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A continuous random variable is a variable that can take an infinite number of possible values within a given range. Unlike discrete random variables that only take distinct, separate values, continuous random variables can take any value in an interval and are often associated with measurable quantities, such as height, weight, or time.

In our exercise, the random variable \(X\) is continuous because it can take any value between 0 and 4. The values are not limited to whole numbers or specific increments, allowing for an infinite number of possibilities within this range. This characteristic is what differentiates continuous random variables from discrete ones.

• They have a continuous probability distribution, denoted by a probability density function (PDF).
• The probability of \(X\) taking an exact value is effectively zero; instead, probabilities are calculated over intervals.
• Continuous distributions require calculus for computation, using integrals to find probabilities.

Understanding that \(X\) is continuous is crucial when transforming it to \(Y\), as it informs the methods used to determine the new distribution.

A probability density function (PDF) is essential when dealing with continuous random variables. The PDF describes the likelihood of a random variable taking on a particular value. It is a function that, when integrated over a range, gives the probability that the variable falls within that range.

The given PDF for the random variable \(X\) in our exercise is \(f_{X}(x) = \frac{x}{8}\) for \(0 \leq x \leq 4\). This function tells us how the probability is distributed over the possible values of \(X\). However, because \(X\) is continuous, the probability of \(X\) being exactly any single value is zero.

• The area under the entire PDF curve equals 1, representing the total probability.
• The shape of the PDF indicates where values of \(X\) are more or less likely to occur.
• To find the probability of \(X\) lying within a specific interval, one would integrate the PDF over that interval.

The concept of the PDF extends to \(Y\) once we transform \(X\) using \(Y = (X-2)^2\). The transformation changes how probability is distributed across \(Y\), requiring us to derive a new PDF for \(Y\).

Transformation of variables is a technique used to find a new random variable from an existing one, often to simplify analysis or obtain a desired form. In this exercise, we transform the random variable \(X\) into \(Y\) with the function \(Y = (X-2)^2\).

This step involves several aspects:

• **Defining the transformation:** We have to clearly express how \(Y\) is formed from \(X\), which here is done using the equation \(Y = (X-2)^2\).
• **Determining the range of transformation:** By analyzing the function and the range of \(X\), we discover that \(Y\) ranges from 0 to 4.
• **Solving mathematically:** Solving the transformation equation gives two possible expressions for \(X\) in terms of \(Y\), \(X = 2 \pm \sqrt{Y}\). These two solutions indicate two branches that satisfy the transformation for each value of \(Y\).
• **Finding the new distribution:** Finally, using the properties of \(X\) and the nature of the transformation, we derive the PDF for \(Y\). This involves considering how probabilities transform, leading to adjusted outcomes for the new random variable's distribution.

The transformation produced \(f_Y(y) = \frac{1}{4}\) for \(0 \leq y \leq 4\). This new PDF reflects the uniform distribution of \(Y\), distilled from the transformation of \(X\)’s non-uniform distribution.