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Magazine Chapter 5: Problem 81
Suppose that \(X\) and \(Y\) are independent, continuous uniform random variables
for \(0
Short Answer
Expert verified
The probability that \(|X-Y|<0.5\) is \(\frac{1}{4}\).
Step by step solution
01
Identify the Distribution
Since both random variables \(X\) and \(Y\) are uniform and independent, each follows a Uniform(0,1) distribution. The joint distribution is therefore a uniform distribution over the unit square \(0 < x < 1\) and \(0 < y < 1\) on the XY-plane.
02
Set Up the Problem
The problem asks for the probability that \(|X-Y| < 0.5\). This can be rewritten as two inequalities: \(-0.5 < X-Y < 0.5\). This represents a band between the lines \(X=Y-0.5\) and \(X=Y+0.5\) on the unit square.
03
Visualize the Integration Area
Sketch the lines \(y=x+0.5\) and \(y=x-0.5\) on the unit square. The valid region is the area where \(-0.5 < y-x < 0.5\). This forms a band crossing through the square from corner to corner. Since part of this region lies outside the boundary of the [0,1] interval, find the effective region within these boundaries.
04
Determine Points of Intersection
Find the intersection points of the lines with the unit square boundaries. The line \(y=x-0.5\) intersects at points (0,0.5) and (0.5,1), while \(y=x+0.5\) intersects at points (0.5,0) and (1,0.5). These points bound the effective integration region.
05
Calculate the Probability as an Area
The effective region within the square is a hexagon which can be split into two triangles. Calculate the area of these triangles within the bounds of the unit square.
06
Calculate Individual Triangle Areas
Calculate each triangle using the points found: The first triangle has vertices (0,0.5), (0.5,1), and (0.5,0), while the second triangle has vertices (1,0.5), (0.5,0), and (0.5,1). Use the area formula for triangles: \(\text{Area} = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\).
07
Compute and Sum the Areas
The area for the first triangle is \(\frac{1}{8}\), and for the second triangle is also \(\frac{1}{8}\), so the total effective area where \(|X-Y| < 0.5\) is \(\frac{1}{8} + \frac{1}{8} = \frac{1}{4}\).
08
Conclusion with Probability
Since the area of the whole region where both \(X\) and \(Y\) lie between 0 and 1 is 1, the probability \(|X-Y| < 0.5\) is exactly the area calculated: \(\frac{1}{4}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuous Uniform Distribution
A continuous uniform distribution is a statistical distribution where all outcomes are equally likely within a specified range. Imagine a spinning wheel that lands anywhere between 0 and 1. - Instead of having more chances landing at certain numbers, every number in the interval is equally probable. - The distribution is characterized by a constant probability density function.In this case, we have two independent continuous uniform random variables, \(X\) and \(Y\), defined on the interval \(0 < x < 1\) and \(0 < y < 1\). This setup indicates that both \(X\) and \(Y\) are equally likely to take any value within this range. Such a scenario covers the entire unit square when visualized on a plane.
Joint Probability Density Function
When observing two random variables together, we use the joint probability density function (joint PDF) to describe the likelihood of different outcomes. This extends the concept of a simple probability distribution to multiple dimensions.- The joint PDF for independent continuous uniform distributions over a unit square can be visualized as a flat, uniform surface. - If random variables \(X\) and \(Y\) are independent and uniform, their joint PDF within the bounds \(0 < x < 1\) and \(0 < y < 1\) is constant.For our exercise, the joint PDF equals 1 over the region since the product of the individual uniform densities is 1 (i.e., the density function for \(X\) and \(Y\) is \(1 \times 1 = 1\)). The specific task of determining \(|X-Y| < 0.5\) means we focus on the area within the unit square where this condition is satisfied, interpreted by the band formed between the lines \(y = x + 0.5\) and \(y = x - 0.5\).
Independent Random Variables
Independence in random variables implies that the outcome of one variable does not influence the outcome of the other. This is a key concept when calculating joint probabilities.- For \(X\) and \(Y\) to be independent, the probability of \(X\) and \(Y\) occurring together is simply the product of their individual probabilities.- This independence simplifies the calculation of joint probabilities since dependencies or correlations do not need to be considered.In our scenario, \(X\) and \(Y\) are independent continuous uniform random variables, which makes the joint PDF calculation straightforward. There is no relationship causing \(X\) to affect \(Y\) or vice versa. As a result, the probability that \(|X-Y| < 0.5\) depends only on the geometric area in the unit square that satisfies this condition, without additional constraints due to dependency between the variables.
