Understanding Inter-Arrival Times in a Poisson Process

The question asks for the fourth raw moment of the inter-arrival time in a system where customers arrive following a Poisson process with parameter \(\lambda\). A key concept here is the relationship between the Poisson process and the exponential distribution.

When events (like customer arrivals) occur randomly over time such that the number of events in a fixed interval follows a Poisson distribution with rate \(\lambda\), the time between consecutive events (the inter-arrival time) follows an exponential distribution with parameter \(\lambda\).

The probability density function (PDF) of an exponentially distributed random variable \(X\) representing the inter-arrival time is given by:

\(f(x; \lambda) = \lambda e^{-\lambda x}\) for \(x \ge 0\), and \(f(x; \lambda) = 0\) for \(x < 0\).

Calculating the Fourth Raw Moment

The \(k\)-th raw moment of a random variable \(X\), denoted by \(\mu_k^{'}\) or \(E[X^k]\), is the expected value of \(X^k\). It is calculated using the formula:

\(E[X^k] = \int_{-\infty}^{\infty} x^k f(x) dx\)

For the inter-arrival time \(X\), which is non-negative and exponentially distributed, the integral becomes:

\(\mu_k^{'} = E[X^k] = \int_{0}^{\infty} x^k (\lambda e^{-\lambda x}) dx\)

We need to find the fourth raw moment, so we set \(k=4\):

\(\mu_4^{'} = E[X^4] = \int_{0}^{\infty} x^4 (\lambda e^{-\lambda x}) dx\)

We can pull the constant \(\lambda\) out of the integral:

\(\mu_4^{'} = \lambda \int_{0}^{\infty} x^4 e^{-\lambda x} dx\)

This integral is a standard form related to the Gamma function. The general formula for the integral \(\int_{0}^{\infty} x^n e^{-ax} dx\) is \(\frac{\Gamma(n+1)}{a^{n+1}}\), where \(\Gamma(z)\) is the Gamma function. For a positive integer \(n\), \(\Gamma(n+1) = n!\).

In our integral, we have \(n=4\) and \(a=\lambda\). Applying the formula:

\(\int_{0}^{\infty} x^4 e^{-\lambda x} dx = \frac{\Gamma(4+1)}{\lambda^{4+1}} = \frac{\Gamma(5)}{\lambda^5}\)

Now we calculate \(\Gamma(5)\). Since 5 is a positive integer, \(\Gamma(5) = 4! = 4 \times 3 \times 2 \times 1 = 24\).

Substituting this back into the integral result:

\(\int_{0}^{\infty} x^4 e^{-\lambda x} dx = \frac{24}{\lambda^5}\)

Finally, substitute this back into the expression for \(\mu_4^{'}\):

\(\mu_4^{'} = \lambda \times \frac{24}{\lambda^5} = \frac{24\lambda}{\lambda^5} = \frac{24}{\lambda^4}\)

Conclusion

The fourth raw moment for the inter-arrival time (which follows an exponential distribution with parameter \(\lambda\)) in a Poisson process with parameter \(\lambda\) is \(\dfrac{24}{\lambda^4}\).

Let's compare this result with the given options:

• Option 1: \(\dfrac{1}{\lambda^4}\)
• Option 2: \(\dfrac{4}{\lambda^4}\)
• Option 3: \(\dfrac{6}{\lambda^4}\)
• Option 4: \(\dfrac{24}{\lambda^4}\)

Our calculated fourth raw moment matches Option 4.

Revision Table: Exponential Distribution Moments

Additional Information: Poisson Process and Exponential Distribution

A Poisson process is a fundamental concept in probability theory and statistics used to model the occurrence of random events over time or space. Key properties include:

• Events occur one at a time.
• The number of events in any given interval is independent of the number of events in any other disjoint interval.
• The probability of an event occurring in a small interval is proportional to the length of the interval, and the probability of more than one event in such an interval is negligible.

If the rate of events is constant (\(\lambda\) events per unit time), then:

• The number of events in a fixed time interval \(t\) follows a Poisson distribution with mean \(\lambda t\).
• The time between successive events (inter-arrival time) follows an exponential distribution with mean \(1/\lambda\).

The exponential distribution is the only continuous distribution with the memoryless property, meaning the future duration does not depend on how much time has already elapsed.