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When dealing with joint probability distributions, understanding marginal probability density functions (pdfs) is essential. These marginal pdfs allow us to focus on one of the random variables, ignoring the others.
To obtain the marginal pdf of a random variable, we integrate the joint pdf over the range of the other variable(s). In this exercise, we have the joint pdf of the lifetimes of two components, denoted as \( f(x, y) = xe^{-x(1+y)} \).
For the random variable \( X \), the marginal pdf is calculated as \( f_X(x) = \int_0^{\infty} xe^{-x(1+y)} dy \). By performing this integration, we simplify the function to \( f_X(x) = e^{-x} \) for \( x \geq 0 \).
Similarly, the marginal pdf of \( Y \) is found using \( f_Y(y) = \int_0^{\infty} xe^{-x(1+y)} dx \). After integrating and simplifying, we get \( f_Y(y) = \frac{1}{(1+y)^2} \) for \( y \geq 0 \).
Knowing the marginal pdfs is crucial since they provide insight into the behavior of individual variables, separate from their interaction in the joint distribution.

Random variables are independent if their joint probability can be decomposed into the product of their marginal probabilities. In simpler terms, knowledge of one variable does not affect the outcome of another.
In the context of this exercise, we have two lifetimes with joint pdf \( f(x, y) = xe^{-x(1+y)} \). To assess independence, we compare this joint pdf with the product of the marginal pdfs, \( f_X(x) \) and \( f_Y(y) \).
For independence, it should hold that \( xe^{-x(1+y)} = e^{-x} \times \frac{1}{(1+y)^2} \). Calculating the left-hand and right-hand expressions shows they do not equal each other. Hence, the random variables \( X \) and \( Y \) are not independent.
Understanding whether variables are independent is vital for making accurate predictions and analyses, as the relationships can significantly impact outcomes and interpretations.

The cumulative distribution function (CDF) quantifies the probability that a random variable takes a value less than or equal to a specific number. For a continuous random variable, it involves integrating the probability density function from negative infinity to the value of interest.
In this exercise, to find the probability that the lifetime \( X \) exceeds 3, we use the CDF of \( X \). The formula used is \( P(X > 3) = 1 - F_X(3) \), where \( F_X(x) = 1 - e^{-x} \) is the CDF of \( X \). Therefore, \( P(X > 3) = e^{-3} \).
The CDF is an essential tool because it allows us to calculate probabilities for ranges of values, not just single points. Whether determining probabilities of being greater than, less than, or between values, the CDF provides a comprehensive measure of likelihood.