Let X
 be a random variable with probability density function

f(x)={2x,0, 0<x<1otherwise.
Find the pdf of Y=8X3
.
Question 2Select one:

a.
f(y)=(1/6)y1/3


b.
f(y)=(1/3)y−1/6


c.
f(y)=(1/6)y−1/3


d.
f(y)=(1/3)y1/6

We are given that \( X \) is a random variable with the probability density function (PDF):

\[
f_X(x) = \begin{cases} 
2x & \text{if } 0 < x < 1, \\
0 & \text{otherwise}.
\end{cases}
\]

We are asked to find the PDF of \( Y = 8X^3 \).

### Step 1: Relate \( Y \) and \( X \)
We have \( Y = 8X^3 \). First, solve for \( X \) in terms of \( Y \):

\[
X = \left(\frac{Y}{8}\right)^{1/3}.
\]

### Step 2: Find the derivative \( \frac{dX}{dY} \)
Next, we find the derivative of \( X \) with respect to \( Y \):

\[
\frac{dX}{dY} = \frac{1}{3} \cdot \left(\frac{Y}{8}\right)^{-2/3} \cdot \frac{1}{8} = \frac{1}{24} \cdot \left(\frac{Y}{8}\right)^{-2/3}.
\]

Simplifying this gives:

\[
\frac{dX}{dY} = \frac{1}{24} \cdot 8^{2/3} \cdot Y^{-2/3} = \frac{1}{6} Y^{-2/3}.
\]

### Step 3: Apply the formula for changing variables
The PDF of \( Y \), \( f_Y(y) \), can be found using the change of variables formula:

\[
f_Y(y) = f_X(x) \cdot \left| \frac{dX}{dY} \right|.
\]

We know that \( X = \left( \frac{Y}{8} \right)^{1/3} \), so

\[
f_X(x) = 2x = 2 \cdot \left( \frac{Y}{8} \right)^{1/3}.
\]

Thus,

\[
f_Y(y) = 2 \cdot \left( \frac{Y}{8} \right)^{1/3} \cdot \frac{1}{6} Y^{-2/3}.
\]

Simplifying this:

\[
f_Y(y) = \frac{1}{6} \cdot Y^{-1/3}.
\]

### Step 4: Determine the support of \( Y \)
Since \( 0 < X < 1 \), it follows that:

\[
Y = 8X^3 \quad \text{implies} \quad 0 < Y < 8.
\]

Thus, the PDF of \( Y \) is:

\[
f_Y(y) = \frac{1}{6} y^{-1/3}, \quad \text{for } 0 < y < 8.
\]

### Answer:
The correct answer is:

\[
\boxed{c. \ f(y) = \frac{1}{6} y^{-1/3}}.
\]

---

Would you like more details or have any questions?

Here are some related questions to consider:
1. How do you derive the PDF transformation for any general variable change?
2. What happens to the PDF when you scale the random variable by a constant?
3. Can you explain why the support of \( Y \) is \( (0, 8) \)?
4. How would the PDF change if the transformation were \( Y = kX^3 \) for some other constant \( k \)?
5. What if we had a different initial PDF for \( X \)?

**Tip**: When transforming random variables, remember to check the support of the new variable, as it can change based on the transformation!

Let X be a random variable with probability density function f(x) = {2x, 0, 0 < x < 1 otherwise}. Find the pdf of Y = 8X^3. Question 2 Select one: a. f(y) = (1/6)y^1/3 b. f(y) = (1/3)y^−1/6 c. f(y) = (1/6)y^−1/3 d. f(y) = (1/3)y^1/6

This problem involves finding the PDF of Y = 8X^3 given a specific probability density function for X. The solution walks through the steps of transforming variables, using the chain rule for derivatives, and applying the change of variables formula to calculate the PDF of Y. Suitable for undergraduate students learning probability theory.

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