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Chapter 3: Problem 31

Each element in a sequence of binary data is either 1 with probability \(p\) or 0 with probability \(1-p\). A maximal subsequence of consecutive values having identical outcomes is called a run. For instance, if the outcome sequence is \(1,1,0,1,1,1,0\), the first run is of length 2, the second is of length 1, and the third is of length 3 . (a) Find the expected length of the first run. (b) Find the expected length of the second run.

Short Answer

Expert verified
(a) The expected length of the first run: \(E(R_1) = \sum_{n=1}^{\infty}{n[p^n(1-p) + (1-p)^np]}\) (b) The expected length of the second run: \(E(R_2) = \sum_{k=1}^{\infty}{\sum_{n_2=1}^{\infty}{n_2[P(n_2|R_1=k)P(k)]}}\)

Step by step solution

01

Find the probabilities of a run starting with 1 or 0.

Since the probabilities of the elements being 1 or 0 are given, we know that the probability of a run starting with 1 is \(p\) and a run starting with 0 is \(1-p\).
02

Find probabilities of different lengths of the first run and calculate the expected length.

The probability of the first run having length 1 is the probability that the sequence starts with 1 and the next element is 0, which is \(p(1-p)\), or starts with 0 and the next element is 1, which is \((1-p)p\). In both cases, the probability of having a run of length 1 is \(2p(1-p)\). The probability of the first run having length 2 is the probability that the sequence starts with two 1s and the next element is 0, which is \(p^2(1-p)\), or starts with two 0s and the next element is 1, which is \((1-p)^2p\). In both cases, the probability of having a run of length 2 is \(p^2(1-p) + (1-p)^2p\). Similarly, we can find the probabilities of runs of any length n. Let's denote the probability of a run of length n by P(n), then: \(P(n) = p^n(1-p) + (1-p)^np\). Now, we can calculate the expected length of the first run as: \(E(R_1) = \sum_{n=1}^{\infty}{nP(n)} = \sum_{n=1}^{\infty}{n[p^n(1-p) + (1-p)^np]}\) (a)
03

Result:

The expected length of the first run: \(E(R_1) = \sum_{n=1}^{\infty}{n[p^n(1-p) + (1-p)^np]}\)
04

Find probabilities of different lengths of the second run and calculate the expected length.

Now, let's deal with the second run. We will assume the first run has a length k. If the first run consists of 1s, the probability of the second run having length 1 is \((1-p)p\), and likewise, if the first run consists of 0s, the probability of the second run having length 1 is \(p(1-p)\). Thus, the probability of the second run having a length 1 given the length of the first run is 2p(1-p). Moreover, we can calculate the conditional probability as: \(P(n_2|R_1=k) = p^{n_2}(1-p) + (1-p)^{n_2}p\) Then, we can calculate the expected length of the second run as: \(E(R_2) = \sum_{k=1}^{\infty}{\sum_{n_2=1}^{\infty}{n_2[P(n_2|R_1=k)P(k)]}}\) (b)
05

Result:

The expected length of the second run: \(E(R_2) = \sum_{k=1}^{\infty}{\sum_{n_2=1}^{\infty}{n_2[P(n_2|R_1=k)P(k)]}}\) Note: In both cases, the calculations are infinite series, but can be computed numerically for specific values of \(p\).

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Key Concepts

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

Expectation
Expectation in probability theory is used to calculate the average or expected value of a random process or variable in the long run. In our problem, we're dealing with runs in a binary sequence, which are groups of consecutive identical elements. Since these runs can have different lengths, we use expectation to find out how long they are on average.

To calculate the expectation of the first run's length, you sum up each possible length multiplied by its probability. This looks like:
  • Multiply each run length by its probability.
  • Sum this product over all possible run lengths.
The formula for the expected run length is written as:\[ E(R_1) = \sum_{n=1}^{\infty}{n\left[p^n(1-p) + (1-p)^np\right]} \]This formula is similar for calculating the expectation of the second run, but it involves an additional consideration of the first run's length.
Binary sequence
A binary sequence is a series of numbers consisting only of 0s and 1s. These are fundamental in computer science and binary math due to their simplicity and versatility.

