Problem 83 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose that fifteen observations are chosen at random from the pdf \(f_{Y}(y)=3 y^{2}, 0 \leq y \leq 1\). Let \(X\) denote the number that lie in the interval \(\left(\frac{1}{2}, 1\right)\). Find \(E(X)\).

Step-by-Step Solution

Answer

The expected value \(E(X)\) is approximately 13.125.

Step 1: Compute the Probability of a Random Observation Falling in the Given Interval

We first need to calculate the probability \(P\left(\frac{1}{2} \leq Y \leq 1\right)\) of a random observation lying in the given interval. This requires integrating the probability density function over the given interval. Thus,\[P\left(\frac{1}{2} \leq Y \leq 1\right) = \int_{1/2}^{1} f_{Y}(y) dy = \int_{1/2}^{1}3 y^{2} dy\]Evaluating the integral gives us the probability.

Step 2: Evaluate the Integral

We know the integral of \(y^{2}\) is \(\frac{1}{3} y^{3}\), so adding the limits to this we obtain:\[\int_{1/2}^{1}3 y^{2} dy = \left[ y^{3} \right]_{1/2}^{1} = 1^3 - \left(\frac{1}{2}\right)^{3} = 1 - \frac{1}{8} = \frac{7}{8}\]This is the probability of an observation lying within the given interval.

Step 3: Compute the Expected Value

The expected value \(E(X)\) is the average result we can expect to get when we repeat the experiment multiple times. This is given by the number of trials (i.e., 15) multiplied by the probability of success (i.e., an observation falling in the given interval). Thus, we have:\[E(X) = n \cdot P = 15 \times \frac{7}{8} = \frac{105}{8} = 13.125\]Hence, on average, we can expect about 13.125 values to fall within the interval \(\left(\frac{1}{2}, 1\right)\) out of the 15 observations.

Key Concepts

Probability Density Function

A Probability Density Function (PDF) is a crucial concept in understanding how probabilities are distributed over a range of outcomes for a continuous random variable. In simple terms, a PDF provides the likelihood of a random variable taking on a specific value or falling within a particular interval. In our problem, we have the probability density function given by \( f_{Y}(y) = 3y^2 \) for \( 0 \leq y \leq 1 \). This function tells us how the values of \( Y \) are distributed between 0 and 1.

The PDF must meet two important conditions:

• It must be non-negative, meaning \( f_{Y}(y) \geq 0 \) for any \( y \) within the domain.
• The total area under the PDF curve over the entire range must equal 1, ensuring that the total probability is 1.

When dealing with PDFs, we're often interested in finding the probability of a random variable falling within a given interval, which involves integrating the PDF over that interval.

Integration in Probability

Integration plays a significant role in probability, particularly when dealing with continuous random variables. Integrating the Probability Density Function (PDF) over a specific interval allows us to find the probability of a random variable falling within that range.

For instance, in the exercise, we want to calculate the probability that a random observation \( Y \) falls within the interval \( (\frac{1}{2}, 1) \). This is achieved by evaluating the integral \( \int_{1/2}^{1} 3y^2 \, dy \). By solving this integral, we find that the probability is \( \frac{7}{8} \) or 0.875.

Understanding integration in probability helps us move beyond individual values and focus on the broader distribution of probabilities. It also connects closely with concepts such as the cumulative distribution function (CDF), which uses integration to encapsulate the probabilities of a variable being less than or equal to a certain value.

Random Variables

Random Variables are a foundational concept in probability and statistics, representing quantities whose outcomes are not deterministic but governed by chance. In our context, the random variable \( Y \) is chosen based on the specified probability density function \( f_{Y}(y) = 3y^2 \).

Random variables can be of two types: discrete or continuous. Our example involves a continuous random variable, since \( Y \) can take any value between 0 and 1. Understanding random variables includes grasping certain related terms:

• Expected Value: This is a measure of the center or "average" outcome we might expect to see over many trials of a random process. It's calculated as the sum of each possible value weighted by its probability.
• Variance: This measures the spread of the random variable's possible outcomes.

In the exercise, we computed the expected value \( E(X) \) by multiplying the number of observations with the probability of them falling within a certain interval. This expected value gives us insight into the average outcome of repeating the experiment several times.