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

Given \( X \) is a continuous random variable with a uniform distribution on the interval \([-3, 4]\), i.e., \( X \sim U(-3, 4) \), we need to find the probability density function (pdf) and the moment-generating function (mgf) of the transformed variable \( Y = X^2 \).

### Step 1: Find the PDF of \( Y = X^2 \)

Since \( Y = X^2 \), \( Y \) takes values between \( 0 \) and \( 16 \) because \( X^2 \) ranges from \( (-3)^2 = 9 \) to \( 4^2 = 16 \). Also, because of the symmetry in squaring, we must handle the transformation carefully.

#### Distribution of \( X \)

The pdf of \( X \), denoted as \( f_X(x) \), for a uniform distribution \( U(a, b) \) is given by:

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

#### Transformation \( Y = X^2 \)

Now, to find the pdf of \( Y = X^2 \), we use the fact that the pdf of a transformed variable \( Y = g(X) \) is given by:

\[
f_Y(y) = \sum_{\{x : g(x) = y\}} \frac{f_X(x)}{|g'(x)|}
\]

Here, \( Y = X^2 \) implies:

\[
x = \pm \sqrt{y}
\]

The derivative \( g'(x) = 2x \), and since we have two solutions for each \( y \), one for \( x = \sqrt{y} \) and another for \( x = -\sqrt{y} \), the pdf of \( Y \) is given by:

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

Substituting \( f_X(x) = \frac{1}{7} \), we get:

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

For \( y \in [0, 9) \), only one solution exists for \( X \), so:

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

Thus, the complete pdf of \( Y \) is:

\[
f_Y(y) =
\begin{cases}
\frac{2}{7\sqrt{y}}, & \text{for } 0 \leq y < 9 \\
\frac{1}{7\sqrt{y}}, & \text{for } 9 \leq y \leq 16 \\
0, & \text{otherwise}
\end{cases}
\]

### Step 2: Find the Moment-Generating Function (MGF) of \( Y \)

The moment-generating function (mgf) of \( Y \), denoted by \( M_Y(t) \), is defined as:

\[
M_Y(t) = \mathbb{E}[e^{tY}] = \int_{0}^{16} e^{ty} f_Y(y) \, dy
\]

Using the piecewise pdf of \( Y \), we split the integral into two parts:

\[
M_Y(t) = \int_0^9 e^{ty} \frac{2}{7\sqrt{y}} \, dy + \int_9^{16} e^{ty} \frac{1}{7\sqrt{y}} \, dy
\]

Each integral can be evaluated, though they are non-elementary and require techniques such as numerical methods or special functions like the exponential integral.

Would you like me to proceed with these detailed calculations or clarify any part of the reasoning so far?

### Relative Questions:
1. How would the solution change if \( X \) were uniformly distributed on a different interval?
2. What are the properties of the uniform distribution that help simplify this problem?
3. How do transformations affect the pdf of continuous random variables?
4. What methods are commonly used to evaluate moment-generating functions when integrals are complex?
5. How does the symmetry in the function \( Y = X^2 \) simplify or complicate finding its pdf?

### Tip: When transforming random variables, always consider the number of possible solutions for each value of the new variable.

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

Explore the step-by-step solution for finding the PDF and MGF of the transformed variable Y = X^2 when X is uniformly distributed on the interval [-3, 4]. This solution involves key concepts from probability theory, including transformation of random variables and moment-generating functions. Suitable for undergraduate students.

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