Suppose X a continous uniform [-3,4] distribution. find the pdf and the cdf of Y=X^2

Given that \( X \) follows a continuous uniform distribution on the interval \([-3, 4]\), we are asked to find the probability density function (PDF) and cumulative distribution function (CDF) of \( Y = X^2 \).

### Step 1: Analyze the distribution of \( X \)
- Since \( X \sim \text{Uniform}[-3, 4] \), the PDF of \( X \) is given by:

\[
f_X(x) = \frac{1}{4 - (-3)} = \frac{1}{7}, \quad \text{for } x \in [-3, 4]
\]

### Step 2: Analyze the transformation \( Y = X^2 \)
- The random variable \( Y = X^2 \) maps the interval \( [-3, 4] \) to the interval \( [0, 16] \).
- The relationship between \( X \) and \( Y \) is not one-to-one, since both \( X = -\sqrt{y} \) and \( X = \sqrt{y} \) correspond to the same value of \( Y \).

### Step 3: Find the PDF of \( Y \)

To find the PDF of \( Y \), we use the method of transformation for continuous random variables. The PDF of \( Y \) is given by:

\[
f_Y(y) = \sum_{x_i \in g^{-1}(y)} \frac{f_X(x_i)}{|g'(x_i)|}
\]

where \( g(X) = X^2 \), so \( g'(x) = 2x \).

#### Case 1: \( y \in [0, 9] \) (since \( X \) ranges from \( -3 \) to \( 3 \))

For \( y \in [0, 9] \), we have two solutions for \( X \): \( X = \sqrt{y} \) and \( X = -\sqrt{y} \).

Thus,

\[
f_Y(y) = \frac{f_X(\sqrt{y})}{|2\sqrt{y}|} + \frac{f_X(-\sqrt{y})}{|2(-\sqrt{y})|}
\]

Since \( f_X(x) = \frac{1}{7} \) for all \( x \in [-3, 4] \), the PDF simplifies to:

\[
f_Y(y) = \frac{1}{7} \cdot \frac{2}{2\sqrt{y}} = \frac{1}{7\sqrt{y}}, \quad \text{for } y \in [0, 9]
\]

#### Case 2: \( y \in [9, 16] \) (since \( X \in [3, 4] \))

For \( y \in [9, 16] \), \( X \geq 0 \) and \( X = \sqrt{y} \). Thus, we only have one solution for \( X \), and the PDF is:

\[
f_Y(y) = \frac{1}{7 \cdot 2\sqrt{y}} = \frac{1}{14\sqrt{y}}, \quad \text{for } y \in [9, 16]
\]

### Step 4: The complete PDF of \( Y \)

\[
f_Y(y) = 
\begin{cases}
\frac{1}{7\sqrt{y}}, & \text{for } y \in [0, 9] \\
\frac{1}{14\sqrt{y}}, & \text{for } y \in [9, 16]
\end{cases}
\]

### Step 5: Find the CDF of \( Y \)

The CDF is the integral of the PDF:

\[
F_Y(y) = P(Y \leq y) = \int_0^y f_Y(t) \, dt
\]

#### Case 1: \( y \in [0, 9] \)

\[
F_Y(y) = \int_0^y \frac{1}{7\sqrt{t}} \, dt = \frac{2\sqrt{y}}{7}, \quad \text{for } y \in [0, 9]
\]

#### Case 2: \( y \in [9, 16] \)

\[
F_Y(y) = F_Y(9) + \int_9^y \frac{1}{14\sqrt{t}} \, dt
\]

We already know \( F_Y(9) = \frac{6}{7} \). So,

\[
F_Y(y) = \frac{6}{7} + \left[ \frac{\sqrt{y}}{7} - \frac{3}{7} \right] = 1 - \frac{4}{7\sqrt{y}}, \quad \text{for } y \in [9, 16]
\]

### Final CDF:

\[
F_Y(y) = 
\begin{cases}
\frac{2\sqrt{y}}{7}, & \text{for } y \in [0, 9] \\
1 - \frac{4}{7\sqrt{y}}, & \text{for } y \in [9, 16]
\end{cases}
\]

---

Would you like further clarifications on this or an explanation of any steps? Here are some related questions you can explore:

1. How would the solution change if the uniform distribution was on a different interval?
2. What if the transformation was \( Y = |X| \) instead of \( X^2 \)?
3. Can you compute the expected value \( E[Y] \) based on this PDF?
4. How does the CDF look graphically for this particular transformation?
5. What other methods can be used to find the PDF of a transformed random variable?

**Tip**: When finding the PDF for a transformation like \( Y = g(X) \), always check if the mapping is one-to-one or not, as it affects the summation over inverse images.

Suppose X a continuous uniform [-3,4] distribution. Find the pdf and the cdf of Y=X^2.

Find the PDF and CDF of Y = X² where X follows a continuous uniform distribution on [-3,4]. This detailed solution involves transforming random variables, calculating the PDF using the method of transformation, and deriving the CDF through integration.

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