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The problem involves finding the marginal densities of random variables \(X\) and \(Y\) given their joint density function.

The given joint density function of \(X\) and \(Y\) is:

\[
f_{X,Y}(x, y) = \begin{cases} 
\frac{25}{23a^3b} \left( \frac{x+b}{2b} \right) \left(1 - \left(\frac{y}{a}\right)^2 \right) & -b \leq x \leq b, \, 0 \leq y \leq a \\
0 & \text{elsewhere}
\end{cases}
\]

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:

\[
f_Y(y) = \frac{25}{23a^3} \left(1 - \left(\frac{y}{a}\right)^2 \right), \quad 0 \leq y \leq a
\]

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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.

Explore how to find the marginal densities of random variables X and Y step-by-step. Understand the joint density function and integrate over the appropriate ranges to derive the marginal densities. Suitable for advanced college students and researchers in probability theory.

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