Question

Suppose a random sample of size 2 is obtained from a distribution which has pdf \(f(x)=2(1-x), 0

Short answer

The probability that one sample observation is at least twice as large as the other is \(\frac{5}{12}\).

Step-by-step solution

Understand the Distribution

The likely PDF function provided is \(f(x) = 2(1-x)\) for \(0 < x < 1\), and zero elsewhere. This means the value of x lies between 0 and 1.

Set up Double Integral

Let X and Y be identical independent random variables with common pdf. The task is to find the probability P(2X < Y) + P(2Y < X). Expressing it mathematically, it becomes a double integral \(\int_{0}^{0.5}\) \(\int_{2x}^{1}\) \(2(1-y)dy dx + \int_{0.5}^{1}\) \(\int_{0}^{y/2}\) \(2(1-y)dy dx\).

Solving the Integrals

Next, solve the integrals separately. After integrating with respect to y in both integrals, you get \(2x - x^2\). After integrating with respect to x in both cases, you get \(\frac{1}{2} - \frac{1}{12} + \frac{1}{3} - \frac{1}{4} = \frac{5}{12}\).

Sum the Results from Both Regions

Since these are distinct intervals, you can simply add up the results. The final probability is \(\frac{5}{12}\).

Key concepts

Probability Density Function

The Probability Density Function (PDF) is a fundamental concept in statistics that describes the likelihood of a random variable taking on a particular value. In the given exercise, the PDF is defined as \(f(x) = 2(1-x)\) for \(0 < x < 1\), and it is zero elsewhere. This indicates that the random variable X can only take values within the interval from 0 to 1.

The function \(f(x) = 2(1-x)\) shapes the likelihood of these values. Since this PDF is continuous, the probability of X being exactly any specific value is zero. Instead, we determine the probability over intervals. The total area under the PDF curve across 0 to 1 should always be equal to 1, ensuring the sum of all probabilities within this range is logically consistent.

PDFs play a crucial role in determining probabilities for continuous random variables. They help us understand the distribution and the chance that the variable will fall into a particular range, forming the bedrock for most probability calculations with continuous data.

Random Variables

Random Variables are used to quantify the outcomes of a random phenomenon. In this context, they represent numerical outcomes from the sample or experiment conducted. In the original exercise, we deal with two random variables, X and Y, which are independent and identically distributed from the given PDF. This means each variable will independently assume a value according to the distribution \(f(x) = 2(1-x)\).

Understanding random variables is key because they allow complex probabilistic systems to be broken down into manageable numerical values. They can be classified into discrete and continuous variables. In our exercise, X and Y are continuous random variables due to their range being an interval (0 to 1).

Our task involves comparing these two variables to determine the probability that one sample observation is at least twice as large as the other. Realizing that these are random variables helps us in employing the correct mathematical approaches such as integration to solve for probabilities.

Integration

Integration is the mathematical process of finding the area under a curve. It is a powerful tool used for working with continuous probability distributions, as seen in the exercise. When dealing with continuous random variables, like in the provided example, we use integration to compute probabilities over certain intervals.

In the problem, we set up a double integral to find the probability that one sample observation is at least twice as large as the other. This involves integrating the given PDF over the specified range for X and Y. Specifically, we calculate the double integral:

• \(\int_{0}^{0.5} \int_{2x}^{1} 2(1-y)dy dx\)
• \(+ \int_{0.5}^{1} \int_{0}^{y/2} 2(1-y)dy dx\)

Here, the limits of integration reflect conditions like \(2X < Y\) and \(2Y < X\), which come from the requirement of one sample being at least twice as large as the other. By solving these integrals, we combine mathematical precision with real-world randomness to find our answer.

Independent Samples

The concept of independent samples is crucial in statistics when dealing with probability distribution functions. Two samples are considered independent if the occurrence of any particular value in one does not affect the probabilities of the other’s outcomes. In essence, their joint probability can be calculated as the product of their individual probabilities.

In the exercise, X and Y are identified as independent samples from the same distribution. This independence is critical because it means any relationship or condition we apply to one variable doesn’t influence the other. This is why the PDF remains the same for both variables.

Understanding the independence of samples helps us use joint probability distributions effectively in our calculations. These kinds of scenarios are often met in real-world applications, ranging from quality control in manufacturing to risk assessments in financial portfolios. By recognizing when samples are independent, we ensure our probability calculations reflect true randomness.