Math Problem Statement
Consider the continuous random variable Y having p.d.f. f(y) = 3y 2 if 0 ≤ y ≤ 1 0 otherwise. Find the mean and variance of Z = √ Y
Solution
Given the continuous random variable with the probability density function (p.d.f.)
\begin{cases} 3y^2 & \text{if } 0 \leq y \leq 1, \\ 0 & \text{otherwise}, \end{cases}$$ we are asked to find the mean and variance of $$Z = \sqrt{Y}$$. ### Step 1: Finding the Mean of $$Z$$ The mean of $$Z$$ is given by: \[ E[Z] = E[\sqrt{Y}] \] The expected value of $$Z$$ can be calculated by integrating the product of $$\sqrt{y}$$ and the p.d.f. of $$Y$$: $$E[Z] = \int_0^1 \sqrt{y} \cdot f(y) \, dy$$ Substituting $$f(y) = 3y^2$$: $$E[Z] = \int_0^1 \sqrt{y} \cdot 3y^2 \, dy = 3 \int_0^1 y^{5/2} \, dy$$ Now, calculate this integral: \[ E[Z] = 3 \left[ \frac{y^{7/2}}{\frac{7}{2}} \right]_0^1 = 3 \cdot \frac{2}{7} = \frac{6}{7} \] So, the mean of $$Z$$ is: $$E[Z] = \frac{6}{7}$$ ### Step 2: Finding the Variance of $$Z$$ The variance of $$Z$$ is given by: \[ \text{Var}(Z) = E[Z^2] - (E[Z])^2 \] First, compute $$E[Z^2]$$: \[ E[Z^2] = E[Y] = \int_0^1 y \cdot f(y) \, dy \] Substituting $$f(y) = 3y^2$$: $$E[Z^2] = \int_0^1 y \cdot 3y^2 \, dy = 3 \int_0^1 y^3 \, dy$$ Now, calculate this integral: \[ E[Z^2] = 3 \left[ \frac{y^4}{4} \right]_0^1 = 3 \cdot \frac{1}{4} = \frac{3}{4} \] Thus, \[ \text{Var}(Z) = E[Z^2] - (E[Z])^2 = \frac{3}{4} - \left(\frac{6}{7}\right)^2 \] Calculating the variance: $$\left(\frac{6}{7}\right)^2 = \frac{36}{49}$$ So, $$\text{Var}(Z) = \frac{3}{4} - \frac{36}{49}$$ Convert both fractions to have a common denominator: $$\frac{3}{4} = \frac{147}{196}, \quad \frac{36}{49} = \frac{144}{196}$$ Thus, $$\text{Var}(Z) = \frac{147}{196} - \frac{144}{196} = \frac{3}{196}$$ Therefore, the variance of $$Z$$ is: $$\text{Var}(Z) = \frac{3}{196}$$ ### Final Answers: - **Mean of $$Z$$**: $$\frac{6}{7}$$ - **Variance of $$Z$$**: $$\frac{3}{196}$$ Would you like further details or have any questions? Here are 5 related questions for further exploration: 1. How would the mean and variance change if $$f(y) = 4y^3$$ instead? 2. What is the cumulative distribution function (C.D.F.) of $$Z = \sqrt{Y}$$? 3. How do you compute the skewness of $$Z$$? 4. What is the interpretation of the mean and variance in the context of a real-world application? 5. How does the transformation $$Z = \sqrt{Y}$$ affect the shape of the distribution? **Tip:** Understanding transformations of random variables can help solve complex problems by simplifying the original variable's distribution.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Expected Value
Variance
Probability Density Function
Formulas
Expected value formula
Variance formula
Theorems
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Suitable Grade Level
Advanced Undergraduate
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