Problem 182 - An Introduction to Mathematical Statistics and Its Applications

Question

A random sample of size \(n=2 k\) is taken from a uniform pdf defined over the unit interval. Calculate \(P\left(X_{1}<\frac{1}{2}, X_{2}>\frac{1}{2}, X_{3}<\frac{1}{2}, X_{4}>\frac{1}{2}, \ldots, X_{2 k}>\frac{1}{2}\right) .\)

Step-by-Step Solution

Answer

The probability that \(X_1\) is less than 0.5, \(X_2\) is greater than 0.5, until \(X_{2k}\) being greater than 0.5 is \(0.5^{2k}\).

Step 1: Understanding the Behavior of a Uniform Distribution

The probability density function (pdf) of a uniform distribution in an interval \( [a, b] \) is defined as \( f(x) = \frac{1}{b - a} \) for \( a ≤ x ≤ b \) and zero elsewhere. In this exercise, the interval is [0, 1], so the pdf is \( f(x) = 1 \) for \( 0 ≤ x ≤ 1 \). This implies that, for a uniform distribution over the interval [0, 1], the probability that a random variable \(X\) takes a value less than or greater than 0.5 is also 0.5.

Step 2: Calculate Probability for Single Random Variable

Given the uniform pdf, the probability that a single \(X_i\) is less than 0.5 is 0.5 (i.e., \(P(X_i<0.5) = 0.5\), the same applies to \(P(X_i>0.5) = 0.5\). Fundamentally, these kind of calculations follow from the symmetry of the uniform distribution around its midpoint.

Step 3: Calculate Combined Probability

In our case we are looking at a sample of size \(2k\), and we want to find the combined probability that \(X_1\) is less than 0.5, \(X_2\) is greater than 0.5, until \(X_{2k}\) being greater than 0.5. As these are independent events (one does not affect the result of another), we can calculate the combined probability by simply multiplying probabilities of each single event together. Therefore, probability of the whole sequence of events is \( 0.5^{2k} \).

Key Concepts

Probability Density Function

To truly understand what's going on in our textbook exercise, we must first delve into the concept of a probability density function (pdf). A pdf helps to model the likelihood of outcomes within a continuous range, and is an essential part of understanding uniform distributions.

The uniform distribution is a type of continuous probability distribution where all intervals of the same length have equal probability. When we talk about a uniform distribution across a unit interval, we're looking at a range from 0 to 1. Imagine this interval as a straight, horizontal line; every point on this line is just as likely to be chosen as any other. Mathematically, the pdf of a uniform distribution on the interval [0, 1] is expressed as \( f(x) = 1 \) for \( 0 \leq x \leq 1 \), and zero for values outside this range.

This uniform pdf is particularly simple because it's just a flat line at height 1 from 0 to 1. The exercise explores a case where we have multiple events, and we use this pdf to determine probabilities of certain outcomes. Remember that the total area under the pdf curve represents the total probability, which is always 1 for any probability distribution.

Independent Events in Probability

Our exercise also takes us through the realm of independent events in probability. But what does it mean for events to be independent? Simply put, we say two events are independent if the occurrence of one event doesn't affect the outcome of the other.

In the context of our uniform distribution example, each random variable \(X_i\) represents an event. Because the selected values for \(X_1, X_2, ..., X_{2k}\) are all from independent trials of the same uniform distribution, their values do not influence each other. For instance, if \(X_1 < 0.5\), this information tells us nothing about the likelihood of \(X_2 > 0.5\), and so on. The independence of these variables is the key factor that allows us to multiply their individual probabilities together to find the joint probability of a series of such events.

Calculating Probabilities

Finally, let's focus on the centerpiece of the exercise - calculating probabilities. The core idea in our example is to determine the combined probability of a sequence of independent events happening together.

Here, each event is described as a random variable either being less than or greater than 0.5. Since the distribution is uniform, each single event has a probability of 0.5. To calculate the probability of the entire sequence of events, which are independent, we simply multiply the probabilities of each event occurring. If we visualize this as a tree diagram where each branch splits into two equally-likely paths (less than or greater than 0.5), it becomes clear why multiplying the probabilities gives us the correct combined probability.

The result, in theory, is beautifully simple: the probability of our sequence of events is \( 0.5^{2k} \). It demonstrates how understanding the individual pieces - uniform distribution, independence, and probability calculations - allows us to solve more complex problems in probability.