Problem 9 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Consider the network example where the dataset is modeled as a realization of a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a Pois \((\mu)\) distribution. We estimate the probability of zero arrivals \(\mathrm{e}^{-\mu}\) by means of \(T=\mathrm{e}^{-\mathscr{X}_{n}}\). Check that $$ \mathrm{E}[T]=e^{-n \mu\left(1-e^{-1 / n}\right)} $$ Hint: write \(T=\mathrm{e}^{-Z / n}\), where \(Z=X_{1}+X_{2}+\cdots+X_{n}\) has a Pois \((n \mu)\) distribution.

Step-by-Step Solution

Answer

The expected value is \(\mathrm{E}[T] = e^{-n \mu (1 - e^{-1/n})}\).

Step 1: Introduction of Random Sample

We begin with a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a Poisson distribution with parameter \(\mu\). The task is to find the expected value of \(T = e^{-\mathscr{X}_{n}}\).

Step 2: Expression for \(T\)

The problem suggests writing \(T\) as \(T = e^{-Z/n}\), where \(Z = X_1 + X_2 + \cdots + X_n\). Notice that \(Z\) is the sum of \(n\) independent Poisson random variables, each with parameter \(\mu\).

Step 3: Distribution of \(Z\)

The sum \(Z\) therefore follows a Poisson distribution with mean \(n\mu\), i.e., \(Z \sim \text{Pois}(n \mu)\). This is because the sum of independent Poisson random variables with parameters \(\mu_i\) is Poisson distributed with parameter equal to the sum of the parameters.

Step 4: Finding \(\mathrm{E}[T]\)

To find \(\mathrm{E}[T]\), calculate \(\mathrm{E}[e^{-Z/n}]\). The moment-generating function (MGF) for a Poisson distribution \(\text{Pois}(n\mu)\) is given by \(M(t) = \exp(n\mu(e^t - 1))\).

Step 5: Apply MGF to \(e^{-Z/n}\)

We substitute \(t = -1/n\) in the MGF of \(Z\) to find \(\mathrm{E}[T]\): \(M(-1/n) = \exp(n\mu(e^{-1/n} - 1))\).

Step 6: Simplification

Simplifying \(M(-1/n)\) gives \(\mathrm{E}[T] = e^{-n \mu (1 - e^{-1/n})}\), which matches the result we need to verify.

Key Concepts

Moment Generating Function

The moment-generating function (MGF) is an essential tool in probability theory and statistics, especially when working with Poisson distributions. It helps us determine the expected value of a function of a random variable, like in our exercise for the expected value of \(e^{-Z/n}\). The MGF of a random variable \(X\) is defined by the formula: \[ M_X(t) = ext{E}[e^{tX}] \] This function, when it exists, offers a lot of information about the distribution of \(X\). For the Poisson distribution, the MGF is given by: \[ M(t) = ext{exp}(n\mu(e^{t} - 1)) \]Understanding the structure of an MGF can drastically simplify complex probability problems, as it provides a systematic method to find moments of the distribution. In our exercise, substituting \(t = -1/n\) into the MGF of \(Z\) allows us to calculate the expected value \(\text{E}[T]\). This is an excellent example of how MGFs can be applied to evaluate complex expressions.

Random Sample

A random sample refers to a set of observations \(X_1, X_2, \ldots, X_n\) taken from a larger population, where each observation follows a specified probability distribution—in this case, the Poisson distribution with parameter \(\mu\). Each observation is made independently, meaning the occurrence or value of one \(X\) has no influence on another.- **Independent observations:** Each observation in the sample is independent of the others.- **Identical distribution:** Every observation in the sample is drawn from the same distribution.These properties are crucial, particularly in our exercise which involves calculating an expectation value. When considering a Poisson distribution, recognizing that our sample is a collection of independent and identically distributed (i.i.d) random variables simplifies the computation of aggregate characteristics like the sum \(Z = X_1 + X_2 + \ldots + X_n\). Realizing these traits allows us to effectively handle problems regarding predictive analytics and inferential statistics.

Expectation Value

Expectation value, also known as the expected value or mean, is a fundamental concept in probability and statistics. It represents the long-term average or central value of a random variable. For any random variable \(X\), the expectation value \(\text{E}[X]\) can be considered the "center" of its distribution.The expectation value is computed using the probability distribution of the variable. For a Poisson distribution with parameter \(\lambda\), the expected value is simply \(\lambda\). In the context of our exercise, finding \(\text{E}[T]\) means we're looking for the expected value of the transformation of \(Z\), which is \( e^{-Z/n} \). To compute this, we utilize the MGF of \(Z\), a crucial tool that makes this calculation more manageable by working with the properties of exponentials and sums. Understanding the expectation value helps in predicting and analyzing data trends in probabilistic models.

Independent Random Variables

Independent random variables are variables whose outcomes do not influence each other. This means that knowing the result of one variable provides no information about the result of the other.In this exercise, each \(X_i\) (where \(i = 1,2,\ldots,n\)) is an independent random variable following a Poisson distribution with parameter \(\mu\). The sum \(Z = X_1 + X_2 + \ldots + X_n\) preserves the independence property, but it also forms a new Poisson random variable with parameter \( n\mu \).Why is this independence significant? - **Simplifies calculation:** When random variables are independent, calculations like finding the joint probability distribution or the sum of random variables become easier.- **Additivity of Poisson parameters:** The property that sums of independent Poisson random variables also have a Poisson distribution is a unique and useful feature, simplifying calculations in our exercise.Understanding these properties is key to solving problems involving multiple random variables and their transformations. It helps statisticians and data analysts work with complex datasets systematically.