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Chapter 6: Problem 28

A series of objects passes a checkpoint. Each object has (independently) probability \(p\) of being defective, and probability \(\alpha\) of being subjected to a check which infallibly detects a defect if it is present. Let \(N\) be the number of objects passing the checkpoint before the first defective is detected, and let \(D\) be the number of these passed objects that were defective (but undetected). Find: (a) The joint p.g.f. of \(D\) and \(N\). (b) \(\mathbf{E}(D / N)\). If the check is not infallible, but errs with probability \(\delta\), find the above two quantities in this case.

Short Answer

Expert verified
(a) Use generating functions; (b) Conditional expectation using modified probability for non-infallible checks.

Step by step solution

01

Define Requirements

Determine the probability distribution that needs to be found. We focus on deriving the joint probability generating function (p.g.f.) of the number of undetected defectives, \(D\), and the number of objects \(N\) before the first defective is detected.
02

Probability of Detected and Undetected Events

Consider that any object that is defective is detected with probability \( \alpha \), and hence undetected with probability \(1 - \alpha\). If an object is not defective, it is not detected and contributes to \(N\).
03

Joint Distribution Function

The joint probability of \(D = d\) and \(N = n\) is characterized by the event that the first \(n-1\) objects are either non-defective or undetected defectives, and the \(n\)-th object is a defective that gets detected. This can be expressed using generating functions.
04

Derive p.g.f. of D and N

The generating function can be expressed as a product of geometric series found by conditioning on the first defective: \[ G(t, u) = \sum_{n=1}^{\infty} \sum_{d=0}^{n-1} \Bigg[ \binom{n-1}{d} p^d (1-p)^{n-1-d} (1-\alpha)^d \alpha \cdot t^d \cdot u^n \Bigg] \]
05

Expectation of D given N

Using the derived generating function, apply differentiation to find the expectation: \[ \mathbf{E}(D|N = n) = \sum_{d=0}^{n-1} d \cdot \Pr(D = d | N = n) \]
06

Adjust for Non-infallible Checks

When the checks err with probability \(\delta\), modify the detection probabilities. The effective detection rate becomes \(\alpha(1-\delta)\), impacting both the joint p.g.f and the conditional expectation.
07

Result Application

Use the results step to interpret your findings and answer any specific questions from the exercise.

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Key Concepts

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

Joint Distribution Function
In probability theory, a joint distribution describes the probability of two random variables occurring together. For the exercise we consider, the variables are the number of undetected defectives, \(D\), and the number of objects, \(N\), passing the checkpoint before the first defective is detected. The joint distribution is represented using their probability generating function (p.g.f.), which is a mathematical tool used to encode probabilities.

The joint p.g.f. in this context is expressed as:\[G(t, u) = \sum_{n=1}^{\infty} \sum_{d=0}^{n-1} \binom{n-1}{d} p^d (1-p)^{n-1-d} (1-\alpha)^d \alpha \cdot t^d \cdot u^n\]
  • \(t\) is associated with the undetected defectives, \(D\).
  • \(u\) relates to the number of objects, \(N\), before the detection of a defective.
  • The series accounts for all possible ways of having \(d\) undetected defects among \(n\) objects, finishing with a detected defective item.
This function captures the probabilities of all combinations of \(D\) and \(N\) as they evolve through the process.
Expectation
Expectation in probability measures the average or expected outcome of a random variable. It gives us an idea about the typical behavior of a variable over numerous trials. In our exercise, we are particularly interested in \(\mathbf{E}(D|N)\), the expected number of undetected defectives given the number of objects that have passed the checkpoint before a defective is detected.

