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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 (26.71) 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
Show the relationship between \( \Phi_X(t) \) and \( \Psi_X(t) \). Verify the expressions for expected value and variance. Demonstrate calculations of given \(P(X > n) = \alpha^{n+1} \).

Step by step solution

01

Understand the Definition of Probability Generating Functions

Recall that for a non-negative integer random variable X, the probability generating function \(\Phi_{X}(t)\) is defined as \(\Phi_{X}(t) = \sum_{n=0}^{\infty} P(X=n)t^n\), and \(\Psi_{X}(t)\) is defined as \(\Psi_{X}(t) = \sum_{n=0}^{\infty} g_{n} t^{n}\), where \(g_{n}\) is the probability that \(X > n\).
02

Prove the Relationship Between \(\Phi_{X}(t)\) and \(\Psi_{X}(t)\)

Observe that \(g_n = P(X > n)\), which can be written as \(1 - P(X \leq n)\). From the definition of \(\Phi_{X}(t)\), we know that \(\sum_{n=0}^{\infty} P(X=n)t^n = \Phi_{X}(t)\). Since \(P(X \leq n) = \sum_{k=0}^{n} P(X=k)\), then \(P(X > n) = 1 - \sum_{k=0}^{n} P(X=k)\). Hence, the series \(\Psi_{X}(t) = \sum_{n=0}^{\infty} (1 - \sum_{k=0}^{n} P(X=k))t^n\). This can be rewritten as \(\Psi_{X}(t) = \sum_{n=0}^{\infty} t^n - \sum_{n=0}^{\infty} (\sum_{k=0}^{n} P(X=k))t^n\), which is equal to \(\frac{1}{1-t} - \Phi_{X}(t)\) derived from geometric series. Thus, \(\Psi_{X}(t) = \frac{1 - \Phi_{X}(t)}{1-t}\).
03

Show \(E[X]\) Using \(\Psi_{X}(t)\)

Set \(t=1\) in \(\Psi_{X}(t)\): \(\Psi_{X}(1)\). Recall that \(\Psi_{X}(1) = \sum_{n=0}^{\infty} g_n\). Since \(g_n = P(X > n)\), the sum equals the expected value of \(X\), hence \(E[X] = \Psi_{X}(1)\).
04

Express Variance of \(X\) Using \(\Psi_{X}(t)\)

From (b), it is given that the variance can be computed as \(\text{Var}[X] = 2 \Psi_{X}'(1) + \Psi_{X}(1) - (\Psi_{X}(1))^2\). Use the derivative \(\Psi_{X}(t)\) to find the expression and evaluate it at \(t=1\).
05

Determine \(\Psi_{X}(t)\) for Specific \(X\)

Given that \(P(X > n) = \alpha^{n+1}\), use this to substitute into the definition of \(\Psi_{X}(t)\): \(\Psi_{X}(t) = \sum_{n=0}^{\infty} \alpha^{n+1} t^n = \alpha \sum_{n=0}^{\infty} (\alpha t)^n = \frac{\alpha}{1-\alpha t}\).
06

Compute and Verify Variance Expression

Including the steps above, differentiating \(\Psi_{X}(t)\) concerning \(t\) and evaluating this at \(t=1\), show that \(2 \Psi_{X}'(1) + \Psi_{X}(1) - (\Psi_{X}(1))^2 = \alpha(1-\alpha)^{-2}\) as the variance of \(X\).

