Problem 79 - An Introduction to Mathematical Statistics and Its Applications

Question

Let the random variable \(Y\) have the uniform distribution over \([a, b]\); that is, \(f_{Y}(y)=\frac{1}{b-a}\) for \(a \leq y \leq b\). Find \(E(Y)\) using Definition 3.5.1. Also, deduce the value of \(E(Y)\), knowing that the expected value is the center of gravity of \(f_{Y}(y)\).

Step-by-Step Solution

Answer

The Expected value E(Y) of the uniform distribution over \([a, b]\) is the midpoint of the interval \([a, b]\), i.e., \(E(Y) = \frac{b+a}{2}\)

Step 1: Write down the Expected Value Formula

The expected value of a random variable \(Y\) with a probability density function \(f_Y(y)\), according to definition 3.5.1, is given by: \(E(Y) = ∫y*f_Y(y) dy\) . Now, substitute the value of \(f_Y(y)\) which is given as \(\frac{1}{b-a}\). So: \(E(Y) = ∫_a^b y * \frac{1}{b-a} dy\)

Step 2: Evaluate the Integral

Integrate the expression to calculate the expected value of \(Y\). \(E(Y) = \frac{1}{b-a} * ∫_a^b y dy = \frac{1}{b-a} * [\frac{1}{2}y^2]_a^b\). Evaluating the square bracket expression at \(b\) and \(a\) gives \(E(Y) = \frac{1}{b-a} * (\frac{1}{2}b^2 - \frac{1}{2}a^2)\)

Step 3: Simplify the Expression

Simplify the above expression to obtain the Expected Value \(E(Y)\): \(E(Y) = \frac{b^2 - a^2}{2(b-a)} = \frac{b+a}{2}\)

Step 4: Deduce the Value of \(E(Y)\)

As per the problem, the expected value is the center of gravity of \(f_Y(y)\). For a uniform distribution like this one, the center of gravity is the midpoint of the interval \([a, b]\). Thus, the expected value we just calculated \(\frac{b+a}{2}\) makes sense as it is indeed the center of the interval.

Key Concepts

Uniform Distribution

When we deal with random variables in statistics, the uniform distribution often comes up as a fundamental concept. It represents a situation where all outcomes are equally likely to occur. Consider a perfectly balanced dice; each of its six faces has an equal chance of landing face up. The same principle applies to a continuous range in the uniform distribution, where any number within a specific interval is equally likely to be observed.

The uniform distribution over an interval \[a, b\] is described mathematically by its probability density function \(f_Y(y)\), which is constant across the interval and zero outside of it. The value of this function is \(\frac{1}{b-a}\) for any point \(y\) within the interval \[a, b\]. The uniformity comes from the fact that the height of the probability density function is consistent, meaning every point in the interval is equally 'dense' with probability.

Probability Density Function

The probability density function (PDF) is a crucial concept in understanding continuous probability distributions. It characterizes the likelihood of a continuous random variable taking on a specific value. Unlike a probability mass function in discrete distributions, the PDF gives us probabilities over intervals, not points, since the probability of a continuous variable taking any exact value is effectively zero.

A PDF, in essence, describes the relative likelihood of the random variable falling within a particular range. For instance, in our uniform distribution scenario, \(f_Y(y) = \frac{1}{b-a}\) indicates a flat PDF over the interval \[a, b\]. It's important to remember that while the PDF can tell us the density of the probability for intervals, to get the probability of the variable falling within a specific range, we'd need to integrate the PDF over that range.

Integrating Probability Functions

Integration plays a pivotal role in calculating probabilities for continuous distributions. When we integrate the probability density function of a continuous random variable, we're essentially summing up an infinite number of infinitesimally small probabilities to find the overall probability within a certain interval. This process is akin to summing the area under the curve of the PDF over the range of interest.

In the context of the uniform distribution, integrating the PDF \(f_Y(y) = \frac{1}{b-a}\) from \(a\) to \(b\) will always yield 1, confirming that the total probability across the entire distribution is 100%. When it comes to finding the expected value, \(E(Y)\), we integrate the product of the random variable \(Y\) and its PDF. This integral gives us the average value or 'center of gravity' for the random variable, embodying the balance point of all possible outcomes weighted by their likelihood.