Question

In Example \(5.3\) if server \(i\) serves at an exponential rate \(\lambda_{i}, i=1,2\), show that \(P\\{\) Smith is not last \(\\}=\left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\right)^{2}+\left(\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\right)^{2}\)

Short answer

The probability that Smith is not last is given by \(P\{\text{Smith is not last}\} = \left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\right)^{2}+\left(\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\right)^{2}\), where \(\lambda_1\) and \(\lambda_2\) are the service rates for server 1 and server 2, respectively. This result is obtained by finding the joint probability density function of the service times, defining the event of interest, calculating the probabilities of the two independent cases, and adding them up.

Step-by-step solution

Setup the problem

Let's denote \(X_1\) and \(X_2\) as the service times for the two servers with rates \(\lambda_1\) and \(\lambda_2\) respectively. According to the problem, both \(X_1\) and \(X_2\) have exponential distributions: \[X_1 \sim Exp(\lambda_1)\] \[X_2 \sim Exp(\lambda_2)\] We are interested in finding the probability that Smith is not last. Let's denote this event as \(A\): \[A: \text{Smith is not last}\] Now let's compute the probability of this event.

Find the joint probability density function (pdf) of X1 and X2

Since \(X_1\) and \(X_2\) are independent, we can find the joint pdf of \(X_1\) and \(X_2\) by multiplying their individual pdfs. - For \(X_1\), the pdf is given by: \(f_1(x_1) = \lambda_1 e^{-\lambda_1 x_1}\), for \(x_1>0\). - For \(X_2\), the pdf is given by: \(f_2(x_2) = \lambda_2 e^{-\lambda_2 x_2}\), for \(x_2>0\). The joint pdf of \(X_1\) and \(X_2\), represented by \(f(x_1, x_2)\), is obtained as follows: \[f(x_1, x_2) = f_1(x_1)f_2(x_2) = \lambda_1 e^{-\lambda_1 x_1} \times \lambda_2 e^{-\lambda_2 x_2} = \lambda_1 \lambda_2 e^{-(\lambda_1 x_1 + \lambda_2 x_2)}\]

Define the event and calculate its probability

The event \(A\) (Smith is not last) can be described as either \(X_1 < X_2\) or \(X_2 < X_1\). The probabilities of these two cases are independent, so we can compute them separately and then add them up: \[P(A) = P(X_1 < X_2) + P(X_2 < X_1)\] We can calculate the probabilities using the joint pdf that we derived and integrate over the appropriate regions: \[P(X_1 < X_2) = \int_{0}^{\infty} \int_{0}^{x_2} \lambda_1 \lambda_2 e^{-(\lambda_1 x_1 + \lambda_2 x_2)} dx_1 dx_2\] \[P(X_2 < X_1) = \int_{0}^{\infty} \int_{0}^{x_1} \lambda_1 \lambda_2 e^{-(\lambda_1 x_1 + \lambda_2 x_2)} dx_2 dx_1\] Now evaluate the integrals.

Evaluate the integrals for both cases

Let's solve both integrals separately: \[P(X_1 < X_2) = \lambda_1 \lambda_2 \int_{0}^{\infty}\left(\int_{0}^{x_2} e^{-(\lambda_1 x_1 + \lambda_2 x_2)} dx_1\right)dx_2\] After integrating \(x_1\): \[P(X_1 < X_2) = \lambda_1 \lambda_2 \int_{0}^{\infty} \left[\frac{e^{-(\lambda_1 x_1 + \lambda_2 x_2)}}{-\lambda_1}\right]_0^{x_2}dx_2 \] \[P(X_1 < X_2) = \int_{0}^{\infty} \frac{\lambda_2}{\lambda_1 + \lambda_2} e^{-\lambda_2 x_2} dx_2 = \left(\frac{\lambda_2}{\lambda_1 + \lambda_2}\right)^2\] Perform similar integration for \(P(X_2 < X_1)\): \[P(X_2 < X_1) = \int_{0}^{\infty} \frac{\lambda_1}{\lambda_1 + \lambda_2} e^{-\lambda_1 x_1} dx_1 = \left(\frac{\lambda_1}{\lambda_1 + \lambda_2}\right)^2\]

Add the probabilities and find the result

Finally, add the probabilities found in Step 4: \[P(A) = P(X_1 < X_2) + P(X_2 < X_1) = \left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\right)^{2}+\left(\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\right)^{2}\] This result verifies that: \[P\{\text{Smith is not last}\} = \left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\right)^{2}+\left(\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\right)^{2}\]

Key concepts

Exponential Distribution

Understanding the exponential distribution is crucial when dealing with certain types of wait times or service times. It's commonly applied in scenarios like the one in the exercise, where the time until a certain event, such as being served by a server, is of interest.

Specifically, the exponential distribution is a continuous probability distribution characterized by a parameter, often denoted by the symbol \(\lambda\), which represents the rate at which events occur. For a positive number \(\lambda\), if a random variable \(X\) follows an exponential distribution with parameter \(\lambda\), it is denoted as \(X \sim Exp(\lambda)\).

A key property of the exponential distribution is its \emph{memorylessness} – the probability of the event occurring in the next instant is the same regardless of how much time has already passed. This property makes it especially useful for modeling situations where the past does not influence the future likelihood of an event occurring.

Probability Density Function (PDF)

Let's dive deeper into the probability density function (PDF), which was referenced in the step-by-step solution of our exercise. A PDF is a fundamental concept when it comes to understanding continuous probability distributions like the exponential distribution.

The purpose of the PDF, often denoted as \(f(x)\), is to describe the likelihood of a continuous random variable falling within a certain range of values. The PDF assigns a probability to every point \(x\) that a random variable can take. In the exponential distribution case, the function is given by \(f(x; \lambda) = \lambda e^{-\lambda x}\) for \(x > 0\).

It's important to note that the value of the PDF at any single point is not a probability itself. Instead, it's the area under the curve of the PDF over an interval that gives us probabilities. Therefore, when we want to find probabilities associated with continuous variables, we use integration to calculate the area.

Integration in Probability

As mentioned in the explanation of the PDF, integration is a crucial process for finding probabilities related to continuous random variables. This aspect of probability calculus was applied in the textbook exercise to compute the probability of events for an exponential distribution.

In the context of the exercise, we used integration to find the probability that one server finishes serving before the other. The method of integrating the PDF over an interval corresponds to summing up all the infinitesimal probabilities across that interval, which provides the cumulative probability of an event occurring.

The beauty of integration in probability lies in its ability to convert the general probability guidelines given by the PDF into specific numerical probabilities that can inform real-world decisions and predictions. This connection between calculus and probability theory is a fundamental aspect of many statistical methods used in fields such as engineering, finance, and the physical sciences.