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The special case of the gamma distribution in which \(\alpha\) is a positive integer \(n\) is called an Erlang distribution. If we replace \(\beta\) by \(1 / \lambda\) in Expression (4.8), the Erlang pdf is $$ f(x ; \lambda, n)=\left\\{\begin{array}{cc} \frac{\lambda(\lambda x)^{n-1} e^{-\lambda x}}{(n-1) !} & x \geq 0 \\ 0 & x<0 \end{array}\right. $$ It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter \(\lambda\), then the total time \(X\) that elapses before all of the next \(n\) events occur has pdf \(f(x ; \lambda, n)\). a. What is the expected value of \(X\) ? If the time (in minutes) between arrivals of successive customers is exponentially distributed with \(\lambda=.5\), how much time can be expected to elapse before the tenth customer arrives? b. If customer interarrival time is exponentially distributed with \(\lambda=.5\), what is the probability that the tenth customer (after the one who has just arrived) will arrive within the next \(30 \mathrm{~min}\) ? c. The event \(\\{X \leq t\\}\) occurs iff at least \(n\) events occur in the next \(t\) units of time. Use the fact that the number of events occurring in an interval of length \(t\) has a Poisson distribution with parameter \(\lambda t\) to write an expression (involving Poisson probabilities) for the Erlang cdf \(F(t ; \lambda, n)=P(X \leq t)\).

Step by step solution

01

Expected Value of Erlang Distribution

The expected value of an Erlang distribution, which is a special case of the gamma distribution with parameter \(\alpha = n\) (a positive integer), is given by \(E[X] = \alpha \beta\). In the Erlang distribution, since \(\beta = \frac{1}{\lambda}\), the expected value becomes \(E[X] = \frac{n}{\lambda}\). Substitute \(n = 10\) and \(\lambda = 0.5\) to find the expected time: \[ E[X] = \frac{10}{0.5} = 20 \text{ minutes} \] This gives the time expected to elapse before the tenth customer arrives.

02

Probability Calculation for the 10th Customer within 30 Minutes

The random variable \(X\) representing the time until the tenth customer follows an Erlang distribution with parameters \(\lambda = 0.5\) and \(n = 10\). We calculate \( P(X \leq 30) \) using the cumulative distribution function (CDF) for the Erlang distribution, which for given \(n\) and \(\lambda\) is: \[ P(X \leq t) = 1 - e^{-\lambda t} \sum_{k=0}^{n-1} \frac{(\lambda t)^k}{k!} \] Substituting \(\lambda = 0.5\), \(n = 10\), and \(t = 30\): \[ P(X \leq 30) = 1 - e^{-15} \sum_{k=0}^{9} \frac{15^k}{k!} \] Calculate this expression to find the required probability.

03

Erlang CDF Expression in Terms of Poisson Probabilities

According to the relationship between Erlang and Poisson distributions, the CDF of Erlang can be given in terms of a Poisson distribution. The number of events \(X\) occurring within a time \(t\) follows a Poisson distribution with parameter \(\lambda t\). Thus, the Erlang CDF can be expressed as: \[ F(t; \lambda, n) = P(X \leq t) = P(Y \geq n) = 1 - P(Y < n) \] where \(Y\) follows a Poisson distribution with mean \(\lambda t\). Therefore, \[ F(t; \lambda, n) = 1 - \sum_{k=0}^{n-1} \frac{(\lambda t)^k e^{-\lambda t}}{k!} \] This is equivalent to the expression used in calculating the probability in Step 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gamma Distribution

The Gamma Distribution is a continuous probability distribution that is crucial in many fields of engineering and statistics. It is defined by two parameters: \( \alpha \) (shape) and \( \beta \) (rate).
These parameters shape the behavior and characteristics of the distribution.
- **Shape Parameter (\( \alpha \))**: Determines the shape of the distribution. When \( \alpha \) is a positive integer, the distribution becomes an Erlang distribution.- **Rate Parameter (\( \beta \)):** Inversely related to the scale; influences how quickly probabilities decline.A special aspect of the Gamma Distribution is how it encompasses other distributions:
- **Erlang Distribution:** A Gamma Distribution with \( \alpha = n \) a positive integer.
- **Exponential Distribution:** A special case when \( \alpha = 1\).Gamma distributions are used to model the time until \( \alpha \) events occur, with applications ranging from finance to reliability analysis.

Exponential Distribution

The Exponential Distribution is a simplified version of the Gamma Distribution that describes the time between events in a Poisson process, where these events occur continuously and independently at a constant average rate.
It's defined by a single parameter \( \lambda \), which is the rate of events occurring.Some key characteristics include:

• **Memoryless Property:** The probability of an event occurring in the future is independent of the past.
• **Mean/Lifetime:** The mean or expected time between events is given by \( 1 / \lambda \).

Exponential distribution models phenomena such as time until failure of a machine part, customer arrival times, and more.

Poisson Distribution

The Poisson Distribution is a discrete probability distribution expressing the probability of a given number of events occurring in a fixed interval of time or space, assuming these events happen with a known constant mean rate and independently of the time since the last event.
It is characterized by the parameter \( \lambda \), the average number of events in an interval.Key points:

• **Uses:** Commonly used in queueing theory, telecommunications, and reliability engineering.
• **Relation to Exponential Distribution:** While Poisson describes the number of events, the Exponential distribution describes the time between events.

The Erlang Distribution can be derived from the Poisson distribution, emphasizing its foundational role in understanding waiting times and arrival processes.

Expected Value

Expected Value, often regarded as the "average" or "mean" of a distribution, is the long-term average outcome of a random variable. For distributions, it guides predictions and decision-making based on probabilistic events.
In the context of the Erlang Distribution, the expected value \( E[X] \) is derived using its parameters:The formula for expected value is \( E[X] = \alpha \beta \). In the case of Erlang, since \( \beta = 1/\lambda \), it simplifies to \( E[X] = \frac{n}{\lambda} \).Why is Expected Value important?

• **Decision Making:** Helps in making informed choices when dealing with uncertainty.
• **Risk Analysis:** Predicts potential losses or gains in scenarios involving risk.

Mastering expected value calculations is crucial in fields ranging from finance to statistics and beyond.