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The joint probability density function (pdf) is a fundamental concept in probability used to describe the likelihood of two independent random variables occurring simultaneously. For independent random variables, such as the lifetimes of two lightbulbs, the joint pdf can be calculated as the product of their individual density functions. In this case, both lifetimes are modeled using an exponential distribution with a parameter \( \lambda = 1 \). The probability density function for an exponentially distributed random variable is given by \( f(x) = \lambda e^{-\lambda x} \).
Thus, when \( \lambda = 1 \), it simplifies to \( f(x) = e^{-x} \). Given that the lightbulbs' lifetimes are independent, the joint pdf \( f(x, y) \) becomes the product of their individual pdfs, resulting in \( f(x, y) = e^{-(x+y)} \). Understanding the joint pdf is crucial when calculating probabilities involving both variables, like finding the total lifetime of several components.

Independence between random variables is a crucial concept when dealing with joint probability density functions. Two random variables, such as the lifetimes \(X\) and \(Y\) of two lightbulbs, are independent if the probability of one variable is unaffected by the state of the other. This means the joint probability can be represented as the product of their individual probabilities.
For exponentially distributed random variables, each possessing its density function given by \( f(x) = e^{-x} \) when \( \lambda = 1 \), independence allows us to compute the joint pdf by multiplying these functions: \( f(x, y) = e^{-x} \cdot e^{-y} = e^{-(x+y)} \). This independence simplifies calculations, as it allows separate integration of each variable when solving for probabilities.

Integration is a mathematical technique often used to find probabilities for continuous random variables. When investigating joint probabilities over a certain region, as shown in this problem, we use integration to determine the probability of specific outcomes.
Take for instance the problem of calculating the probability that the sum of two lightbulbs’ lifetimes is at most 2. This requires evaluating an integral over the defined region where \(x + y \leq 2\). By setting up a double integral \( \int_0^2 \int_0^{2-x} e^{-(x+y)} \; dy \; dx \), we capture the entire area where the condition holds. The order of integration typically starts with the inner integral over one variable, while keeping the other constant, and then moving to the outer integral.

• Inner integration deals with limits specific to the region of interest, such as \(\int_0^{2-x} e^{-(x+y)} \; dy\).
• Outer integration sums up contributions over the entire region \(\int_0^2 (...) \; dx\).

Applying correct limits and understanding the integration process facilitate calculating cumulative probabilities across defined intervals.

Calculating probabilities involves finding the likelihood of random variables falling within a certain range or satisfying particular conditions. In continuous probability problems, like those involving exponential distributions, formulas and integrals are used for precise probability determinations.
For example, calculating \( P(X \leq 1, Y \leq 1) \) involves utilizing the exponential distribution formula \( P(X \leq a) = 1 - e^{-a} \) for each independent variable. Since \(X\) and \(Y\) are independent, their joint probability is the product of both, \( (1 - e^{-1})^2 \).
Similarly, determining complex probabilities, such as the total lifetime between 1 and 2, involves finding the cumulative probability over an interval by integrating the joint pdf across the region and subtracting to exclude unwanted regions. These calculations often require both understanding of theoretical concepts and practical application of integrals and arithmetic to derive exact probability values.