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To solve this problem without using the Jacobian, we will first determine the cumulative distribution function (CDF) and probability density function (PDF) of the transformed random variable \( Y = X(2 - X) \) given the PDF of \( X \).

### Given:
- The PDF of \( X \) is:
\[
f_X(x) = \begin{cases}
\frac{1}{2}x & \text{for } 0 < x < 2, \\
0 & \text{otherwise}.
\end{cases}
\]
- We need to find the CDF and PDF of \( Y = X(2 - X) \).

#### Step 1: Range of \( Y \)
We first determine the range of \( Y \) by analyzing the transformation \( Y = X(2 - X) \).

This function is quadratic in \( X \), and can be rewritten as:
\[
Y = 2X - X^2.
\]
To find the range of \( Y \), we determine the maximum and minimum values of \( Y \) over the domain \( 0 < x < 2 \).

- The derivative of \( Y \) with respect to \( X \) is:
\[
\frac{dY}{dX} = 2 - 2X.
\]
- Setting the derivative to 0 to find the critical points:
\[
2 - 2X = 0 \implies X = 1.
\]
At \( X = 1 \), the value of \( Y \) is:
\[
Y(1) = 1(2 - 1) = 1.
\]

Now we compute \( Y \) at the endpoints of the interval \( 0 < x < 2 \):
- For \( X = 0 \), we get:
\[
Y(0) = 0(2 - 0) = 0.
\]
- For \( X = 2 \), we get:
\[
Y(2) = 2(2 - 2) = 0.
\]

So the range of \( Y \) is \( 0 \leq Y \leq 1 \).

#### Step 2: Finding the CDF of \( Y \)
The CDF of \( Y \), denoted as \( F_Y(y) \), is given by:
\[
F_Y(y) = P(Y \leq y) = P(X(2 - X) \leq y).
\]
This requires solving for \( X \) in terms of \( Y \). The quadratic equation \( Y = X(2 - X) \) simplifies to:
\[
X^2 - 2X + Y = 0.
\]
Using the quadratic formula:
\[
X = \frac{2 \pm \sqrt{4 - 4Y}}{2} = 1 \pm \sqrt{1 - Y}.
\]
Since \( 0 < X < 2 \), we take the solution \( X = 1 - \sqrt{1 - Y} \). So,
\[
F_Y(y) = P\left( X \leq 1 - \sqrt{1 - y} \right).
\]

We now find the probability for \( X \) in the interval \( (0, 2) \). Using the PDF of \( X \), the CDF of \( Y \) becomes:
\[
F_Y(y) = \int_0^{1 - \sqrt{1 - y}} f_X(x) \, dx = \int_0^{1 - \sqrt{1 - y}} \frac{1}{2}x \, dx.
\]

This is the integral of \( \frac{1}{2}x \), which evaluates as follows:
\[
F_Y(y) = \left[ \frac{1}{4}x^2 \right]_0^{1 - \sqrt{1 - y}} = \frac{1}{4}(1 - \sqrt{1 - y})^2.
\]

#### Step 3: Finding the PDF of \( Y \)
The PDF \( f_Y(y) \) is the derivative of the CDF \( F_Y(y) \) with respect to \( y \):
\[
f_Y(y) = \frac{d}{dy} \left( \frac{1}{4}(1 - \sqrt{1 - y})^2 \right).
\]

Using the chain rule to differentiate, we get:
\[
f_Y(y) = \frac{1}{4} \cdot 2(1 - \sqrt{1 - y}) \cdot \left( \frac{1}{2}(1 - y)^{-\frac{1}{2}} \right).
\]
Simplifying this expression:
\[
f_Y(y) = \frac{1 - \sqrt{1 - y}}{2\sqrt{1 - y}}.
\]

Thus, the PDF of \( Y \) is:
\[
f_Y(y) = \frac{1 - \sqrt{1 - y}}{2\sqrt{1 - y}} \quad \text{for } 0 < y < 1.
\]

### Summary:
- The CDF of \( Y \) is:
\[
F_Y(y) = \frac{1}{4}(1 - \sqrt{1 - y})^2.
\]
- The PDF of \( Y \) is:
\[
f_Y(y) = \frac{1 - \sqrt{1 - y}}{2\sqrt{1 - y}} \quad \text{for } 0 < y < 1.
\]

Let me know if you'd like further clarifications or details on any step! Here are some follow-up questions for deeper understanding:

1. What is the significance of the quadratic transformation of \( X \) to \( Y \)?
2. How does the support of \( X \) influence the support of \( Y \)?
3. Why did we choose the solution \( X = 1 - \sqrt{1 - y} \) and not \( X = 1 + \sqrt{1 - y} \)?
4. How would the PDF change if the range of \( X \) were different?
5. Can you explain the importance of the CDF when finding the PDF of a transformed variable?

**Tip:** When solving for the distribution of a transformed variable, it is often helpful to first understand the range and behavior of the transformation before diving into calculations.

Suppose that the p.d.f. of a random variable X is given by f(x) = (1/2)x for 0 < x < 2, and f(x) = 0 otherwise. Also, suppose that Y = X(2 - X). Determine the CDF and the PDF of Y.

Learn how to calculate the cumulative distribution function (CDF) and probability density function (PDF) of a transformed variable Y = X(2 - X) from a given PDF of X. This problem involves probability transformations, quadratic equations, and advanced calculus.

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