91Ó°ÊÓ

Since \(X_1\) and \(X_2\) are independent exponential random variables with rate \(\mu\), their pdfs are given by: $$ f_{X_{1}}(x) = \mu e^{-\mu x} \text{ and } f_{X_{2}}(x) = \mu e^{-\mu x}. $$ First, let's find the cumulative distribution function (cdf) of \(X_{(1)}\), which represents the probability that the minimum of the two random variables is less than or equal to some value \(x\). Using the cdf for independent random variables, we get: $$ F_{X_{(1)}}(x) = P(X_{(1)} \le x) = 1 - P(X_{(1)} > x) = 1 - P(X_1 > x) P(X_2 > x) = 1 - (1 - F_{X_{1}}(x))(1 - F_{X_{2}}(x)). $$ Since the cdf for exponential random variables is: $$ F_{X_1}(x) = F_{X_2}(x) = 1 - e^{-\mu x}, $$ then the cdf for \(X_{(1)}\) becomes: $$ F_{X_{(1)}}(x) = 1 - (e^{-\mu x})^2 = 1 - e^{-2\mu x}. $$ Now, we can find the pdf of \(X_{(1)}\), which is the derivative of the cdf: $$ f_{X_{(1)}}(x) = \frac{dF_{X_{(1)}}(x)}{dx} = 2\mu e^{-2\mu x}. $$ Similarly, to find the cdf of \(X_{(2)}\) using the cdf for independent random variables, we get: $$ F_{X_{(2)}}(x) = P(X_{(2)} \le x) = P(X_1 \le x) P(X_2 \le x) = F_{X_{1}}(x) F_{X_{2}}(x) = (1 - e^{-\mu x})^2. $$ Now, we can find the pdf of \(X_{(2)}\), which is the derivative of the cdf: $$ f_{X_{(2)}}(x) = \frac{dF_{X_{(2)}}(x)}{dx} = 2\mu e^{-\mu x}(1 - e^{-\mu x}). $$

To compute the expected values, we will use the formula for the expected value of a continuous random variable: $$ E[X_{(1)}] = \int_{0}^{\infty} x f_{X_{(1)}}(x) \, dx = \int_{0}^{\infty} x (2\mu e^{-2\mu x}) \, dx $$ Integration by parts can be used to evaluate the integral: $$ u = x, \quad dv = 2\mu e^{-2\mu x} dx $$ $$ du = dx, \quad v = -\frac{1}{\mu} e^{-2\mu x} $$ Now, we substitute and apply the integration by parts formula: $$ E[X_{(1)}] = -\frac{1}{\mu} xe^{-2\mu x}|_{0}^{\infty} + \frac{1}{\mu}\int_{0}^{\infty} e^{-2\mu x} dx = 0 + \frac{1}{\mu} \left[-\frac{1}{2\mu} e^{-2\mu x}\right]_{0}^{\infty} = \frac{1}{2\mu} $$ Similarly, for \(E[X_{(2)}]\): $$ E[X_{(2)}] = \int_{0}^{\infty} x f_{X_{(2)}}(x) \, dx = \int_{0}^{\infty} x (2\mu e^{-\mu x}(1 - e^{-\mu x})) \, dx $$ The integral can be split and evaluated separately: $$ E[X_{(2)}] = 2\mu\int_{0}^{\infty} x e^{-\mu x} dx - 2\mu\int_{0}^{\infty} x e^{-2\mu x} dx $$ Using the previous integration by parts for the latter part and the same method for the other part, we get: $$ E[X_{(2)}] = \frac{3}{2\mu} $$

Similar to step 2, integrate \(x^2\) multiplied by the corresponding pdf: $$ E[X^{2}_{(1)}] = \int_{0}^{\infty} x^2 f_{X_{(1)}}(x) \, dx = \int_{0}^{\infty} x^2 (2\mu e^{-2\mu x}) \, dx $$ Using integration by parts twice (similar to step 2), we get: $$ E[X^{2}_{(1)}] = \frac{2}{(2\mu)^2} $$ Now for \(E[X^{2}_{(2)}]\): $$ E[X^{2}_{(2)}] = \int_{0}^{\infty} x^2 f_{X_{(2)}}(x) \, dx = \int_{0}^{\infty} x^2 (2\mu e^{-\mu x}(1 - e^{-\mu x})) \, dx $$ Split the integral and evaluate separately using integration by parts twice, similar to step 2: $$ E[X^{2}_{(2)}] = \frac{11}{6\mu^2} $$

Finally, we will compute the variances using the formula: $$ \operatorname{Var}[X] = E[X^2] - E[X]^2 $$ For \(X_{(1)}\): $$ \operatorname{Var}[X_{(1)}] = \frac{2}{(2\mu)^2} - \left(\frac{1}{2\mu}\right)^2 = \frac{1}{(2\mu)^2} $$ For \(X_{(2)}\): $$ \operatorname{Var}[X_{(2)}] = \frac{11}{6\mu^2} - \left(\frac{3}{2\mu}\right)^2 = \frac{1}{2\mu^2} $$ In conclusion: (a) \(E[X_{(1)}] = \frac{1}{2\mu}\) (b) \(\operatorname{Var}[X_{(1)}] = \frac{1}{(2\mu)^2}\) (c) \(E[X_{(2)}] = \frac{3}{2\mu}\) (d) \(\operatorname{Var}[X_{(2)}] = \frac{1}{2\mu^2}\)