Question

If \(X_{1}\) and \(X_{2}\) are independent random variables such that \(X_{1} \sim \operatorname{Exp}(\lambda)\) and \(X_{2} \sim \operatorname{Exp}(\lambda)\), show that \(Z=\min \left\\{X_{1}, X_{2}\right\\}\) follows \(\operatorname{Exp}(2 \lambda) .\) Hence, generalize this result for \(n\) independent exponential random variables.

Short answer

The random variable Z, defined as the minimum of two independent random variables following the exponential distribution with parameter \(\lambda\), follows an exponential distribution with parameter \(2\lambda\). This result can be generalized for 'n' independent exponential random variables, in which case Z follows an exponential distribution with parameter \(n\lambda\).

Step-by-step solution

Identify the Cumulative Distribution Function (CDF) for Exponential Distribution

Given that \(X_1\) and \(X_2\) are independent random variables with exponential distributions, it's known that the CDF of an exponential random variable \(X \sim \operatorname{Exp}(\lambda)\) is \(F_X(x) = 1 - e^{-\lambda x}\), for \(x \geq 0\).

Derive the CDF for Z

Variable \(Z\) equals to the minimum of \(X_1\) and \(X_2\). The CDF of \(Z\), \(F_Z(z)\), is the probability that \(Z\) is less or equal to \(z\), which equals to the probability that both \(X_1\) and \(X_2\) are less or equal to \(z\). Due to independence, it is calculated as \(F_{Z}(z) = 1 - P\left(X_{1}>z\right) \cdot P\left(X_{2}>z\right) = 1 - (1 - F_{X_{1}}(z)) \cdot (1 - F_{X_{2}}(z)) = 1 - [1 - (1 - e^{-\lambda z})][1 - (1 - e^{-\lambda z})] = 1 - (e^{-\lambda z})^2 = 1 - e^{-2\lambda z}\). Hence, \(Z\) follows \(\operatorname{Exp}(2 \lambda)\).

Generalize the Result for n Variables

For 'n' independent exponential random variables \(X_{1}, X_{2}, ..., X_{n}\) where each follows \(\operatorname{Exp}(\lambda)\), the random variable \(Z\) being the minimum of these 'n' variables would follow \(\operatorname{Exp}(n \lambda)\). Because, similar to the derivation process above, the CDF \(F_Z(z)\) reaches to \(1 - e^{-n\lambda z}\), which shows that \(Z\) follows \(\operatorname{Exp}(n \lambda)\).

Key concepts

Cumulative Distribution Function

Understanding the Cumulative Distribution Function (CDF) is crucial when studying probability distributions. For an exponential distribution, which is a continuous distribution describing the time between events in a Poisson process, the CDF helps determine the probability that a random variable is less than or equal to a certain value.

For an exponentially distributed random variable \(X\) with rate parameter \(\lambda\), the CDF is given by the formula \(F_X(x) = 1 - e^{-\lambda x}\), where \(x \geq 0\).

- **\(1 - e^{-\lambda x}\)**: This represents the probability that the random variable \(X\) takes a value less than or equal to \(x\).

The CDF starts at zero when \(x = 0\) and approaches 1 as \(x\) increases, indicating that the likelihood of \(X\) taking on very large values decreases exponentially.

This concept is critical in the given problem, where independent exponential random variables are involved, and we are considering the minimum of such variables.

Independent Random Variables

Independent random variables are foundational in probability theory and statistics. They represent variables that do not affect one another, meaning the outcome of one variable offers no insight into the outcome of another.

In the context of our exercise, \(X_1\) and \(X_2\) are independent random variables, both following an exponential distribution with the same rate parameter \(\lambda\). The independence allows us to use a specific rule for calculating joint probabilities.

- **Independence Rule**: For two independent events \(A\) and \(B\), the probability of both \(A\) and \(B\) occurring is the product of their probabilities: \(P(A \cap B) = P(A) \cdot P(B)\).

Applying this to our problem, to find the CDF of \(Z = \min(X_1, X_2)\), we calculate \(1 - P(X_1 > z) \cdot P(X_2 > z)\). Independence justifies this product, offering a streamlined way to derive the distribution of \(Z\).

Thus, understanding the independent behavior of \(X_1\) and \(X_2\) is key to solving for the probability characteristics of \(Z\).

Minimum of Random Variables

The concept of the minimum of random variables is instrumental in various probability and statistical contexts. Here, we focus on how the minimum of independent exponential random variables behaves.

Taking the minimum, such as \(Z = \min(X_1, X_2)\), involves finding the smallest value among the given random variables. Since \(X_1\) and \(X_2\) are independent and identically distributed as Exponential(\(\lambda\)), \(Z\) will also follow an exponential distribution but with a new rate parameter.

To determine \(Z\)'s distribution, we use the CDF approach, which calculates \(F_{Z}(z) = 1 - (1 - F_{X_1}(z))(1 - F_{X_2}(z)) = 1 - e^{-2\lambda z}\). This result indicates that the minimum of two Exponential(\(\lambda\)) random variables is an Exponential(\(2\lambda\)) random variable.

Generalizing further, if we're finding the minimum of \(n\) independent \(X_i\)'s, each \(X_i\) distributed as Exponential(\(\lambda\)), \(Z\) follows Exponential(\(n\lambda\)).

• **Minimum Calculation**: The CDF for \(Z\) becomes \(1 - e^{-n\lambda z}\), illustrating how the rate parameter scales with the number of variables.

This property is particularly useful in applications such as survival analysis, queues, and reliability engineering.