To determine the value of $c$ such that $f_{X,Y}(x,y) = c(1 - x^2)(1 - y^2)$ is a valid joint probability density function, we need to ensure that the total probability integrates to 1 over the entire support of the random variables $X$ and $Y$.

Given the support is $0 \leq x \leq 1$ and $0 \leq y \leq 1$, we need to solve the following integral:

$\int_{0}^{1} \int_{0}^{1} f_{X,Y}(x,y) \, dx \, dy = 1$

Substituting the expression for $f_{X,Y}(x,y)$:

$\int_{0}^{1} \int_{0}^{1} c(1 - x^2)(1 - y^2) \, dx \, dy = 1$

This can be separated into two integrals:

$c \int_{0}^{1} (1 - x^2) \, dx \int_{0}^{1} (1 - y^2) \, dy = 1$

First, evaluate the integral with respect to $x$:

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

Next, evaluate the integral with respect to $y$:

$\int_{0}^{1} (1 - y^2) \, dy = \left[ y - \frac{y^3}{3} \right]_{0}^{1} = \left( 1 - \frac{1}{3} \right) - (0 - 0) = \frac{2}{3}$

Now multiply these results together:

$c \left( \frac{2}{3} \right) \left( \frac{2}{3} \right) = 1$

$c \cdot \frac{4}{9} = 1$

Solving for $c$:

$c = \frac{9}{4}$

Thus, the value of $c$ that ensures $f_{X,Y}(x,y)$ is a valid joint probability density function is:

$\boxed{c = \frac{9}{4}}$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you find the marginal PDF from a joint PDF?
2. What is the interpretation of a joint probability density function?
3. How would you determine the expected value of $X$ using this joint PDF?
4. What conditions must a function satisfy to be a valid joint PDF?
5. How can you calculate the correlation coefficient between $X$ and $Y$ given their joint PDF?

Tip: Always check the limits of integration when working with joint PDFs to ensure they cover the entire support of the random variables.