Question

Let \(X_{1}\) and \(X_{2}\) be independent exponential random variables, each having rate \(\mu .\) Let $$ X_{(1)}=\operatorname{minimum}\left(X_{1}, X_{2}\right) \text { and } X_{(2)}=\operatorname{maximum}\left(X_{1}, X_{2}\right) $$ Find (a) \(E\left[X_{(1)}\right]\) (b) \(\operatorname{Var}\left[X_{(1)}\right]\) (c) \(E\left[X_{(2)}\right]\) (d) \(\operatorname{Var}\left[X_{(2)}\right]\)

Short answer

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) \(\operatorname{Var}[X_{(2)}] = \frac{1}{2\mu^2}\)

Step-by-step solution

Find pdf of \(X_{(1)}\) and \(X_{(2)}\)

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: $$ f_{X_{(2)}}(x) = \frac{dF_{X_{(2)}}(x)}{dx} = 2\mu e^{-\mu x}(1 - e^{-\mu x}). $$

Compute \(E[X_{(1)}]\) and \(E[X_{(2)}]\)

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: $$ E[X_{(2)}] = \frac{3}{2\mu} $$

Compute \(E[X^{2}_{(1)}]\) and \(E[X^{2}_{(2)}]\)

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: $$ E[X^{2}_{(2)}] = \frac{11}{6\mu^2} $$

Compute \(\operatorname{Var}[X_{(1)}]\) and \(\operatorname{Var}[X_{(2)}]\)

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) \(\operatorname{Var}[X_{(2)}] = \frac{1}{2\mu^2}\)

Key concepts

Cumulative Distribution Function

The cumulative distribution function (CDF) plays a pivotal role in understanding probabilistic behavior of random variables. It essentially provides the probability that a random variable is less than or equal to a certain value. For an exponential random variable, the CDF is defined as:\begin{align*}F_X(x) = 1 - e^{-\text{rate} \times x}\begin{align*}In our exercise, we explored the CDF for the minimum and maximum of two exponential random variables. The CDF for the minimum, designated by \(X_{(1)}\), is particularly straightforward, as it represents the probability that both independent variables exceed a certain value, thus producing:\begin{align*}F_{X_{(1)}}(x) = 1 - (e^{-\text{rate} \times x})^2\begin{align*}Understanding the CDF is vital for grasping exponential random variables which inherently describes the time until an event occurs, ensuring everything from service times to lifespans are statistically predictable.

Expected Value

The expected value, or mean, of a random variable is a measure of the central tendency, akin to a long-term average. For an exponential random variable with rate parameter \(\mu\), the general formula for expected value is the reciprocal of the rate, or \(\frac{1}{\mu}\).

In the context of the exercise involving \(X_{(1)}\) and \(X_{(2)}\), the expected values correspond to the average of the minimum and maximum times, respectively, until an event happens. The solution we provided demonstrates the use of integration by parts, a technique in calculus that facilitates the computation of the expected values for more complex scenarios. This process confirms the analytical outcomes that the expected time for the minimum is \(\frac{1}{2\mu}\) and for the maximum is \(\frac{3}{2\mu}\), reflecting the intuitive notion that the maximum wait is indeed longer than the minimum.

Variance of Random Variables

Variance measures the spread or variability of a random variable’s possible values. It sheds light on the extent to which a random variable diverges from its expected value. We calculate it using the formula:\begin{align*}\text{Var}(X) = E[X^2] - (E[X])^2\begin{align*}In our solution, after finding the expectations for \(X_{(1)}\) and \(X_{(2)}\), we applied this formula. The variance tells us about the reliability or consistency of the time it takes for an event to happen. It is especially informative for processes described by exponential random variables, indicating the predictability of such events. The results for \(X_{(1)}\) and \(X_{(2)}\), \(\frac{1}{(2\mu)^2}\) and \(\frac{1}{2\mu^2}\) respectively, indicate that the range of outcomes is wider for the maximum when compared to the minimum.

Probability Density Function

A probability density function (PDF) pinpoints the likelihood of random variables taking on specific values. For exponential random variables, the PDF is expressed as:\begin{align*}f(x) = \text{rate} \times e^{-\text{rate} \times x}\begin{align*}With \(X_{(1)}\) and \(X_{(2)}\), the solution illustrated how to manipulate PDFs to capture the minimum and maximum of two independent exponentially distributed variables. The PDFs for the minimum and maximum are derived by differentiating their respective CDFs. These functions, specifically \(2\mu e^{-2\mu x}\) for \(X_{(1)}\) and \(2\mu e^{-\mu x}(1 - e^{-\mu x})\) for \(X_{(2)}\), outline how likely it is for a certain time to pass before the first and second occurrences of events, respectively. Understanding these PDFs is crucial, as they serve as the building blocks for more advanced calculations, such as expected value and variance.