Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a continuous type distribution. (a) Find \(P\left(X_{1} \leq X_{2}\right), P\left(X_{1} \leq X_{2}, X_{1} \leq X_{3}\right), \ldots, P\left(X_{1} \leq X_{i}, i=2,3, \ldots, n\right)\) (b) Suppose the sampling continues until \(X_{1}\) is no longer the smallest observation, (i.e., \(\left.X_{j}

Short answer

The probabilities of \(X_1\) being less than or equal to \(X_i\) is always \(0.5\). The probability that \(Y=y\) is \(1/(y(y+1))\), representing the number of trials needed until \(X_1\) is no longer the smallest observation. The mean and variance of \(Y\) can be calculated using the given formulas, however, it might not be possible if the sums do not converge.

Step-by-step solution

Finding \(P(X_1 \leq X_i)\)

Each \(X_i\) is independent and identically distributed, so the probability that one of them is the greatest is equal for all. Hence, the probability that \(X_1\) is smaller or equal to any other \(X_i\) is \(0.5\). For part (a), \(P(X_1 \leq X_i) = 0.5\) for \(i=2,3, \ldots, n\).

Finding the distribution of \(Y\)

In part (b), the task is to find the probability distribution of \(Y\), which represents the number of trials needed until \(X_1\) is no longer the smallest observation. The probability that \(Y=y\) means that \(X_1\) was the smallest observation in the first \(y\) trials but it was larger than the \((y+1)\)-th observation. Since all \(X_i\) are independent, this is equal to the event that among \((y+1)\) observations, the first is the smallest, so \(P(Y=y) = 1/(y(y+1))\).

Computing mean and variance of \(Y\)

Part (c) requires finding the mean and variance of \(Y\). The mean is calculated as \(\mu = \sum_{y=1}^{\infty} y * P(Y=y)\). Using the calculated \(P(Y=y)\) from step 2, the mean could be calculated (if the series converges). The variance is calculated as \(\sigma^2 = \sum_{y=1}^{\infty} (y-\mu)^2 * P(Y=y)\), however calculation of these might not be possible if the series do not converge.

Key concepts

Random Variables

In probability theory, a random variable is essentially a numerical description of the outcome of a random event. You can think of it as a variable that can take different values, each with a certain probability. This is important in the context of our exercise because each observation, like \(X_1, X_2, \ldots, X_n\), is considered as a random variable.
Random variables can be either discrete or continuous. Discrete random variables have specific outcomes you can count (like tossing a die), whereas continuous random variables, like those in our exercise, can take any value within a range.

• Discrete example: Number of heads in 10 flips of a coin.
• Continuous example: Heights of students in a classroom.

Understanding random variables is essential because they allow us to model and analyze randomness in the real world. They provide a foundation for predicting and understanding probabilities within given constraints.

Mean and Variance

The mean and variance of a probability distribution give us an understanding of the distribution's central tendency and spread. The mean, sometimes called the expected value, is simply the average of all possible outcomes. In our exercise, we calculate the mean of the random variable \(Y\), which represents the trials needed until \(X_1\) is no longer the smallest observation.
The formula for the mean \(\mu\) of a discrete random variable \(Y\) is:\[\mu = \sum_{y} y \cdot P(Y=y)\]This helps us determine the average number of trials needed until \(X_1\) is no longer the smallest.
Variance, on the other hand, measures the variability or spread of a distribution. It tells us how much the outcomes are spread out from the mean. The formula for variance \(\sigma^2\) is:\[\sigma^2 = \sum_{y} (y - \mu)^2 \cdot P(Y=y)\]This equation turns the abstract concept of spread into a calculation. In our context, it would help us understand how variable the number of trials is from the mean.
Both mean and variance are fundamental in statistical analysis to summarize and interpret the properties of probability distributions.

Independent and Identically Distributed

When we say a set of random variables are independent and identically distributed (i.i.d), it implies that each variable has the same probability distribution as the others and they are all mutually independent.
This is a major simplification that allows statisticians to analyze data more easily. For example, if \(X_1, X_2, \ldots, X_n\) are i.i.d, it means each \(X_i\) is independent of the others, so the outcome of one does not affect the outcome of another. It also means that each \(X_i\) follows the same probability distribution.

• Independence: Knowing the value of one variable does not help us predict the others.
• Identically Distributed: All variables share the same probability distribution.

This underpinning concept was used in the solution to determine that each \(X_i\) had equal probability. That's why in the exercise, the probability that \(X_1\) is the smallest is equal across all trials.