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Key Concepts

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

Random Variables
In probability theory, a random variable is a variable whose possible values are outcomes of a random phenomenon. It is a way to map outcomes of a random process to numbers. There are two main types of random variables: discrete and continuous.
Discrete random variables take on countable values, like rolling a dice or the number of heads in a series of coin tosses. For example, if we roll a 6-sided die, the outcome might be 1, 2, 3, 4, 5, or 6.
Continuous random variables, on the other hand, take on an infinite number of possible values within a given range. For instance, the time it takes for a computer to process a request or the height of students in a class can be modeled as continuous random variables.
Understanding random variables is crucial as we often seek to determine the probability of different outcomes associated with these variables. They are fundamental to calculating other statistical measures and to defining the probability generating functions.
Probability generating functions help in characterizing the distribution of a discrete random variable. For a non-negative integer random variable X, the probability generating function \(\Phi_{X}(t)\) is defined as: \(\Phi_{X}(t) = \sum_{n=0}^{\infty} P(X=n)t^n\).
Think of probability generating functions as a compact way to encapsulate the entire distribution of a random variable. By manipulating this function, we can extract important information about the random variable, such as probabilities, expectations, and variances.
Expectation
Expectation, also known as the expected value or mean, is a key concept in probability. It provides a measure of the 'central' value of a random variable.
Mathematically, the expectation of a discrete random variable X is given by: \(\mathbb{E}[X] = \sum_{x} x \cdot P(X=x) \).
This represents the average value of X if we were to repeat an experiment many times. It helps us understand what we can 'expect' from the random variable on average.
In the context of probability generating functions, the expectation can be derived through: \(\mathbb{E}[X] = \Psi_{X}(1)\). The function \(\Psi_{X}(t) = \sum_{n=0}^{\infty} g_{n} t^{n} \), where \(\Psi_{X}(1) = \sum_{n=0}^{\infty} g_n\) sums up probabilities that X is greater than a certain value, which leads to the expected value.
Variance
Variance is another critical concept which measures how much the values of a random variable deviate from the expected value. It quantifies the spread of the random variable’s possible values.
Mathematically, variance is given by: \(\text{Var}[X] = \mathbb{E}[X^2] - (\mathbb{E}[X])^2\). This equation subtracts the square of the mean from the expected value of the square of the random variable.
In simpler terms, it gives us an idea about the 'spread' of the data. A higher variance means that the data points are spread out more widely. Conversely, a low variance means they are closely clustered around the mean.
For our generating function, the variance of X can be expressed as: \(\text{Var}[X] = 2 \Psi_{X}'(1) + \Psi_{X}(1) - (\Psi_{X}(1))^2\). This more advanced formula involves the first derivative of the generating function \(\Psi_{X}(t)\).
Probability Theory
Probability theory is the branch of mathematics that deals with the analysis of random phenomena. At its core, it involves the study of how likely events are to occur.
Some fundamental concepts in probability theory include:
- **Sample Space**: The set of all possible outcomes of a random experiment.
- **Event**: A subset of the sample space that we are interested in. For example, getting an even number when rolling a die.
- **Probability**: A measure that quantifies the likelihood of an event, typically ranging from 0 (impossible event) to 1 (certain event).
Understanding probability theory helps us predict the likelihood of various outcomes and make informed decisions. For instance, if we know the probability of rain tomorrow, we can decide whether to carry an umbrella.
In our context, we extensively use probability theory to define and manipulate probability generating functions. These functions are pivotal in deriving critical properties of random variables, like their mean (expectation) and their spread (variance).
By using concepts from probability theory such as generating functions, we can simplify complex problems and derive meaningful insights about random variables and their distributions.

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Most popular questions from this chapter

A shopper buys 36 items at random in a supermarket where, because of the sales tax imposed, the final digit (the number of pence) in the price is uniformly and randomly distributed from 0 to \(9 .\) Instead of adding up the bill exactly she rounds each item to the nearest 10 pence, rounding up or down with equal probability if the price ends in a ' 5 '. Should she suspect a mistake if the cashier asks her for 23 pence more than she estimated?

\(26.3 A\) and \(B\) each have two unbiased four-faced dice, the four faces being numbered \(1,2,3,4\). Without looking, \(B\) tries to guess the sum \(x\) of the numbers on the bottom faces of \(A\) 's two dice after they have been thrown onto a table. If the guess is correct \(B\) receives \(x^{2}\) euros, but if not he loses \(x\) euros. Determine \(B\) 's expected gain per throw of \(A\) 's dice when he adopts each of the following strategies: (a) he selects \(x\) at random in the range \(2 \leq x \leq 8\); (b) he throws his own two dice and guesses \(x\) to be whatever they indicate; (c) he takes your advice and always chooses the same value for \(x\). Which number would you advise? 26.4 Use the method of induction to prove equation (26.16), the probability addition law for the union of \(n\) general events.

A particle is confined to the one-dimensional space \(0 \leq x \leq a\) and classically it can be in any small interval \(d x\) with equal probability. However, quantum mechanics gives the result that the probability distribution is proportional to \(\sin ^{2}(n \pi x / a)\), where \(n\) is an integer. Find the variance in the particle's position in both the classical and quantum mechanical pictures and show that, although they differ, the latter tends to the former in the limit of large \(n\), in agreement with the correspondence principle of physics.

(a) Gamblers \(A\) and \(B\) each roll a fair six-faced die, and \(B\) wins if his score is strictly greater than \(A\) 's. Show that the odds are 7 to 5 in \(A\) 's favour. (b) Calculate the probabilities of scoring a total \(T\) from two rolls of a fair die for \(T=2,3, \ldots, 12 .\) Gamblers \(C\) and \(D\) each roll a fair die twice and score respective totals \(T_{C}\) and \(T_{D}, D\) winning if \(T_{D}>T_{C} .\) Realising that the odds are not equal, \(D\) insists that \(C\) should increase her stake for each game. \(C\) agrees to stake \(£ 1.10\) per game, as compared to \(D\) 's \(£ 1.00\) stake. Who will show a profit?

The random variables \(X\) and \(Y\) take integer values \(\geq 1\) such that \(2 x+y \leq 2 a\) where \(a\) is an integer greater than \(1 .\) The joint probability within this region is given by $$ \operatorname{Pr}(X=x, Y=y)=c(2 x+y) $$ where \(c\) is a constant, and it is zero elsewhere. Show that the marginal probability \(\operatorname{Pr}(X=x)\) is $$ \operatorname{Pr}(X=x)=\frac{6(a-x)(2 x+2 a+1)}{a(a-1)(8 a+5)} $$ and obtain expressions for \(\operatorname{Pr}(Y=y),(\mathrm{a})\) when \(y\) is even and \((\mathrm{b})\) when \(y\) is odd. Show further that $$ E[Y]=\frac{6 a^{2}+4 a+1}{8 a+5} $$ (You will need the results about series involving the natural numbers given in subsection 4.2.5.)

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