Problem 14 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Suppose we choose arbitrarily a point from the square with corners at \((2,1),(3,1),(2,2)\), and \((3,2)\). The random variable \(A\) is the area of the triangle with its corners at \((2,1),(3,1)\), and the chosen point. (See also Exercise \(5.9\) and Figure 7.5.) Compute \(\mathrm{E}[A]\).

Step-by-Step Solution

Answer

The expected area \(\mathrm{E}[A]\) is \(\frac{7}{4}\).

Step 1: Understand the Problem

We need to find the expected area of a triangle formed by the points \((2,1)\), \((3,1)\), and a randomly chosen point within the square whose vertices are \((2,1)\), \((3,1)\), \((2,2)\), and \((3,2)\). The expected value of a random variable represents its average value over many trials.

Step 2: Define the Area Function

The area of the triangle with vertices at \((2,1)\), \((3,1)\), and a point \((x, y)\) is given by the formula for the area of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\): \[\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\]. Substituting the values, the area \(A\) is: \[A = \frac{1}{2} |2(y-1) + 3(1-y)| = \frac{1}{2} |5 - x|\].

Step 3: Set Up the Expected Value Integral

The expected value of \(A\) can be calculated using the double integral over the area of the square: \[\mathrm{E}[A] = \int_{2}^{3} \int_{1}^{2} \frac{1}{2} |5 - x| \, dy \, dx.\] Since the function only involves \(x\), the integration across \(y\) will simply scale the area by 1 (since the range for \(y\) is 1 to 2).

Step 4: Integrate over y

Since there's no \(y\) in \(\frac{1}{2} |5-x|\), the integral over \(y\) just gives us a factor of \(1\): \[\int_{1}^{2} \, dy = [y]_{1}^{2} = 1.\] This simplifies the integral to: \[\mathrm{E}[A] = \int_{2}^{3} \frac{1}{2} |5-x| \, dx.\]

Step 5: Integrate over x

To evaluate \(\int_{2}^{3} \frac{1}{2} |5-x| \, dx\), observe that \(5-x\) is always positive over the interval \([2, 3]\). Thus, \[\int_{2}^{3} \frac{1}{2} (5-x) \, dx = \frac{1}{2} \int_{2}^{3} (5-x) \, dx.\]Integrate: \[\frac{1}{2} \left[ 5x - \frac{x^2}{2} \right]_{2}^{3}= \frac{1}{2} \left((15 - \frac{9}{2}) - (10 - 2)\right) = \frac{1}{2} (\frac{7}{2}) = \frac{7}{4}.\]

Step 6: Compute the Final Expected Value

We obtained: \[\mathrm{E}[A] = \frac{7}{4}.\] This is the expected area of the triangle formed by points \((2,1)\), \((3,1)\), and a random point within the given square.

Key Concepts

Area of a Triangle

The area of a triangle is one of the fundamental concepts in geometry. It quantifies the space occupied by a triangle in a plane. Calculating the area involves understanding how the triangle's vertices, or corners, dictate its size and shape. A common formula used to find the area of a triangle given its vertex coordinates

• \((x_1, y_1)\)
• \((x_2, y_2)\)
• \((x_3, y_3)\)

is the determinant method: \[\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|.\]This formula is handy because it only requires the coordinates of the triangle's vertices. In the context of the exercise, two of the triangle's vertices are fixed at

• \((2,1)\)
• \((3,1)\)

, and the third vertex is a random point within a square. Understanding and applying the area-formula helps you find how the area dynamically changes depending on the position of the third vertex.

Double Integral

Double integrals are a powerful tool in calculus. They allow us to calculate accumulated quantities over a two-dimensional area, such as the total mass of a thin plate or the average value of a function over a region. In the context of the original exercise, we are dealing with the expected area of a triangle, for which you need to sum up small areas to find a total average value.To compute the expected area, we need to integrate over the entire region where the random point could be located — in this case, the square defined by the vertices

• \((2,1)\)
• \((3,1)\)
• \((2,2)\)
• \((3,2)\)

.The double integral of the function \(\frac{1}{2} |5-x|\) over this region is written as:\[\int_{2}^{3} \int_{1}^{2} \frac{1}{2} |5-x| \, dy \, dx.\]Because the function itself does not depend on \(y\), this simplifies the process as the integration over \(y\) simply amounts to scaling by the width of the region in the \(y\)-direction.The integration over \(x\) then calculates the expected value of the function within the limits dictated by the square.

Random Variable

A random variable is a concept used in probability and statistics to describe a variable whose possible values are determined by a random phenomenon. In the exercise at hand, the random variable is represented by the area \(A\) of a triangle. Here, \(A\) isn't fixed; instead, it depends on where a point chosen randomly within a square falls. Each possible point location gives a different triangle and therefore a different area.The role of an expected value in the context of a random variable is crucial. It provides a central measure of the "average" outcome of the random variable over numerous realizations. Simply put, the expected value sums up all possible values the random variable can take, weighted by their probabilities, to give a single "average" value. For this exercise, the expected value, \(\mathrm{E}[A]\), effectively averages the areas that arise from all possible positions of the random point within the square.Thus, understanding random variables and expected values can significantly enrich your comprehension of various stochastic processes, allowing you to make predictions about complex, variable-driven phenomena.