Math Problem Statement
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
Solution
We are given that is a random variable with the probability density function (PDF):
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!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Random Variables
Probability Density Function (PDF)
Transformation of Variables
Formulas
f_Y(y) = f_X(x) * |dX/dY|
Y = 8X^3
X = (Y/8)^(1/3)
dX/dY = (1/6)Y^(-2/3)
Theorems
Change of Variables Formula for PDFs
Suitable Grade Level
Undergraduate level (Probability Theory or Statistics Course)