To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

Question

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Answer 1

Question

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If $i=2b−1$ then it shows an overflow error. Otherwise, it increases the counter by 1 with probability $1/(ni+1−ni)$ and leaves the counter with probability $1−1/(ni+1−ni)$ .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from $ni$ to $ni+1$ with a probability of $1ni+1−ni$ and leaving the value unchanged.

The expected increase is calculated as,

$ni+1−nini+1−ni=1$

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: $ni=100i$ for all $i≥0$ and variance of the counter depends on the $ni$ .

Explanation:

Consider that $Vn$ and $Xj$ are the variant and pairwise independent events. The variance $Vn$ is calculated as,

$Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]$

Since $ni=100i$ and $ni+1−ni=100(i+1)−100i=100$ . Therefore, the value represented by the counter with probability $99/100$ due to the jth INCREMENT is 0 and with the probability of $1/100$ .

The value represented increases by 100. Therefore, by the equation (C.27)

$Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99$

Now, adding the variances of the $Xj$ gives $Var[Vn]=99n$ .

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Answer 2

Question

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If $i=2b−1$ then it shows an overflow error. Otherwise, it increases the counter by 1 with probability $1/(ni+1−ni)$ and leaves the counter with probability $1−1/(ni+1−ni)$ .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from $ni$ to $ni+1$ with a probability of $1ni+1−ni$ and leaving the value unchanged.

The expected increase is calculated as,

$ni+1−nini+1−ni=1$

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: $ni=100i$ for all $i≥0$ and variance of the counter depends on the $ni$ .

Explanation:

Consider that $Vn$ and $Xj$ are the variant and pairwise independent events. The variance $Vn$ is calculated as,

$Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]$

Since $ni=100i$ and $ni+1−ni=100(i+1)−100i=100$ . Therefore, the value represented by the counter with probability $99/100$ due to the jth INCREMENT is 0 and with the probability of $1/100$ .

The value represented increases by 100. Therefore, by the equation (C.27)

$Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99$

Now, adding the variances of the $Xj$ gives $Var[Vn]=99n$ .

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Answer 3

Question

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If $i=2b−1$ then it shows an overflow error. Otherwise, it increases the counter by 1 with probability $1/(ni+1−ni)$ and leaves the counter with probability $1−1/(ni+1−ni)$ .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from $ni$ to $ni+1$ with a probability of $1ni+1−ni$ and leaving the value unchanged.

The expected increase is calculated as,

$ni+1−nini+1−ni=1$

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: $ni=100i$ for all $i≥0$ and variance of the counter depends on the $ni$ .

Explanation:

Consider that $Vn$ and $Xj$ are the variant and pairwise independent events. The variance $Vn$ is calculated as,

$Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]$

Since $ni=100i$ and $ni+1−ni=100(i+1)−100i=100$ . Therefore, the value represented by the counter with probability $99/100$ due to the jth INCREMENT is 0 and with the probability of $1/100$ .

The value represented increases by 100. Therefore, by the equation (C.27)

$Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99$

Now, adding the variances of the $Xj$ gives $Var[Vn]=99n$ .

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Answer 4

Question

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If $i=2b−1$ then it shows an overflow error. Otherwise, it increases the counter by 1 with probability $1/(ni+1−ni)$ and leaves the counter with probability $1−1/(ni+1−ni)$ .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from $ni$ to $ni+1$ with a probability of $1ni+1−ni$ and leaving the value unchanged.

The expected increase is calculated as,

$ni+1−nini+1−ni=1$

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: $ni=100i$ for all $i≥0$ and variance of the counter depends on the $ni$ .

Explanation:

Consider that $Vn$ and $Xj$ are the variant and pairwise independent events. The variance $Vn$ is calculated as,

$Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]$

Since $ni=100i$ and $ni+1−ni=100(i+1)−100i=100$ . Therefore, the value represented by the counter with probability $99/100$ due to the jth INCREMENT is 0 and with the probability of $1/100$ .

The value represented increases by 100. Therefore, by the equation (C.27)

$Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99$

Now, adding the variances of the $Xj$ gives $Var[Vn]=99n$ .

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Answer 5

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If $i=2b−1$ then it shows an overflow error. Otherwise, it increases the counter by 1 with probability $1/(ni+1−ni)$ and leaves the counter with probability $1−1/(ni+1−ni)$ .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from $ni$ to $ni+1$ with a probability of $1ni+1−ni$ and leaving the value unchanged.

The expected increase is calculated as,

$ni+1−nini+1−ni=1$

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: $ni=100i$ for all $i≥0$ and variance of the counter depends on the $ni$ .

Explanation:

Consider that $Vn$ and $Xj$ are the variant and pairwise independent events. The variance $Vn$ is calculated as,

$Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]$

Since $ni=100i$ and $ni+1−ni=100(i+1)−100i=100$ . Therefore, the value represented by the counter with probability $99/100$ due to the jth INCREMENT is 0 and with the probability of $1/100$ .

The value represented increases by 100. Therefore, by the equation (C.27)

$Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99$

Now, adding the variances of the $Xj$ gives $Var[Vn]=99n$ .

Answer 6

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If $i=2b−1$ then it shows an overflow error. Otherwise, it increases the counter by 1 with probability $1/(ni+1−ni)$ and leaves the counter with probability $1−1/(ni+1−ni)$ .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from $ni$ to $ni+1$ with a probability of $1ni+1−ni$ and leaving the value unchanged.

The expected increase is calculated as,

$ni+1−nini+1−ni=1$

Hence, the expected increment represented by the counter is 1.

Answer 7

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If $i=2b−1$ then it shows an overflow error. Otherwise, it increases the counter by 1 with probability $1/(ni+1−ni)$ and leaves the counter with probability $1−1/(ni+1−ni)$ .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from $ni$ to $ni+1$ with a probability of $1ni+1−ni$ and leaving the value unchanged.

The expected increase is calculated as,

$ni+1−nini+1−ni=1$

Hence, the expected increment represented by the counter is 1.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: $ni=100i$ for all $i≥0$ and variance of the counter depends on the $ni$ .

Explanation:

Consider that $Vn$ and $Xj$ are the variant and pairwise independent events. The variance $Vn$ is calculated as,

$Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]$

Since $ni=100i$ and $ni+1−ni=100(i+1)−100i=100$ . Therefore, the value represented by the counter with probability $99/100$ due to the jth INCREMENT is 0 and with the probability of $1/100$ .

The value represented increases by 100. Therefore, by the equation (C.27)

$Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99$

Now, adding the variances of the $Xj$ gives $Var[Vn]=99n$ .

Answer 13

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: $ni=100i$ for all $i≥0$ and variance of the counter depends on the $ni$ .

Explanation:

Consider that $Vn$ and $Xj$ are the variant and pairwise independent events. The variance $Vn$ is calculated as,

$Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]$

Since $ni=100i$ and $ni+1−ni=100(i+1)−100i=100$ . Therefore, the value represented by the counter with probability $99/100$ due to the jth INCREMENT is 0 and with the probability of $1/100$ .