Problem 13 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(Z_{n}\) have an Erlang- \(n\) distribution with parameter \(\lambda\). a. Use integration by parts to show that for \(a \geq 0\) and \(n \geq 2\) : $$ \mathrm{P}\left(Z_{n} \leq a\right)=\int_{0}^{a} \frac{\lambda^{n} z^{n-1} \mathrm{e}^{-\lambda z}}{(n-1) !} \mathrm{d} z=-\frac{(\lambda a)^{n-1}}{(n-1) !} \mathrm{e}^{-\lambda a}+\mathrm{P}\left(Z_{n-1} \leq a\right) $$ b. Use a to show that for \(a \geq 0\) : $$ \mathrm{P}\left(Z_{n} \leq a\right)=-\sum_{i=1}^{n-1} \frac{(\lambda a)^{i}}{i !} \mathrm{e}^{-\lambda a}+\mathrm{P}\left(Z_{1} \leq a\right) $$ c. Conclude that for \(a \geq 0\) : $$ \mathrm{P}\left(Z_{n} \leq a\right)=1-\mathrm{e}^{-\lambda a} \sum_{i=0}^{n-1} \frac{(\lambda a)^{i}}{i !} $$

Step-by-Step Solution

Answer

The given Erlang distribution formulas are verified using recursive integration by parts.

Step 1: Recall the Erlang Distribution PDF

The probability density function (PDF) for an Erlang- distribution with shape parameter \( n \) and rate parameter \( \lambda \) is given by:\[ f(z; n, \lambda) = \frac{\lambda^n z^{n-1} e^{-\lambda z}}{(n-1)!}\]This is valid for \( z \geq 0 \).

Step 2: Set Up the Integration for CDF

The cumulative distribution function (CDF) \( \mathrm{P}(Z_n \leq a) \) is defined by the integral:\[\mathrm{P}(Z_n \leq a) = \int_0^a \frac{\lambda^n z^{n-1} e^{-\lambda z}}{(n-1)!} \, dz\]

Step 3: Apply Integration by Parts

Integration by parts formula is:\[\int u \, dv = uv - \int v \, du\]Let \( u = z^{n-1} \) and \( dv = \frac{\lambda^n e^{-\lambda z}}{(n-1)!} \, dz \).Then \( du = (n-1)z^{n-2} \, dz \) and \( v = -\frac{e^{-\lambda z}}{\lambda} \).Now substitute into the integration by parts formula to get:\[\int_0^a \frac{\lambda^n z^{n-1} e^{-\lambda z}}{(n-1)!} \, dz = \left[-\frac{\lambda^{n-1} z^{n-1} e^{-\lambda z}}{(n-1)!}\right]_0^a + \int_0^a \frac{(n-1) \lambda^{n-1} z^{n-2} e^{-\lambda z}}{(n-2)!} \, dz\]

Step 4: Evaluate and Simplify the CDF Expression

Evaluate the expression at \( z = a \) and \( z = 0 \):\[-\frac{\lambda^{n-1} a^{n-1} e^{-\lambda a}}{(n-1)!} + 0\]So the result becomes:\[\mathrm{P}(Z_n \leq a) = -\frac{(\lambda a)^{n-1}}{(n-1)!} e^{-\lambda a} + \mathrm{P}(Z_{n-1} \leq a)\]

Step 5: Generalize to Sum for (b)

From part (a), recursively apply the formula for \( \mathrm{P}(Z_{n-1} \leq a) \), \( \mathrm{P}(Z_{n-2} \leq a) \), ..., down to \( \mathrm{P}(Z_1 \leq a) \):\[\mathrm{P}(Z_n \leq a) = --- = -\sum_{i=1}^{n-1} \frac{(\lambda a)^i}{i!} e^{-\lambda a} + \mathrm{P}(Z_1 \leq a)\]

Step 6: Establish Final Expression for (c)

Use the fact that \( \mathrm{P}(Z_1 \leq a) = 1 - e^{-\lambda a} \):\[\mathrm{P}(Z_n \leq a) = 1 - e^{-\lambda a} \sum_{i=0}^{n-1} \frac{(\lambda a)^i}{i!}\]Combine part (a) and (b) to derive the final expression.

Key Concepts

Probability Density Function

The Probability Density Function (PDF) is a fundamental concept in probability and statistics. It describes how the values of a random variable are distributed. For continuous random variables, like those in the Erlang distribution, a PDF helps us to understand the likelihood of a variable taking a specific value within a certain range.

In the case of an Erlang distribution with parameters \( n \) and \( \lambda \), the PDF is expressed as: \[ f(z; n, \lambda) = \frac{\lambda^n z^{n-1} e^{-\lambda z}}{(n-1)!} \] This PDF defines the probability of the variable \( Z \) occurring between a range of values for \( z \). It integrates over the range to provide the probability of the variable being less than or equal to a certain threshold. It's crucial to note that the PDF for the Erlang distribution encompasses expressions that involve exponentials and gamma functions, reflecting the distribution's characteristics following a specific shape and rate.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is another essential tool in understanding random variables. It provides the probability that a random variable \( Z \) will be less than or equal to a given value \( a \).

In the context of the Erlang distribution, the CDF is defined as the integral of the PDF from 0 to \( a \): \[ \mathrm{P}(Z_n \leq a) = \int_0^a \frac{\lambda^n z^{n-1} e^{-\lambda z}}{(n-1)!} \, dz \] This definition captures the area under the curve of the probability density function up to the value \( a \). The CDF is especially important for quantifying cumulative probabilities over an interval. In simpler terms, it helps us understand the total probability that the variable will fall below a particular threshold, providing a more complete understanding of the distribution's behavior over continuous ranges.

Integration by Parts

Integration by Parts is a powerful technique derived from the product rule of differentiation. It is frequently used to solve complex integration problems where the direct integration of two multiplied functions isn't straightforward. The formula for Integration by Parts is: \[ \int u \, dv = uv - \int v \, du \] Here, the goal is to choose \( u \) and \( dv \) wisely so that the resulting integral \( \int v \, du \) is simpler to evaluate.

In dealing with the CDF of an Erlang distribution, Integration by Parts comes into play by helping us simplify the integral defining the CDF: - Here, let \( u = z^{n-1} \) and \( dv = \frac{\lambda^n e^{-\lambda z}}{(n-1)!} \, dz \).
- Differentiate and integrate to get \( du = (n-1)z^{n-2} \, dz \) and \( v = -\frac{e^{-\lambda z}}{\lambda} \).
- This application allows breaking down the complex integration into manageable parts, ultimately simplifying the evaluation through recursive breakdowns.

Exponential Distribution

The Exponential Distribution is a simpler special case of the Erlang distribution, often used to model the time between events in a process that follows a constant average rate. It is characterized by a single rate parameter \( \lambda \), similar to the rate parameter in the Erlang distribution.

The probability density function for an Exponential Distribution is: \[ f(x; \lambda) = \lambda e^{-\lambda x} \] This function describes the likelihood of time between events being exactly \( x \).

In our context, understanding the Exponential Distribution helps to derive simpler conclusions about the behavior of random variables, particularly when dealing with the sum or combinations of independent exponential variables. This distribution is crucial when we discuss the Erlang-1 distribution \( Z_1 \), which is exactly an Exponential distribution. Understanding this relationship helps in reinterpreting and applying the CDF findings in contexts where exponential variables are at play.