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Chapter 3: Problem 71

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"?

Short Answer

Expert verified
The memoryless property arises because geometric trials are independent; past failures don't affect future probabilities.

Step by step solution

01

Understanding the Problem

We need to demonstrate certain properties of a geometric random variable with success probability \( p \). Remember, the geometric random variable \( Y \) gives the number of trials until the first success. We will solve each part step by step.
02

Part a: Calculating \( P(Y > a) \)

The probability \( P(Y > a) \) is the probability of having no success in the first \( a \) trials. Since each trial is independent and the probability of failure in each trial is \( q = 1-p \), the probability of failing \( a \) times is \( q \times q \times \cdots \times q = q^a \). Thus, \[ P(Y > a) = q^a. \]
03

Step b: Calculating\( P(Y > a + b | Y > a) \)

To find \( P(Y > a + b | Y > a) \), we consider the conditional probability definition: \[ P(Y > a + b | Y > a) = \frac{P((Y > a + b) \cap (Y > a))}{P(Y > a)}. \] Since \( (Y > a + b) \subseteq (Y > a) \), this simplifies to: \[ P(Y > a + b | Y > a) = \frac{P(Y > a + b)}{P(Y > a)}. \] We know from Step 2 that \( P(Y > a) = q^a \) and \( P(Y > a + b) = q^{a+b} \), hence: \[ P(Y > a + b | Y > a) = \frac{q^{a+b}}{q^a} = q^b. \] This is equal to \( P(Y > b) \).
04

Part b: Understanding the Memoryless Property

The result \( P(Y > a + b | Y > a) = P(Y > b) \) implies that the probability of success after already having \( a \) failures does not depend on \( a \). The future probability of success behaves as if no previous trials occurred; hence, we say the geometric distribution is memoryless.
05

Part c: Why the Memoryless Property is 'Obvious'?

The assumption underlying a geometric random variable is independent and identical trials until the first success. Each trial is like starting the experiment anew, unaffected by past trials. Therefore, regardless of the number of prior failures, the probability \( P(Y > b) \) remains the same, making the result obvious due to the nature of independent trials.

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Key Concepts

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

Memoryless Property of Geometric Distribution
The memoryless property is one of the most intriguing aspects of the geometric distribution. It suggests that, regardless of any prior failures, the probability of success remains unchanged going forward.
This is a bit like resetting the experiment every time we perform a trial.

This means that once we know a task has already failed a set number of times, the probability of success does not rely on those failures. Formally, this can be stated as:
  • The probability that the next success will occur in more than an additional \( b \) trials given that you have already failed for \( a \) trials is the same as it would be if you were starting anew.
  • Mathematically, this is represented as \( P(Y > a + b \mid Y > a) = P(Y > b) \).
Because of this property, we say the geometric distribution "forgets" past failures, giving it the apt name of the memoryless property.
Understanding Conditional Probability in Geometric Context
Conditional probability allows us to calculate the probability of an event given that another event has occurred. In the context of geometric distribution, it investigates the probability of future success based on already known specifications.

With the geometric distribution, conditional probability examines outcomes based on a subset of the trials:
  • For example, \( P(Y > a + b \mid Y > a) \) is determined by considering only outcomes where \( Y > a \). It tells us about the probability of an additional \( b \) trials without success after already experiencing \( a \) trials without success.
  • This simplifies in the geometric context due to its memoryless nature, meaning the additional \( b \) trials behave the same as if they were the very first trials.
These calculations help us see how the behavior after specific conditions aligns with the properties of a geometric distribution.
Independent Trials: The Basis of Geometric Distribution
The concept of independent trials is crucial for understanding geometric distribution. Each trial is independent, meaning the outcome of one trial does not affect the next.
This feature is foundational to why the geometric distribution exhibits its other properties, including being memoryless.
When dealing with independent trials:
  • The likelihood of success or failure in a trial is unaffected by the results of any prior trials. This means each trial acts as if it's completely fresh and unaffected by preceding ones.
  • In terms of the geometric distribution, it emphasizes that each trial leading to the first success is separate from the previous trials until that success occurs.
This separation of each trial helps in predicting and interpreting results without any bias introduced from previous outcomes, offering a pure statistical perspective.

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Most popular questions from this chapter

Let \(Y\) have a hypergeometric distribution $$p(y)=\frac{\left(\begin{array}{l} r \\ y \end{array}\right)\left(\begin{array}{l} N-r \\ n-y \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)} \quad y=0,1,2, \ldots, n$$ a. Show that $$P(Y=n)=p(n)=\left(\frac{r}{N}\right)\left(\frac{r-1}{N-1}\right)\left(\frac{r-2}{N-2}\right) \cdots\left(\frac{r-n+1}{N-n+1}\right)$$ b. Write \(p(y)\) as \(p(y | r) .\) Show that if \(r_{1}\frac{p\left(y+1 | r_{1}\right)}{p\left(y+1 | r_{2}\right)}$$ c. Apply the binomial expansion to each factor in the following equation: $$(1+a)^{N_{1}}(1+a)^{N_{2}}=(1+a)^{N_{1}+N_{2}}$$ Now compare the coefficients of \(a^{n}\) on both sides to prove that $$\left(\begin{array}{c} N_{1} \\ 0 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n \end{array}\right)+\left(\begin{array}{c} N_{1} \\ 1 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n-1 \end{array}\right)+\cdots+\left(\begin{array}{c} N_{1} \\ n \end{array}\right)\left(\begin{array}{c} N_{2} \\ 0 \end{array}\right)=\left(\begin{array}{c} N_{1}+N_{2} \\ n \end{array}\right)$$ d. Using the result of part ( \(\underline{\mathrm{c}}\) ), conclude that $$\sum_{y=0}^{n} p(y)=1$$

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