To find \(\mathbf{E}(D|N = n)\), we derive it from the joint p.g.f. using differentiation:\[\mathbf{E}(D|N = n) = \sum_{d=0}^{n-1} d \cdot \Pr(D = d | N = n)\]
  • This calculation requires understanding the distribution of \(D\) conditioned on \(N\).
  • It reveals how many defective items tend to pass undetected, relying on \(N\).
This insight helps in characterizing the reliability of the detection mechanism.
Conditional Probability
Conditional probability is the probability of one event occurring with some relationship to one or more other events. In this context, it describes how the probability of \(D\) being a certain value is affected when the value of \(N\) is known.

The concept is essential for understanding how the number of defectives that remain undetected relates to other factors, such as the number of items \(N\) passing the checkpoint. The relationship is expressed as:\[\Pr(D = d | N = n)\]
  • This formula provides the probability of \(D\) taking on a specific value given \(N = n\).
  • It changes the perspective from general probability to a more localized measure, which can be critical in applications like quality control.
The full analysis of these conditional probabilities demands careful interpretation of both statistical and real-world implications, particularly when detection isn't infallible.

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Most popular questions from this chapter

A gambler repeatedly plays the game of guessing whether a fair coin will fall heads or tails when tossed. For each correct prediction he wins \(拢 1\), and for each wrong one he loses \(拢 1\). At the start of play, he holds \(拢 n\) (where \(n\) is a positive integer), and he has decided to stop play as soon as either (i) he has lost all his money, or (ii) he possesses \(拢 K\), where \(K\) is a given integer greater than \(n\). Let \(p(n)\) denote for \(1 \leq n \leq K-1\) the probability that he loses all his money, and let \(p(0)=1, p(K)=0\). Show that \(p(n)=\frac{1}{2}(p(n-1)+p(n+1)) ;(1 \leq n \leq K-1)\). $$ G(s)=\sum_{n=0}^{k-1} p(n) s^{n} $$ then, provided \(s \neq 1\), $$ G(s)=\frac{1}{(1-s)^{2}}\left(1-(2-p(1)) s+p(K-1) s^{K+1}\right) . $$ Hence, or otherwise, show that \(p(1)=1-1 / K, p(K-1)=1 / K\) and that, in general, \(p(n)=\) \(1-n / K\).

Each packet of a certain breakfast cereal contains one token, coloured either red, blue, or green. The coloured tokens are distributed randomly among the packets, each colour being equally likely. Let \(X\) be the random variable that takes the value \(j\) when I find my first red token in the \(j\) th packet which I open. Obtain the probability generating function of \(X\), and hence find its expectation. More generally, suppose that there are tokens of \(m\) different colours, all equally likely. Let \(Y\) be the random variable that takes the value \(j\) when I first obtain a full set, of at least one token of each colour, when I open my \(j\) th packet. Find the generating function of \(Y\), and show that its expectation is \(m\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{m}\right)\).

Show that for \(\alpha>0, \beta>0, \alpha+\beta<1\), $$ G(s, t)=\frac{\log (1-\alpha s-\beta t)}{\log (1-\alpha-\beta)} $$ is a bivariate p.g.f. Find the marginal p.g.f.s and the covariance.

Let the number of tosses required for a fair coin to show a head be \(T\). An integer \(X\) is picked at random from \(\\{1, \ldots, T\\}\) with equal probability \(\frac{1}{T}\) of picking any one. Find \(G_{X}(s)\).

Define the probability generating function of an integer valued random variable \(X\), and show how it may be used to obtain the mean \(\mu_{X}\), variance \(\sigma_{X}^{2}\), and third moment about the mean \(\gamma_{X}\). (a) Let \(Y=\sum_{i=1}^{N} X_{i}\), where the \(X_{i}\) are independent integer valued random variables identically distributed as \(X\). Let \(\mu_{X}=0\), and let \(N\) be an integer valued random variable distributed independently of the \(X_{i}\). Show that \(\sigma_{Y}^{2}=\mu_{N} \sigma_{X}^{2}\), and \(\gamma_{Y}=\mu_{N} \gamma_{X}\). (b) Find \(\sigma_{Y}^{2}\) when \(\mu_{X} \neq 0\).
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