Math Problem Statement
Solve this ques step by step
Solution
The problem involves finding the marginal densities of random variables and given their joint density function.
The given joint density function of and is:
\frac{25}{23a^3b} \left( \frac{x+b}{2b} \right) \left(1 - \left(\frac{y}{a}\right)^2 \right) & -b \leq x \leq b, \, 0 \leq y \leq a \\ 0 & \text{elsewhere} \end{cases}$$ where $$a > 0$$ and $$b > 0$$ are constants. ### 1. Marginal Density of $$X$$ The marginal density $$f_X(x)$$ is obtained by integrating the joint density function over all possible values of $$Y$$: $$f_X(x) = \int_{0}^{a} f_{X,Y}(x, y) \, dy$$ Substituting the given joint density function: $$f_X(x) = \int_{0}^{a} \frac{25}{23a^3b} \left( \frac{x+b}{2b} \right) \left(1 - \left(\frac{y}{a}\right)^2 \right) dy$$ This can be factored and simplified: $$f_X(x) = \frac{25(x+b)}{46a^3b^2} \int_{0}^{a} \left(1 - \frac{y^2}{a^2}\right) dy$$ Now, compute the integral: $$\int_{0}^{a} \left(1 - \frac{y^2}{a^2}\right) dy = \left[ y - \frac{y^3}{3a^2} \right]_{0}^{a}$$ Evaluating the bounds: $$\int_{0}^{a} \left(1 - \frac{y^2}{a^2}\right) dy = \left(a - \frac{a^3}{3a^2}\right) = \left(a - \frac{a}{3}\right) = \frac{2a}{3}$$ Thus: $$f_X(x) = \frac{25(x+b)}{46a^3b^2} \cdot \frac{2a}{3} = \frac{25(x+b)}{69a^2b^2}$$ So, the marginal density $$f_X(x)$$ is: $$f_X(x) = \frac{25(x+b)}{69a^2b^2}, \quad -b \leq x \leq b$$ ### 2. Marginal Density of $$Y$$ The marginal density $$f_Y(y)$$ is obtained by integrating the joint density function over all possible values of $$X$$: $$f_Y(y) = \int_{-b}^{b} f_{X,Y}(x, y) \, dx$$ Substituting the given joint density function: $$f_Y(y) = \int_{-b}^{b} \frac{25}{23a^3b} \left( \frac{x+b}{2b} \right) \left(1 - \left(\frac{y}{a}\right)^2 \right) dx$$ Factor out the terms not dependent on $$x$$: $$f_Y(y) = \frac{25 \left(1 - \left(\frac{y}{a}\right)^2 \right)}{46a^3b^2} \int_{-b}^{b} (x+b) \, dx$$ The integral of $$x+b$$ over $$[-b, b]$$ is: $$\int_{-b}^{b} (x+b) \, dx = \left[\frac{x^2}{2} + bx \right]_{-b}^{b}$$ Evaluating the bounds: \[ \int_{-b}^{b} (x+b) \, dx = \left[\frac{b^2}{2} + b^2\right] - \left[\frac{(-b)^2}{2} - b^2 \right] = \frac{3b^2}{2} + \frac{b^2}{2} = 2b^2 \] Thus: $$f_Y(y) = \frac{25 \cdot 2b^2 \left(1 - \left(\frac{y}{a}\right)^2 \right)}{46a^3b^2} = \frac{25(1 - \left(\frac{y}{a}\right)^2)}{23a^3}$$ So, the marginal density $$f_Y(y)$$ is: $$f_Y(y) = \frac{25}{23a^3} \left(1 - \left(\frac{y}{a}\right)^2 \right), \quad 0 \leq y \leq a$$ ### Summary: - The marginal density of $$X$$ is: $$f_X(x) = \frac{25(x+b)}{69a^2b^2}, \quad -b \leq x \leq b$$ - The marginal density of $$Y$$ is: $$f_Y(y) = \frac{25}{23a^3} \left(1 - \left(\frac{y}{a}\right)^2 \right), \quad 0 \leq y \leq a$$ --- Do you want details or have any questions? Here are some relative questions: 1. How does the range of integration affect the marginal densities? 2. Why is it necessary to integrate over the entire range of the other variable to find the marginal density? 3. What would change if the range of $$Y$$ was different, say $$-a \leq y \leq a$$? 4. How do the constants $$a$$ and $$b$$ affect the shape of the marginal densities? 5. What is the significance of the joint density being zero outside the given range? **Tip:** When solving for marginal densities, always check the limits of integration to ensure they correspond to the defined support of the random variables.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Joint Density Functions
Marginal Densities
Formulas
Joint density function
Marginal density function
Theorems
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Suitable Grade Level
Advanced College
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