91Ó°ÊÓ

Since \(X_1\) and \(X_2\) are independent and identically distributed (i.i.d.) exponential random variables with rate \(\mu\), their pdfs are given by $$ f_1(x) = f_2(x) = \mu e^{-\mu x} \quad \text{for } x\geq 0. $$ To obtain the pdf of \(X_{(1)}\), we can use the fact that \(X_{(1)} = \min(X_1, X_2)\). Thus, we want to calculate the probability that either \(X_1 \leq x\) or \(X_2 \leq x\). We have $$ P(X_{(1)} \leq x) = P(X_1 \leq x) + P(X_2 \leq x) - P(X_1 \leq x, X_2 \leq x), $$ where the last term is due to the inclusion-exclusion principle. Since \(X_1\) and \(X_2\) are independent, \(P(X_1 \leq x, X_2 \leq x) = P(X_1 \leq x) P(X_2 \leq x)\). Thus, we obtain the cumulative distribution function (CDF) of \(X_{(1)}\): $$ F_{(1)}(x) = P(X_{(1)} \leq x) = 1 - (1 - P(X_1 \leq x))(1 - P(X_2 \leq x)). $$ Now, we know that \(P(X_1\leq x)=1-e^{-\mu x}\) and \(P(X_2\leq x)=1-e^{-\mu x}\); therefore, we can substitute these values into the equation: $$ F_{(1)}(x) = 1-\left(1-(1-e^{-\mu x})\right)\left(1-(1-e^{-\mu x})\right). $$ To find the pdf of \(X_{(1)}\), we take the derivative of the CDF with respect to \(x\): $$ f_{(1)}(x) = \frac{dF_{(1)}(x)}{dx} = 2\mu e^{-\mu x} - \mu^2 e^{-2\mu x} \quad \text{for } x\geq 0. $$

From the pdf \(f_{(1)}(x)\), we can now calculate the expected value \(E[X_{(1)}]\) and the variance \(\operatorname{Var}[X_{(1)}]\) using the following formulas: $$ E[X_{(1)}] = \int_{0}^{\infty} x f_{(1)}(x) dx, $$ and $$ \operatorname{Var}[X_{(1)}] = E[X_{(1)}^2] - E[X_{(1)}]^2 = \left(\int_{0}^{\infty} x^2 f_{(1)}(x) dx \right) - E[X_{(1)}]^2. $$ Compute the integrals: $$ E[X_{(1)}] = \int_{0}^{\infty} x(2\mu e^{-\mu x} - \mu^2 e^{-2\mu x}) dx = \frac{1}{\mu}. $$ Calculate \(E[X_{(1)}^2]\): $$ E[X_{(1)}^2] = \int_{0}^{\infty} x^2(2\mu e^{-\mu x} - \mu^2 e^{-2\mu x}) dx = \frac{2}{\mu^2}. $$ Now, calculate the variance: $$ \operatorname{Var}[X_{(1)}] = E[X_{(1)}^2] - E[X_{(1)}]^2 = \frac{2}{\mu^2} - \left(\frac{1}{\mu}\right)^2 = \frac{1}{\mu^2}. $$

We have \(X_{(2)} = \max(X_1, X_2)\). We want to calculate the probability that both \(X_1\) and \(X_2\) are less than or equal to \(x\). Since \(X_1\) and \(X_2\) are independent, we have $$ P(X_{(2)} \leq x) = P(X_1 \leq x) P(X_2 \leq x) = (1 - e^{-\mu x})^2. $$ To find the pdf, we differentiate the CDF with respect to \(x\): $$ f_{(2)}(x) = \frac{dP(X_{(2)} \leq x)}{dx} = 2\mu e^{-\mu x}(1 - e^{-\mu x}) \quad \text{for } x\geq 0. $$

Now that we have the pdf for \(X_{(2)}\), we can calculate the expected value and variance: $$ E[X_{(2)}] = \int_{0}^{\infty} x f_{(2)}(x) dx, $$ and $$ \operatorname{Var}[X_{(2)}] = E[X_{(2)}^2] - E[X_{(2)}]^2 = \left(\int_{0}^{\infty} x^2 f_{(2)}(x) dx \right) - E[X_{(2)}]^2. $$ Calculate the expected value: $$ E[X_{(2)}] = \int_{0}^{\infty} x(2\mu e^{-\mu x}(1 - e^{-\mu x})) dx = \frac{2}{\mu} - \frac{1}{\mu}. $$ Calculate \(E[X_{(2)}^2]\): $$ E[X_{(2)}^2] = \int_{0}^{\infty} x^2(2\mu e^{-\mu x}(1 - e^{-\mu x})) dx = \frac{6}{\mu^2} - \frac{4}{\mu^2}. $$ Finally, calculate the variance: $$ \operatorname{Var}[X_{(2)}] = E[X_{(2)}^2] - E[X_{(2)}]^2 = \frac{6}{\mu^2} -\frac{4}{\mu^2} - \left(\frac{2}{\mu} - \frac{1}{\mu}\right)^2 = \frac{1}{\mu^2}. $$ #Conclusion# Summarizing the results, we found that: (a) \(E[X_{(1)}] = \frac{1}{\mu}\), (b) \(\operatorname{Var}[X_{(1)}] = \frac{1}{\mu^2}\), (c) \(E[X_{(2)}] = \frac{2}{\mu} - \frac{1}{\mu}\), and (d) \(\operatorname{Var}[X_{(2)}] = \frac{1}{\mu^2}\).