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Chapter 3: Problem 19

Let \(p\) be the probability that the tip points downward after a person throws a drawing pin once. A person throws a drawing pin until it points downward for the first time. Let \(X\) be the number of throws for this to happen. She then throws the drawing pin another \(X\) times. Let \(Y\) be the number of times the drawing pin points downward in the latter series of throws. Find the distribution of \(Y\) (cf. Problem II. \(7.21\) ).

Short Answer

Expert verified
The distribution of \(Y\) is derived using conditional binomial distributions, summing over geometric possibilities.

Step by step solution

01

Define the Random Variable X

The random variable \(X\) represents the number of throws until the drawing pin points downward for the first time. This is the definition of a geometric random variable with parameter \(p\), the probability of success (pointing downward). The probability of \(X\) being equal to \(k\) is given by \(P(X = k) = (1-p)^{k-1} p\) for \(k = 1, 2, 3, \ldots\).
02

Define the Random Variable Y

The random variable \(Y\) represents the number of times the pin points downward in the next \(X\) throws. Given \(X = k\), there will be \(k\) additional throws. Each throw is an independent Bernoulli trial with probability \(p\) of pointing downward (success). Thus, given \(X = k\), \(Y\) follows a Binomial distribution with parameters \(k\) and \(p\), i.e., \(Y | X = k \sim \text{Binomial}(k, p)\).
03

Use the Law of Total Probability

To find the overall distribution of \(Y\), use the law of total probability which involves the sum over all possible values of \(X\). Thus, the probability of \(Y = y\) is given by the sum over the conditional probabilities, weighted by the probability of each \(X\): \(P(Y = y) = \sum_{k=y}^{\infty} P(Y = y | X = k) P(X = k)\).
04

Compute P(Y = y | X = k) and P(X = k)

Using \(Y | X = k \sim \text{Binomial}(k, p)\), we have \(P(Y = y | X = k) = \binom{k}{y} p^y (1-p)^{k-y}\). Also, \(P(X = k) = (1-p)^{k-1} p\).
05

Derive the Distribution of Y

Plugging in the conditional and marginal probabilities, the distribution of \(Y\) is:\[P(Y = y) = \sum_{k=y}^{\infty} \binom{k}{y} p^y (1-p)^{k-y} (1-p)^{k-1} p = p^{y+1} \sum_{k=y}^{\infty} \binom{k}{y} (1-p)^{2k-y-1}.\]To solve this sum requires recognizing a negative binomial form or additional transformations, indicating an aggregated probabilistic pattern dependent on these parameters.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a fundamental probability distribution used to model the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. It's denoted as \( \text{Binomial}(n, p) \), where \( n \) is the number of trials, and \( p \) is the probability of success in each trial.
  • Example: Flipping a coin 10 times and counting how many times it lands heads up, where head is considered a success, is a binomial distribution scenario. Here, the number of trials \( n = 10 \) and the probability of flipping heads \( p = 0.5 \).
  • The probability mass function (PMF) for a binomially distributed random variable \( Y \) is given by:
    \[P(Y = y) = \binom{n}{y} p^y (1-p)^{n-y},\]where \( y \) is the number of successes. \( \binom{n}{y} \) is the binomial coefficient, which counts the number of ways to choose \( y \) successes out of \( n \) trials.
In the context of geometric distribution, if after \( X \) trials you perform another series of trials, \( Y \) follows a binomial distribution. That鈥檚 why, given \( X = k \), \( Y | X = k \sim \text{Binomial}(k, p) \). This reflects how many times the pin points downward in the additional throws.
Law of Total Probability
The law of total probability provides a way to calculate the probability of an event based on a series of conditional probabilities. It's incredibly useful when dealing with composite events that can be broken down into simpler components.
  • Application: If you know the probabilities of an event occurring under several mutually exclusive scenarios, you can combine these probabilities to find the overall probability. In this exercise, it's used to find the distribution of \( Y \) by considering all possible values of \( X \).
  • The law is formally expressed as:
    \[P(A) = \sum_{i} P(A | B_i) P(B_i),\]where \( A \) is the event of interest, \( B_i \) are mutually exclusive events that cover all possibilities, and \( P(A | B_i) \) are the conditional probabilities of \( A \) given each \( B_i \).
In the given problem, using the law involves summing over all feasible outcomes of \( X \) to derive the probability distribution of \( Y \). It aggregates information systematically, yielding a comprehensive solution.
Random Variables
Random variables are a core concept in probability and statistics, representing outcomes of stochastic processes. They can take on different values depending on the result of a random process.
  • Discrete Random Variables: These can take on specific values, often integers, such as the number of heads when flipping a coin multiple times. \( X \) and \( Y \) in this problem are both discrete random variables.
  • Geometric Random Variable: This is a specific type of discrete random variable. It describes the number of trials needed to get the first success in a series of independent Bernoulli trials, characterized by the geometric distribution. For \( X \), each trial (throw of a drawing pin) is independent of the others.
Identifying and understanding the behavior of different random variables, like \( X \sim \text{Geometric}(p) \) and \( Y | X = k \sim \text{Binomial}(k, p) \), is crucial to solving complex probability distribution problems.
Independent Bernoulli Trials
Independent Bernoulli trials form the basis of geometric and binomial distributions. They are a series of experiments where each trial results in a success or failure, and each trial is independent of others.
  • Independence: The outcome of one trial does not affect the outcome of another. This is key in calculating probabilities as it simplifies the modeling of such trials.
  • Bernoulli Trial: Each trial has exactly two outcomes: success with probability \( p \), or failure with probability \( 1-p \). This is the simplest possible random experiment.
By considering independent Bernoulli trials, you can effectively model scenarios like drawing pin throws in the original problem. These trials underpin the geometric distribution of \( X \) and the conditional binomial distribution of \( Y \), ensuring both depend only on their respective parameters without interference from prior outcomes.

