Question

If \(X_{i}, i=1,2,3\), are independent exponential random variables with rates \(\lambda_{i}\), \(i=1,2,3\), find (a) \(\left.P \mid X_{1}

Short answer

\(P(X_1 < X_2 < X_3) = \frac{\lambda_1\lambda_2\lambda_3}{(\lambda_1 + \lambda_2 + \lambda_3)^3} \\\) \(P(X_1 < X_2 \mid \max(X_1, X_2, X_3)=X_3)=\frac{1}{2} \\\) \(E[\max X_i \mid X_1 < X_2 < X_3] = \frac{1}{\lambda_1 + \lambda_2 + \lambda_3} \\\) \(E[\max X_i]=\frac{1}{\lambda_1} + \frac{1}{2} (\frac{1}{\lambda_2} - \frac{1}{\lambda_1}) + \frac{1}{3} (\frac{1}{\lambda_3} - \frac{1}{\lambda_2})\)

Step-by-step solution

Define joint probability density function (PDF)

Since \(X_1, X_2\), and \(X_3\) are independent exponential random variables, their joint PDF is the product of their individual PDFs: \(f(x_1,x_2,x_3) = f_{X_1}(x_1)f_{X_2}(x_2)f_{X_3}(x_3) = \lambda_{1}e^{-\lambda_{1}x_{1}}\lambda_{2}e^{-\lambda_{2}x_{2}}\lambda_{3}e^{-\lambda_{3}x_{3}}\)

Compute the probability with the joint PDF

We want to compute the probability \(P(X_1 < X_2 < X_3)\). This can be computed by integrating the joint PDF over the region where \(x_1 < x_2 < x_3\): \(P\left(X_{1}<X_{2}<X_{3}\right)=\int_{0}^{\infty} \int_{x_{1}}^{\infty} \int_{x_{2}}^{\infty} f(x_1,x_2,x_3) \, dx_3 dx_2 dx_1\)

Solve the integral

Plugging in the joint PDF and solving the integral: \(P\left(X_{1} \max(X_1,X_2))\). Notice that the events \(X_1 \max(X_1,X_2)\) are independent, hence we have: \(P(X_1 < X_2 \mid \max(X_1, X_2, X_3)=X_3)=\frac{P(X_1 \max(X_1,X_2))}{P(X_3 > \max(X_1,X_2))}=\frac{P(X_1 \max(X_1,X_2))}{P(X_3 > \max(X_1,X_2))}=P(X_1<X_2)=\frac{1}{2}\) (c) Expected value \(E[\max X_i \mid X_1 < X_2 < X_3]\) In this case, we know that \(X_3 = \max(X_1, X_2, X_3)\), so: \(E[\max X_i \mid X_1 < X_2 < X_3] = E[X_3 \mid X_1 < X_2 < X_3]\) And from the results obtained in part (a): \(E[X_3 \mid X_1 < X_2 < X_3]=\frac{1}{\lambda_1 + \lambda_2 + \lambda_3}\) (d) Expected value \(E[\max X_i]\) In general, for an exponential random variable \(X\) with rate \(\lambda\), we have \(E[X] = \frac{1}{\lambda}\). Now we want to find \(E[\max X_i]\): \(E[\max X_i]=E[X_1]+\frac{1}{2}(E[X_2] - E[X_1])+ \frac{1}{3}(E[X_3] - E[X_2])\) Plugging in the expected values of \(X_1, X_2\) and \(X_3\), we get: \(E[\max X_i]=\frac{1}{\lambda_1} + \frac{1}{2} (\frac{1}{\lambda_2} - \frac{1}{\lambda_1}) + \frac{1}{3} (\frac{1}{\lambda_3} - \frac{1}{\lambda_2})\)

Key concepts

Probability Density Function

Understanding the probability density function (PDF) is crucial when dealing with continuous random variables, such as exponential random variables. In the context of the given problem, each random variable, denoted as \(X_i\), has an associated PDF that characterizes the likelihood of the variable taking on a certain value. The exponential distribution's PDF for a rate parameter \(\lambda\) is given by \(f(x) = \lambda e^{-\lambda x}\) for \(x \ge 0\).

The PDF essentially details how the probability is distributed across the range of possible values. For exponential random variables, the PDF reflects that values closer to 0 are more likely than values further away, consistent with the 'memoryless' property of the exponential distribution. Since the question involves independent random variables, their joint PDF is simply the product of their individual PDFs — a fact used in the solution to express the collective behavior of multiple variables.

Conditional Probability

Conditional probability is the probability of an event occurring provided that another event has already occurred. This is a cornerstone concept in probability and statistics that allows us to update our predictions based on new information.

As per the example, calculating the conditional probability of \(X_1 < X_2\) given that \(X_3\) is the maximum involves understanding that the events \(X_1 \max(X_1,X_2)\) are independent. Here, knowing that \(X_3\) is the largest value doesn't affect the likelihood of \(X_1\) being less than \(X_2\), hence the simplicity of the solution leading to a probability of 1/2. This reinforces the concept of independence within the context of conditional probabilities.

Expected Value

The expected value, or mean, of a random variable is a fundamental measure that provides the 'long-term average' — or the center of gravity for the probability distribution. For exponential random variables, the expected value is given by \(E[X] = \frac{1}{\lambda}\), where \(\lambda\) is the rate parameter of the distribution.

In the provided problem, we used the concept of the expected value to find the average of the maximum value given a condition — a useful measure in real-world scenarios to predict averages under certain constraints. Additionally, the expected value's linear property was applied to find the expected maximum of the three independent variables by weighting the individual expected values accordingly.

Integration of Probability

The integration of the probability density function over a given range yields the probability of a random variable falling within that range. When we integrate a joint PDF across specific limits — as in the case of finding \(P(X_1 < X_2 < X_3)\) — we're essentially summing up the probabilities across a volume where the condition holds true.

This process of integrating the exponential PDFs of independent random variables highlights how probabilities are computed for variables interacting under certain conditions. The techniques of integration are fundamental to finding probabilities in continuous distributions and are prominently showcased in the solution to the exercise, where multiple integrals indicate layers of conditional interactions.