Question

(a) In two sets of binomial trials \(T\) and \(t\) the probabilities that a trial has a successful outcome are \(P\) and \(p\) respectively, with corresponding probabilites of failure of \(Q=1-P\) and \(q=1-p .\) One 'game' consists of a trial \(T\) followed, if \(T\) is successful, by a trial \(t\) and then a further trial \(T .\) The two trials continue to alternate until one of the \(T\) trials fails, at which point the game ends. The score \(S\) for the game is the total number of successes in the t-trials. Find the PGF for \(S\) and use it to show that $$ E[S]=\frac{P p}{Q}, \quad V[S]=\frac{P p(1-P q)}{Q^{2}} $$ (b) Two normal unbiased six-faced dice \(A\) and \(B\) are rolled alternately starting with \(A\); if \(A\) shows a 6 the experiment ends. If \(B\) shows an odd number no points are scored, if it shows a 2 or a 4 then one point is scored, whilst if it records a 6 then two points are awarded. Find the average and standard deviation of the score for the experiment and show that the latter is the greater.

Short answer

The PGF for S is \[\frac{Pp}{Q - PQx}\]. Thus, \(E[S] = \frac{Pp}{Q}\) and \(V[S] = \frac{Pp(1-Pq)}{Q^2}\). The average score for the dice game is 2/5, and the standard deviation is greater.

Step-by-step solution

Understand the Setup of the Game

Each game consists of trials from two sets: trial set T with success probability P and failure probability Q, and trial set t with success probability p and failure probability q. The trials alternate between these two sets until a failure in trial set T ends the game. The score S is the total number of successes in trials from set t.

Define the Probability Generating Function (PGF)

The PGF of a random variable S, G_S(x), is defined as: \[G_S(x) = E[x^S]\]The game continues with alternating trials until a failure occurs in trial set T. If the game starts with a successful T-trial, the PGF can be written as the sum of series encapsulating each possible outcome sequence.

Derive the PGF for S

Given the game rules, the PGF G_S(x) can be derived by considering the alternation of trials and their probabilities. Specifically, if the first T-trial succeeds (with probability P), and then each following t-trial success accumulates. Therefore, \[G_S(x) = \frac{Ppgx + PQx \times PG_S(x)}{Q}\], where the first term is the case where only one t-trial succeeds, and the second term is the recursive part where the game continues. Solving for G_S(x), we find:\[G_S(x) = \frac{Pp}{Q - PQx}\]

Find the Expectation (E[S])

To find the expectation E[S], the first derivative of the PGF G_S(x) at x=1 is required:\[E[S] = G'_S(1)\].Differentiating G_S(x) with respect to x, we get:\[G'_S(1) = \frac{Pp}{(Q-PQ)^{2}} = \frac{Pp}{Q^2}\].

Find the Variance (V[S])

The variance V[S] can be found using the second derivative of the PGF G''_S(x) at x=1 or using the relation:\[V[S] = G''_S(1) + G'_S(1) - (G'_S(1))^2\].Differentiating G''_S(x) with respect to x, and evaluating at x=1, we get:\[V[S] = \frac{Pp(1-PQ)}{Q^2}\].

Understand the Dice Game Setup

For the dice game, two dice, A and B, are rolled alternatively starting with A. If A rolls a 6, the game ends. For B, scores depend on the number rolled: 0 points for odd numbers, 1 point for 2 or 4, and 2 points for a 6. We need to calculate the average and standard deviation of the score.

Define the Expected Score for Dice B

Determining the expected score for dice B requires calculating the weighted average of the points. The probabilities for each face are: P(B=odd) = 1/2, P(B=2 or 4) = 1/3, P(B=6) = 1/6. So, E[Score_B] = 0*(1/2) + 1*(1/3) + 2*(1/6) = 2/6 = 1/3.

Define the Expected Number of B Rolls

The expected number of B rolls is the reciprocal of the probability that A does not roll a 6, which is 5/6. The expected number is 1/(5/6) = 6/5.

Calculate the Average Score

The average score is the product of the expected score for dice B and the expected number of rolls of B: \[E[Score] = E[Score_B] * E[Number_B] = (1/3) * (6/5) = 2/5\].

Calculate the Variance and Standard Deviation of the Score

To find the variance, we use the law of total variance: \[Var(Score) = E[Var(Score | B)] + Var(E[Score | B])\]. We calculate the variance for B and the total formula: \[Var(Score) = 1/3(1-1/3) + (1/3 * 6/5)^2 = 7/15\]. The standard deviation is the square root: \[SD(Score) = \sqrt{7/15}\], which is greater than the average score of 2/5.

Key concepts

Expectation

Expectation, or expected value, is a fundamental concept in probability that represents the average outcome of a random variable over a large number of trials. It is often used to predict the long-term results of random processes. For a discrete random variable X with outcomes \(x_1, x_2, \ldots, x_n\) and probabilities \(p_1, p_2, \ldots, p_n\), the expectation \(E[X]\) is given by:
\[E[X] = \sum_{i=1}^{n} x_i p_i \]
Expectation can describe various types of probability distributions, helping to quantify the center of the distribution. For example, in the context of a binomial distribution where we consider sequences of independent trials, the expectation gives us a sense of how often successes occur in the long run.
It's a key measure for summarizing the performance or outcome of an algorithm, game, or any other stochastic process.

Variance

Variance measures the spread or divergence of a set of values, indicating how far the values lie from the expectation (mean). For a discrete random variable X, with expectation \(E[X]\), the variance \(V[X]\) is defined as:
\[V[X] = E[(X - E[X])^2] \]
This formula signifies that variance is the expected value of the squared difference between the random variable and its mean. Another way of expressing variance is:
\[ V[X] = E[X^2] - (E[X])^2 \]
Variance is crucial in understanding the degree of uncertainty or volatility in the outcomes of the random variable. A higher variance indicates that the data points are more spread out from the mean, whereas a lower variance signifies that they are clustered closely around the mean. In addition to expectation, variance is also key in analyzing stochastic processes, as seen in games involving binomial trials, aiding in the assessment of risk and performance stability.

Binomial Trials

Binomial trials, or Bernoulli trials, are a sequence of independent experiments where each experiment or trial has exactly two possible outcomes: success or failure. Each trial is characterized by:

• Probability of success (P): The likelihood that a single trial results in a success.
• Probability of failure (Q): The likelihood that a single trial results in a failure, with \ Q = 1 - P \.

The number of successes in a fixed number of trials (n) follows a binomial distribution, which is described by the parameters n and P. The probability mass function for x successes in n trials is given by:
\[ P(X = x) = \binom{n}{x} P^x Q^{n-x} \]
Binomial trials are widely used to model scenarios in various fields such as quality control, finance, and sports. For instance, in the textbook exercise, binomial trials are utilized to determine the expected number and variance of successes in alternating trials of two different probability setups.

Probability

Probability is the measure of the likelihood that an event will occur. It quantifies uncertainty and is expressed as a number between 0 and 1, where 0 represents impossibility and 1 represents certainty. Fundamental concepts of probability include:

• Sample Space (S): The set of all possible outcomes of an experiment.
• Event (E): A subset of the sample space S, which can consist of one or more outcomes.
• Probability of an event (P(E)): The ratio of the number of favorable outcomes to the number of possible outcomes in the sample space, for a finite sample space.

Probabilities are used in various methods of statistical analysis. They are fundamental in calculating more complex measures such as expectation and variance. Moreover, understanding probability is essential for reasoning about risk and making decisions in the presence of uncertainty. In the context of binomial trials mentioned in the exercise, probability directly influences the expected number of successes and their variability.