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Magazine Chapter 4: Problem 21
A random square has a side length that is a uniform [0,1] random variable. Find the expected area of the square.
Short Answer
Expert verified
The expected area of the square is \( \frac{1}{3} \).
Step by step solution
01
Understand the Problem
The problem asks us to find the expected value of the area of a square whose side length is a uniformly distributed random variable between 0 and 1.
02
Express Area as a Function of Side Length
The area of a square is given by the formula \( A = x^2 \), where \( x \) is the side length. In this problem, \( x \) is a random variable uniformly distributed between 0 and 1.
03
Write the Expected Area Formula
The expected value \( E[A] \) of the area \( A \) for a continuous random variable is given by the integral of \( A(x) \cdot f(x) \), where \( f(x) \) is the probability density function (PDF) of \( x \). For a uniform distribution over [0, 1], \( f(x) = 1 \).
04
Set Up the Integral for Expected Value
Substitute \( A(x) = x^2 \) and \( f(x) = 1 \) into the expected value formula: \[ E[A] = \int_0^1 x^2 \cdot 1 \, dx = \int_0^1 x^2 \, dx \].
05
Evaluate the Integral
Calculate the integral: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \, \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \].
06
Conclude the Calculation
The expected area of the square is \( \frac{1}{3} \). We have found that the mean area of a square, given uniformly random side length between 0 and 1, is one-third of a unit squared.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Uniform Distribution
A uniform distribution is a type of probability distribution where all outcomes are equally likely to occur. Imagine spinning a perfectly balanced spinner numbered from 0 to 1. Any point on this spinner is equally likely to land as any other. That's the essence of a uniform distribution.
In mathematical terms, for a random variable that is uniformly distributed over an interval like [0, 1], each value in this range has the same probability. This leads to a Probability Density Function (PDF) that is constant over this range. For a uniform distribution over [0, 1], the PDF is 1, indicating that all values are equally likely.
This concept helps simplify calculations, such as finding the expected value of a variable using integration, because the constant PDF means the focus is simply on the function of interest, like the square's area in this problem.
In mathematical terms, for a random variable that is uniformly distributed over an interval like [0, 1], each value in this range has the same probability. This leads to a Probability Density Function (PDF) that is constant over this range. For a uniform distribution over [0, 1], the PDF is 1, indicating that all values are equally likely.
This concept helps simplify calculations, such as finding the expected value of a variable using integration, because the constant PDF means the focus is simply on the function of interest, like the square's area in this problem.
Area Calculation
The area of a square is calculated by squaring its side. If you know the side length, say 'x', then the area 'A' is given by the formula:
To determine the expected area, you need to calculate the average value this random variable will take. This involves integrating the square of the side length over the possible range [0, 1]. By integrating \( x^2 \) from 0 to 1, we effectively account for all possible sizes of the square and their associated probabilities, resulting in the expected value of the area.
- \[ A = x^2 \]
To determine the expected area, you need to calculate the average value this random variable will take. This involves integrating the square of the side length over the possible range [0, 1]. By integrating \( x^2 \) from 0 to 1, we effectively account for all possible sizes of the square and their associated probabilities, resulting in the expected value of the area.
Probability Density Function
A Probability Density Function (PDF) describes the likelihood of different outcomes for a continuous random variable. Unlike discrete variables where specific outcomes have distinct probabilities, a PDF involves a curve over a continuum of values.
For a uniform distribution, like in this exercise, the PDF is straightforward. It is a constant value since every outcome is equally likely. Specifically, for a uniform distribution over the interval [0, 1], the PDF is 1. The formula for the expected value of a random variable involves integrating the product of the variable's value and its PDF.
For a uniform distribution, like in this exercise, the PDF is straightforward. It is a constant value since every outcome is equally likely. Specifically, for a uniform distribution over the interval [0, 1], the PDF is 1. The formula for the expected value of a random variable involves integrating the product of the variable's value and its PDF.
- In our exercise: To find the expected area of the square, we set up the integral of the area formula \( x^2 \) times the PDF (which is 1) across the interval from 0 to 1.
- This simplification (due to a constant PDF) focuses the calculation solely on the function \( x^2 \), as seen in the integration \[ E[A] = \int_0^1 x^2 \, dx \], which results in the expected area.
