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For a non-negative integer random variable \(X\), in addition to the probability generating function \(\Phi_{X}(t)\) defined in equation (30.71), it is possible to define the probability generating function $$ \Psi_{X}(t)=\sum_{n=0}^{\infty} g_{n} t^{n} $$ where \(g_{n}\) is the probability that \(X>n\) (a) Prove that \(\Phi_{X}\) and \(\Psi_{X}\) are related by $$ \Psi_{X}(t)=\frac{1-\Phi_{X}(t)}{1-t} $$ (b) Show that \(E[X]\) is given by \(\Psi_{X}(1)\) and that the variance of \(X\) can be expressed as \(2 \Psi_{X}^{\prime}(1)+\Psi_{X}(1)-\left[\Psi_{X}(1)\right]^{2}\) (c) For a particular random variable \(X\), the probability that \(X>n\) is equal to \(\alpha^{n+1}\), with \(0<\alpha<1\). Use the results in \((\mathrm{b})\) to show that \(V[X]=\alpha(1-\alpha)^{-2}\).

Short Answer

Expert verified
Prove that \( \Psi_X(t) = \frac{1 - \Phi_X(t)}{1 - t} \). Use \( E[X] \ and \ variance formulas for for \( X \). Using \( \ g_n \ and \ P(X)\alpha^{n+1}, \ the variance \ X\.

Step by step solution

01

- Understanding the Probability Generating Functions

Given two different probability generating functions (PGF) for a non-negative integer random variable X: 1. Standard PGF: \(\Phi_X(t) = \mathbb{E}[t^X] \), 2. Alternative PGF: \( \Psi_X(t) = \sum_{n=0}^{\infty} g_n t^n \), where \( g_n \) is the probability that \( X > n \).
02

- Relationship between the PGFs

We need to prove the relationship: \( \Psi_X(t) = \frac{1 - \Phi_X(t)}{1 - t} \) Recall that: \( g_n = P(X > n) = \sum_{k=n+1}^{\infty} p_k \), where \(p_k\) is the probability that X equals k.
03

- Expansion of \( \Psi_X(t) \)

Consider the summation definition of \(\Psi_X(t)\): \( \Psi_X(t) = \sum_{n=0}^{\infty} g_n t^n \) Substitute \(g_n \) into the sum: \( \Psi_X(t) = \sum_{n=0}^{\infty} \sum_{k=n+1}^{\infty} p_k t^n \)
04

- Interchange the Order of Summation

By swapping the order of summation, we get: \( \Psi_X(t) = \sum_{k=1}^{\infty} p_k \sum_{n=0}^{k-1} t^n \) The inner sum is a geometric series: \( \sum_{n=0}^{k-1} t^n = \frac{1 - t^k}{1 - t} \)
05

- Substitution

Distribute \( p_k \) across the terms: \( \Psi_X(t) = \frac{1}{1 - t} \sum_{k=1}^{\infty} p_k - \sum_{k=1}^{\infty} p_k t^k \)
06

- Simplify Using \( \Phi_X(t) \)

Recognize that \( \sum_{k=1}^{\infty} p_k \) sums to 1 by the definition of the probability mass function: \( \sum_{k=1}^{\infty} p_k t^k = \Phi_X(t) - p_0 \), since \(\sum_{k=0}^{\infty} p_k t^k = \Phi_X(t)\) and \( p_0 = P(X = 0) \). Substitute and simplify: \( \Psi_X(t) = \frac{1}{1 - t} (1 - \Phi_X(t)) \)
07

- Expectation \( E[X] \) from \( \Psi_X(t) \)

Note that \( \Psi_X(1) = \sum_{n=0}^{\infty} g_n \). Recognize that \(g_n\) is the complement of the cumulative distribution function F(n). Thus, \(E[X] = \Psi_X(1)\).
08

- Variance \(V[X]\) from \(\Psi_X(t)\)

