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To solve the problem, we will first derive the Cumulative Distribution Function (CDF) for the random variable \( Y_1 = X^3 \) given the Probability Density Function (PDF) of \( X \), and then find the PDF of \( Y_1 \).

**Step 1: Derive the CDF of \( Y_1 \)**
The CDF of \( Y_1 \), denoted as \( F_{Y_1}(y) \), is the probability that \( Y_1 \leq y \). Given that \( Y_1 = X^3 \), this is equivalent to calculating \( P(X^3 \leq y) \). We start by considering \( y \) values and their corresponding \( x \) intervals.

For \( Y_1 = X^3 \) where \( X \in [-1, 2] \):
- If \( y < -1 \), then there are no \( x \) values such that \( x^3 \leq y \) because the minimum cube value \( x \) can take is \( -1^3 = -1 \).
- If \( -1 \leq y \leq 1 \), then \( -\sqrt[3]{y} \leq x \leq \sqrt[3]{y} \) (considering \( x \) values between -1 and 1).
- If \( y > 1 \), then \( -1 \leq x \leq \sqrt[3]{y} \) (since \( x \leq 2 \) and \( 2^3 = 8 \)).

**CDF Calculation**:
\[ F_{Y_1}(y) = P(X^3 \leq y) \]
\[ = \int_{-\sqrt[3]{y}}^{\sqrt[3]{y}} f_X(x) \, dx \] when \( -1 \leq y \leq 1 \)
\[ = \int_{-1}^{\sqrt[3]{y}} f_X(x) \, dx \] when \( 1 < y \leq 8 \)

Given \( f_X(x) = \frac{x^2}{3} \) for \( x \in [-1, 2] \), we compute the integrals:

For \( -1 \leq y \leq 1 \):
\[ F_{Y_1}(y) = \int_{-\sqrt[3]{y}}^{\sqrt[3]{y}} \frac{x^2}{3} \, dx \]
\[ = \frac{2}{3} \int_{0}^{\sqrt[3]{y}} x^2 \, dx \]
\[ = \frac{2}{3} \cdot \frac{x^3}{3} \Big|_{0}^{\sqrt[3]{y}} \]
\[ = \frac{2}{9} y \]

For \( 1 < y \leq 8 \):
\[ F_{Y_1}(y) = \int_{-1}^{\sqrt[3]{y}} \frac{x^2}{3} \, dx \]
\[ = \frac{1}{3} \left( \frac{x^3}{3} \right) \Big|_{-1}^{\sqrt[3]{y}} \]
\[ = \frac{1}{9} \left( (\sqrt[3]{y})^3 + 1 \right) \]
\[ = \frac{y + 1}{9} \]

**Step 2: Derive the PDF of \( Y_1 \)**
The PDF, \( f_{Y_1}(y) \), is the derivative of the CDF:
\[ f_{Y_1}(y) = \frac{d}{dy} F_{Y_1}(y) \]

For \( -1 \leq y \leq 1 \):
\[ f_{Y_1}(y) = \frac{d}{dy} \left( \frac{2}{9} y \right) = \frac{2}{9} \]

For \( 1 < y \leq 8 \):
\[ f_{Y_1}(y) = \frac{d}{dy} \left( \frac{y + 1}{9} \right) = \frac{1}{9} \]

**Summary**:
\[ f_{Y_1}(y) = \begin{cases} 
0 & \text{if } y < -1 \\
\frac{2}{9} & \text{if } -1 \leq y \leq 1 \\
\frac{1}{9} & \text{if } 1 < y \leq 8 \\
0 & \text{if } y > 8
\end{cases} \]

Would you like any further explanations on this topic? Here are some related questions:
1. How does the transformation of a variable affect the support of the PDF?
2. Why does the cubic transformation result in a piecewise function for the PDF?
3. What happens if we transform \( X \) using another function, like \( Y_2 = X^2 \)?
4. How do you compute the expected value of \( Y_1 \)?
5. Could you explain the concept of Jacobian in variable transformation for PDFs?

**Tip**: Always check the bounds and interval changes when dealing with transformations of random variables, as this will impact the PDF and CDF calculations.

Let X be a random variable with PDF f_X(x) = x^2/3 for x in [-1, 2], and Y_1 = X^3. Derive the CDF and PDF of Y_1.

This problem explores the transformation of a random variable X with a given PDF to Y_1 = X^3. Learn how to derive the cumulative distribution function (CDF) and probability density function (PDF) of the new random variable Y_1 using integration techniques.

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