Problem 86 - Applied Statistics and Probability for Engineers

Question

The random variable \(X\) has the probability distribution $$ f_{X}(x)=\frac{x}{8}, \quad 0 \leq x \leq 4 $$ Determine the probability distribution of \(Y=(X-2)^{2}\).

Step-by-Step Solution

Answer

The distribution of Y is given by \(f_{Y}(y) = \frac{2 + \sqrt{y}}{16 \sqrt{y}}\) for \(0 \leq y \leq 4\).

Step 1: Understand the Functions

We are given the probability distribution function (PDF) of random variable \(X\) as \(f_{X}(x)=\frac{x}{8}\) for the interval \(0 \leq x \leq 4\). We need to find the distribution of \(Y = (X-2)^2\). This requires finding how the transformation \(Y = g(X)\) affects the distribution.

Step 2: Find the Inverse Function

The transformation \(Y = (X-2)^{2}\) can be solved for \(X\) to express \(X\) in terms of \(Y\). Solving for \(X\), we get two branches: \(X = 2 + \sqrt{Y}\) and \(X = 2 - \sqrt{Y}\). Since \(0 \leq x \leq 4\), \(X = 2 + \sqrt{Y}\) is the valid transformation in this context.

Step 3: Determine the Range of Y

To determine the range for \(Y\), we consider the range of \(X\). With \(0 \leq X \leq 4\), substituting these into \((X-2)^2\), we have \(0 \leq Y \leq 4\). However, by examining smaller segments, we see \(X\) ranges from 0 to 2 results in \(0 \leq Y \leq 4\), and from 2 to 4 results in a different interpretation under the transformation. Hence, \(0 \leq Y \leq 4\) but notice that as \(X = 2 + \sqrt{Y}\), the effective range is limited to \([0, 4]\).

Step 4: Change of Variables Formula

We utilize the change of variables formula for transformation: \(f_{Y}(y) = f_{X}(g^{-1}(y)) \cdot \left| \frac{d}{dy} g^{-1}(y) \right|\). Here, \(g^{-1}(y) = 2 + \sqrt{y}\). Compute the derivative \( \frac{d}{dy}g^{-1}(y) = \frac{1}{2 \sqrt{y}}\).

Step 5: Compute the Probability Distribution of Y

Compute the distribution function \(f_{Y}(y)\) by substituting \(g^{-1}(y)\) into \(f_{X}(x)\): \(f_{Y}(y) = \frac{2 + \sqrt{y}}{8} \times \frac{1}{2 \sqrt{y}}\) for \(0 \leq y \leq 4\). Simplifying, we get: \(f_{Y}(y) = \frac{2 + \sqrt{y}}{16 \sqrt{y}}\).

Key Concepts

Random Variables

A random variable is a numerical value that results from a random event. In essence, it's a way of quantifying randomness. For example, think about flipping a coin. If we let heads be 1 and tails be 0, the result is a random variable.

Random variables can be continuous or discrete:

• **Discrete random variables** take on a finite or countable number of values, like the number of heads in a series of coin flips.
• **Continuous random variables** can take any value within an interval, such as the height of a randomly chosen person.

In the given exercise, we deal with a continuous random variable, denoted by \(X\). Its values range between 0 and 4. By using the probability density function, we understand the likelihood of \(X\) taking on a given value within this range. This concept is crucial when predicting outcomes based on random events.

Probability Density Function

The probability density function (PDF) is a key component in probability theory for continuous random variables. It describes how the values of a random variable are distributed.

The given PDF is \(f_{X}(x) = \frac{x}{8}\) for \(0 \le x \le 4\). The PDF provides the relative likelihood that a random variable \(X\) will take on a value \(x\). Here’s how it works:

• The PDF is not the probability; rather, it indicates where the values are most densely packed.
• The area under the PDF curve gives the probability. For continuous random variables, this requires integration.
• For the PDF to be valid, the total area under the curve (from \(-\infty\) to \(+\infty\)) must equal 1.

In this example, the PDF indicates that values of \(X\) closer to 4 are more likely than values closer to 0. This PDF is used to determine how \(X\) behaves and, subsequently, how \(Y\), a function of \(X\), distributes its values.

Transformation of Variables

Transformation of variables is a technique used to find the probability distribution of a new random variable, which is a function of another random variable. Consider \(Y = (X-2)^2\), a transformation of \(X\). This method allows us to explore how changes in \(X\) affect \(Y\).

Here’s the process:

• **Expressing the Transformation:** To find \(Y\)’s distribution, express \(X\) in terms of \(Y\). Solving \(Y = (X-2)^2\), we obtain \(X = 2 + \sqrt{Y}\). Given the original domain, we ensure we use the correct branch of the function.
• **Validation of Range:** We check the range of \(Y\) using \(X\)'s bounds, resulting in \(0 \leq Y \leq 4\).
• **Applying Change of Variables:** Calculate the PDF of \(Y\) using the formula \(f_{Y}(y) = f_{X}(g^{-1}(y)) \cdot \left| \frac{d}{dy} g^{-1}(y) \right|\). This involves simplifying the expression to find whether the transformation maintains a valid PDF for \(Y\).

The transformation helps in shifting our perspective from \(X\) to analyzing outcomes of scenarios dependent on \(Y\), illustrating the probability landscape of the transformed variable.