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Most popular questions from this chapter

Suppose that the offspring distribution in a branching process is the \(\operatorname{Ge}(p)\)-distribution, and let \(X(n)\) be the number of individuals in generation \(n, n=0,1,2, \ldots .\)Suppose that the offspring distribution in a branching process is the \(\operatorname{Ge}(p)\)-distribution, and let \(X(n)\) be the number of individuals in generation \(n, n=0,1,2, \ldots .\)(a) What is the probability of extinction? Now suppose that \(p=\frac{1}{2}\), and set \(g_{n}(t)=g_{X(n)}(t)\). (b) Show that $$ g_{n}(t)=\frac{n-(n-1) t}{n+1-n t}, \quad n=1,2, \ldots $$ (c) Show that $$ P(X(n)=k)= \begin{cases}\frac{n}{n+1}, & \text { for } k=0, \\\ \frac{n^{k-1}}{(n+1)^{k+1}}, & \text { for } k=1,2, \ldots .\end{cases} $$ (d) Show that $$ P(X(n)=k \mid X(n)>0)=\frac{1}{n+1}\left(\frac{n}{n+1}\right)^{k-1}, \text { for } k=1,2, \ldots, $$ that is, show that the number of individuals in generation \(n\), given that the population is not yet extinct, follows an \(\mathrm{Fs}\left(\frac{1}{n+1}\right)\)-distribution. Suppose the population becomes extinct at generation number \(N\). (e) Show that $$ P(N=n)=g_{n-1}\left(\frac{1}{2}\right)-g_{n-1}(0), \quad n=1,2, \ldots $$ (f) Show that \(P(N=n)=\frac{1}{n(n+1)}, \quad n=1,2, \ldots\) (and hence that \(P(N<\infty)=1\), i.e., \(\eta=1\) ). (g) Compute \(E N\). Why is this a reasonable answer?

A lazy person collects his mail once a week; in the afternoon. Every Sunday he decides, by throwing a regular die, which day (Monday, Tuesday, ..., Saturday) he is going to collect his mail the following week. The number of letters he obtains each weekday follows a Po(1)-distribution, and the number of letters obtained different days are independent. Find the expected value and variance of the number of letters in the mailbox when he comes collecting. Remark. Note that the mailbox may contain letters from the previous week.

Let \(\\{X(t), t \geq 0\\}\) be a family of random variables, and let \(T\) be a nonnegative random variable with density \(f_{T}(t)\), which is independent of \(\\{X(t), t \geq 0\\}\). Furthermore, let \(\varphi_{X(t)}(u)=\varphi(t, u)\) be the characteristic function of \(X(t), t \geq 0 .\) Show that $$ \varphi_{X(T)}(u)=\int_{0}^{\infty} \varphi(t, u) f_{T}(t) d t, \quad-\infty

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Consider a branching process with a \(\operatorname{Po}(m)\)-distributed offspring. Let \(X(1)\) and \(X(2)\) be the number of individuals in generations 1 and 2 , respectively. Determine the generating function of (a) \(X(1)\), (b) \(X(2)\), (c) \(X(1)+X(2)\), and (d) Determine \(\operatorname{Cov}(X(1), X(2))\).
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