, the integral \( \mu_n = \int_{-a}^{a} x^n (2a)^{-1} \, dx \) is antisymmetric about the y-axis, which means positive and negative areas cancel each other out. Thus, \( \mu_n \) is equal to 0 for all odd n.

Key concepts

Probability Density Function

A Probability Density Function (PDF), represented as P(x), is an essential concept in statistics. It describes how the probability is distributed over the values of a continuous random variable. Think of it as a graph that tells you how likely different outcomes are: the greater the value of the PDF at a particular point, the higher the likelihood that the random variable will be near that point.

For the square distribution described in the exercise, the PDF is uniquely simple. It assigns a constant probability density of (2a)^{-1} within the interval from -a to a and a zero probability outside of that interval. This form of a PDF is characteristic of uniform distributions because each outcome in the interval is equally likely. The square distribution is a specific type of uniform distribution with a neat, boxy shape when plotted.

N-th Moment Calculation

The concept of moments in the context of probability distributions is akin to the idea of 'averages.' While the first moment is the mean of the distribution, higher moments capture more nuanced characteristics like variation, skewness (asymmetry), and kurtosis ('tailedness'). Calculating these moments involves integrating the power of the variable multiplied by the PDF.

The n-th moment of a distribution, mathematically noted as \( \mu_n \), is calculated using the integral \( \mu_n = \int_{-\infty}^{\infty} x^n P(x) \, dx \). However, because the square distribution's PDF is zero outside the interval [-a, a], the integral simplifies, and you only have to consider the interval where the PDF has a positive value. It's like ignoring the parts of a picture that are just blank space, focusing on where the actual image is.

Symmetric Probability Distribution

A symmetric probability distribution is one where both sides of the distribution mirror each other across a central point, called the axis of symmetry. It's like the distribution has taken a selfie - its left and right look exactly the same from the point of symmetry!

In the case of our square distribution, the y-axis serves as the axis of symmetry. This symmetry implies that all odd moments (1st, 3rd, 5th, etc.) are zero because the equal and opposite values on either side of the y-axis cancel each other out. Envision a seesaw with perfectly equal weights on both ends - it stays balanced, and no side dips down. This cancellation effect is a handy characteristic, especially when dealing with complex integrals, as it simplifies the process of finding the moments of the distribution.