Differentiate \( \Psi_X(t)\) with respect to t and evaluate at t=1: \( \Psi_X'(t) = \frac{d}{dt} [\frac{1 - \Phi_X(t)}{1 - t}] \). Applying the quotient rule, solve to find expressions involving \( \Phi_X(t)\) and its derivatives. Then compute \( 2 \Psi_X'(1) + \Psi_X(1) - (\Psi_X(1))^2 \) for the variance.
09

- Apply Relationship to Given \( X \)

With \( P(X > n) = \alpha^{n+1} \): \( \Psi_X(t) = \sum_{n=0}^{\infty} \alpha^{n+1} t^n = \frac{\alpha}{1-\alpha t} \).\( E[X] = \Psi_X(1) \ = \alpha/(1-{\alpha}) \) and the variance is obtained by differentiating \( \Psi_X(t) \).
10

- Compute the Variance

Differentiate \( \Psi_X(t) = \frac{\alpha}{1 - \alpha t} \), evaluate at t=1, then use the variance formula:\( V[X] = 2 \Psi_X'(1) + \Psi_X(1) - (\Psi_X(1))^2 = \alpha (1 - \alpha)^{-2} \).

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Key Concepts

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

Non-Negative Integer Random Variable
A non-negative integer random variable serves as a fundamental concept in probability theory. It is a random variable, denoted as \(X\), that can only take non-negative integer values (0, 1, 2, etc.). This means \(X\) cannot be negative or a fraction.
Understanding non-negative integer random variables is crucial for solving many probability problems, as they frequently come up in scenarios involving counts and sequences.
A key feature of these variables is their probability mass function (PMF), which gives the probabilities that the variable is equal to specific values. Mathematically, the PMF is defined as \(P(X = k)\) for \(k = 0, 1, 2, \text{etc.}\), where each of these probabilities is non-zero and their sum totals to one.
Relationship Between PGFs
Probability Generating Functions (PGFs) are powerful tools used to encapsulate the entire distribution of a non-negative integer random variable \(X\) in a compact form.
The standard PGF, denoted by \(\Phi_X(t)\), is defined as:
$$\Phi_X(t) = \mathbb{E}[t^X] = \sum_{k=0}^{\infty} P(X = k) t^k $$
The alternative PGF, \(\Psi_X(t)\), is given by:
$$\Psi_X(t) = \sum_{n=0}^{\infty} P(X > n) t^n $$
To understand the relationship between these two PGFs, note that \(P(X > n)\) is equivalent to the tail sum of probabilities starting from \(n+1\):
$$P(X > n) = \sum_{k=n+1}^{\infty} P(X=k) $$
By examining their series expansions and manipulating sums, it’s found that these PGFs are related by:
$$\Psi_X(t) = \frac{1 - \Phi_X(t)}{1 - t} $$
This relationship shows that one PGF can be derived from the other, and highlights the interconnection of different representations of the probability of a random variable.
Expectation and Variance Calculations
Expectation (mean) and variance are two crucial statistical measures for any random variable. They provide insights into the average value and the spread of the values taken by the random variable.
For a non-negative integer random variable \(X\) with PGF \(\Psi_X(t)\):
To find the expectation \(\mathbb{E}[X]\), evaluate \(\Psi_X\) at \(t=1\):
$$ \mathbb{E}[X] = \Psi_X(1) $$
For variance, a more involved calculation is required. Using \(\Psi_X(t)\), the variance \(V[X]\) is given by:
$$ V[X] = 2 \Psi_X'(1) + \Psi_X(1) - (\Psi_X(1))^2 $$
Here's how each component contributes:
  • \(\Psi_X'(t)\) is the derivative of \(\Psi_X(t)\) with respect to \(t\).
  • Evaluating \(\Psi_X'(1)\) tells us about the rate of change of the PGF at \(t=1\).
  • The variance formula then combines these terms to measure the spread of \(X\) around its mean.

Understanding these formulas allows us to extract meaningful statistics from the PGFs and provides deeper insight into the behavior of the random variable.

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