To calculate the means \( \mu_X \) and \( \mu_Y \) of the random variables \( X \) and \( Y \), we use the definitions of the expected values:

\[
\mu_X = \mathbb{E}[X] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x \, f_{X,Y}(x,y) \, dy \, dx
\]
\[
\mu_Y = \mathbb{E}[Y] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} y \, f_{X,Y}(x,y) \, dx \, dy
\]

Given the joint PDF \( f_{X,Y}(x,y) = \frac{9}{4}(1 - x^2)(1 - y^2) \) for \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \), and \( f_{X,Y}(x,y) = 0 \) otherwise, the means can be calculated as follows:

### Mean \( \mu_X \):

\[
\mu_X = \int_{0}^{1} \int_{0}^{1} x \cdot \frac{9}{4}(1 - x^2)(1 - y^2) \, dy \, dx
\]

The integral with respect to \( y \) can be separated:

\[
\mu_X = \frac{9}{4} \int_{0}^{1} x(1 - x^2) \left( \int_{0}^{1} (1 - y^2) \, dy \right) dx
\]

We already computed \( \int_{0}^{1} (1 - y^2) \, dy = \frac{2}{3} \):

\[
\mu_X = \frac{9}{4} \cdot \frac{2}{3} \int_{0}^{1} x(1 - x^2) \, dx
\]

Now, evaluate the integral with respect to \( x \):

\[
\int_{0}^{1} x(1 - x^2) \, dx = \int_{0}^{1} (x - x^3) \, dx = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}
\]

Thus:

\[
\mu_X = \frac{9}{4} \cdot \frac{2}{3} \cdot \frac{1}{4} = \frac{9}{24} = \frac{3}{8}
\]

### Mean \( \mu_Y \):

By symmetry, \( \mu_Y \) will have the same form as \( \mu_X \) because \( f_{X,Y}(x,y) \) is symmetric in \( x \) and \( y \):

\[
\mu_Y = \frac{9}{4} \cdot \frac{2}{3} \int_{0}^{1} y(1 - y^2) \, dy = \frac{3}{8}
\]

So, the means are:

\[
\mu_X = \mu_Y = \frac{3}{8}
\]

Would you like more details or have any questions? 

Here are 5 related questions:
1. How would you calculate the variance \( \text{Var}(X) \) and \( \text{Var}(Y) \) using this joint PDF?
2. What does the symmetry of the joint PDF tell us about the relationship between \( X \) and \( Y \)?
3. How would you compute the covariance between \( X \) and \( Y \)?
4. How does the joint PDF being nonzero only in the first quadrant affect the means \( \mu_X \) and \( \mu_Y \)?
5. What would the marginal PDFs of \( X \) and \( Y \) look like?

**Tip:** Symmetry in the joint PDF often simplifies the calculations for means and variances, as the results can sometimes be inferred rather than computed separately.

Learn how to calculate the means μX and μY of random variables X and Y using their joint probability density function (PDF). This problem involves integration techniques and understanding joint probability distributions. Suitable for undergraduate students studying probability theory and mathematical statistics.

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