/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 If \(Y\) has a geometric distrib... [FREE SOLUTION] | 91Ó°ÊÓ

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If \(Y\) has a geometric distribution with success probability \(p,\) show that $$P(Y=\text { an odd integer })=\frac{p}{1-q^{2}}$$.

Short Answer

Expert verified
The probability that \( Y \) is an odd integer is \( \frac{p}{1-q^2} \).

Step by step solution

01

Understanding the Geometric Distribution

The geometric distribution models the number of Bernoulli trials needed to get the first success. The probability mass function is given by \( P(Y = k) = (1-p)^{k-1} p \), where \( k \) is the number of trials needed to get the first success and \( p \) is the probability of success on each trial.
02

Express Probability for Odd Integer Outcomes

Odd integers can be represented as \( Y = 2n - 1 \) for \( n = 1, 2, 3, ... \). The probability that \( Y = 2n - 1 \) is \( P(Y = 2n - 1) = (1-p)^{2n-2} p \).
03

Sum the Series for Odd Integer Outcomes

Consider the series \( \sum_{n=1}^{\infty} (1-p)^{2n-2} p \). Factor out \( p \), leading to \( p \sum_{n=1}^{\infty} q^{2n-2} \), where \( q = 1-p \). This simplifies to \( p \sum_{n=0}^{\infty} q^{2n} \) with \( n-1 = 0 \), a geometric series.
04

Find the Sum of the Infinite Geometric Series

The sum of the infinite geometric series \( \sum_{n=0}^{\infty} q^{2n} \) is given by \( \frac{1}{1-q^2} \), since it is a geometric series with first term 1 and common ratio \( q^2 \).
05

Combine Results

Substitute the series sum back to obtain the probability: \( p \times \frac{1}{1-q^2} = \frac{p}{1-q^2} \). Therefore, the probability that \( Y \) is an odd integer is \( \frac{p}{1-q^2} \).

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Key Concepts

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

Probability Mass Function
In statistics, the Probability Mass Function (PMF) is a crucial concept, especially when dealing with discrete random variables like those found in a geometric distribution. It provides the probabilities of different possible outcomes. For a geometrically distributed variable, the formula for the PMF is given by:
  • \( P(Y = k) = (1-p)^{k-1} p \)
This equation tells us that for a random variable \( Y \), which follows a geometric distribution, the probability that \( Y \) takes a specific value \( k \) is a product of two components:
  • \( (1-p)^{k-1} \): Represents the probability of having \( k-1 \) failures before the first success.
  • \( p \): Represents the probability of success on the \( k^{th} \) trial.
The parameter \( p \) denotes the success probability in each trial, while \( 1-p \) is the failure probability. Understanding the PMF helps in recognizing patterns in data and predicting probabilities of different outcomes in a sequence of Bernoulli trials.
Infinite Series
An infinite series is the sum of an infinite sequence of numbers. It is a foundational topic in calculus and has expansive applications in mathematics. In our geometric distribution example, dealing with odd integer outcomes involves summing probabilities over infinitely many terms. The series in question is:
  • \( \sum_{n=1}^{\infty} (1-p)^{2n-2} p \)
By understanding infinite series, we realize that even though individual probabilities for each odd integer diminish, the sum of these probabilities can converge to a specific value. In this case, we transform the series so that it becomes a more manageable geometric series.
The art of working with infinite series involves recognizing patterns and simplifying expressions. This helps in finding closed-form solutions, which are essential for determining summed probabilities succinctly.
Geometric Series
A geometric series is an infinite series in which each term after the first is formed by multiplying the previous one by a fixed, non-zero number called the common ratio. The general form of a geometric series is:
  • \( a + ar + ar^2 + ar^3 + \ldots \)
  • Sum \( = \frac{a}{1-r} \), for \( |r| < 1 \)
In our situation, after recognizing the pattern of terms \( q^{2n-2} \) in the series, it matches the template of a geometric series with:
  • First term \( a = 1 \) (once adjusted for \( n = 0 \))
  • Common ratio \( r = q^2 \)
This allows us to sum it easily as \( \frac{1}{1-q^2} \), providing the critical element needed to solve our original probability problem. Geometric series simplify the calculation by reducing complex infinite sums to straightforward formulas.

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Most popular questions from this chapter

Let \(Y\) denote a geometric random variable with probability of success \(p\) a. Show that for a positive integer \(a\) $$P(Y>a)=q^{a}$$, b. Show that for positive integers \(a\) and \(b\) ,$$P(Y>a+b | Y>a)=q^{b}=P(Y>b)$$ This result implies that, for example, \(P(Y>7 | Y>2)=P(Y>5) .\) Why do you think this property is called the memoryless property of the geometric distribution? c. In the development of the distribution of the geometric random variable, we assumed that the experiment consisted of conducting identical and independent trials until the first success was observed. In light of these assumptions, why is the result in part (b) "obvious"?

How many times would you expect to toss a balanced coin in order to obtain the first head?

A fire-detection device utilizes three temperature-sensitive cells acting independently of each other in such a manner that any one or more may activate the alarm. Each cell possesses a probability of \(p=.8\) of activating the alarm when the temperature reaches \(100^{\circ}\) Celsius or more. Let Y equal the number of cells activating the alarm when the temperature reaches \(100^{\circ}\). a. Find the probability distribution for \(Y\). b. Find the probability that the alarm will function when the temperature reaches \(100^{\circ}\).

This exercise demonstrates that, in general, the results provided by Tchebysheff's theorem cannot be improved upon. Let \(Y\) be a random variable such that $$p(-1)=\frac{1}{18}, \quad p(0)=\frac{16}{18}, \quad p(1)=\frac{1}{18}$$ a. Show that \(E(Y)=0\) and \(V(Y)=1 / 9\) b. Use the probability distribution of \(Y\) to calculate \(P(|Y-\mu| \geq 3 \sigma) .\) Compare this exact probability with the upper bound provided by Tchebysheff's theorem to see that the bound provided by Tchebysheff's theorem is actually attained when \(k=3\) *.c. In part.(b) we guaranteed \(E(Y)=0\) by placing all probability mass on the values \(-1,0,\) and1, with \(p(-1)=p(1) .\) The variance was controlled by the probabilities assigned to \(p(-1)\) and \(p(1) .\) Using this same basic idea, construct a probability distribution for a random variable \(X\) that will yield \(P\left(\left|X-\mu_{X}\right| \geq 2 \sigma_{X}\right)=1 / 4\) * d. If any \(k>1\) is specified, how can a random variable \(W\) be constructed so that \(P\left(\left|W-\mu_{W}\right| \geq k \sigma_{W}\right)=1 / k^{2} ?\)

Suppose that \(Y\) is a random variable with moment-generating function \(m(t)\). a. What is \(m(0) ?\) b. If \(W=3 Y,\) show that the moment-generating function of \(W\) is \(m(3 t)\) c. If \(X=Y-2,\) show that the moment-generating function of \(X\) is \(e^{-2 t} m(t)\)

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