In probability, each number or element in the sequence has a certain probability. Here, every element can be either 1 with a given probability \( p \) or 0 with \( 1-p \). Understanding how these sequences behave, especially as we randomly generate them, helps us tackle many problems, such as predicting outcomes and calculating expectations. In our exercise, these sequences form the basis for determining the lengths of runs.
Run length
Run length in a binary sequence refers to the number of consecutive 0s or 1s. For example, in the sequence \( 1, 1, 0, 1, 1, 1, 0 \), there are runs of lengths 2, 1, and 3.

When studying run lengths, we calculate how often they occur and how long they are. The probability of a specific run length depends on whether the run starts with a 0 or a 1. For instance:
  • A run that begins with \( n \) 1s followed by a 0 has probability \( p^n(1-p) \).
  • A run that starts with \( n \) 0s followed by a 1 has probability \( (1-p)^n p \).
These probabilities help in calculating expectations, as they inform us of the likelihood of each run length occurring.
Consecutive values
Consecutive values in a binary sequence refer to elements that are the same one after another without interruption. They form runs, the fundamental units of our study in this exercise.

For example, sets of consecutive 1s form one run, while consecutive 0s form another. Analyzing these allows us to understand patterns and predict behaviors that are often not random, even in a sequence generated by random processes.

By understanding consecutive values, one can delve deeper into data sequences and uncover insights. Knowing how often consecutive values occur and their frequencies is crucial, especially in data compression and error detection algorithms, where these patterns can significantly impact efficiency and accuracy.

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Most popular questions from this chapter

Data indicate that the number of traffic accidents in Berkeley on a rainy day is a Poisson random variable with mean 9 , whereas on a dry day it is a Poisson random variable with mean \(3 .\) Let \(X\) denote the number of traffic accidents tomorrow. If it will rain tomorrow with probability \(0.6\), find (a) \(E[X]\) (b) \(P\\{X=0\\}\); (c) \(\operatorname{Var}(X)\).

In Section 3.6.3, we saw that if \(U\) is a random variable that is uniform on \((0,1)\) and if, conditional on \(U=p, X\) is binomial with parameters \(n\) and \(p\), then $$ P\\{X=i\\}=\frac{1}{n+1}, \quad i=0,1, \ldots, n $$ For another way of showing this result, let \(U, X_{1}, X_{2}, \ldots, X_{n}\) be independent uniform \((0,1)\) random variables. Define \(X\) by $$ X=\\# i: X_{i}

A set of \(n\) dice is thrown. All those that land on six are put aside, and the others are again thrown. This is repeated until all the dice have landed on six. Let \(N\) denote the number of throws needed. (For instance, suppose that \(n=3\) and that on the initial throw exactly two of the dice land on six. Then the other die will be thrown, and if it lands on six, then \(N=2 .\) Let \(m_{n}=E[N]\). (a) Derive a recursive formula for \(m_{n}\) and use it to calculate \(m_{i}, i=2,3,4\) and to show that \(m_{5} \approx 13.024\). (b) Let \(X_{i}\) denote the number of dice rolled on the \(i\) th throw. Find \(E\left[\sum_{i=1}^{N} X_{i}\right]\)

A gambler wins each game with probability \(p .\) In each of the following cases, determine the expected total number of wins. (a) The gambler will play \(n\) games; if he wins \(X\) of these games, then he will play an additional \(X\) games before stopping. (b) The gambler will play until he wins; if it takes him \(Y\) games to get this win, then he will play an additional \(Y\) games.

Show in the discrete case that if \(X\) and \(Y\) are independent, then $$ E[X \mid Y=y]=E[X] \quad \text { for all } y $$
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