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Chapter 3: Q122E (page 140)

Consider a communication source that transmits packets containing digitized speech. After each transmission, the receiver sends a message indicating whether the transmission was successful or unsuccessful. If a transmission is unsuccessful, the packet is re-sent. Suppose a voice packet can be transmitted a maximum of \({\rm{10}}\) times. Assuming that the results of successive transmissions are independent of one another and that the probability of any particular transmission being successful is \({\rm{p}}\), determine the probability mass function of the rv \({\rm{X = }}\)the number of times a packet is transmitted. Then obtain an expression for the expected number of times a packet is transmitted.

Short Answer

Expert verified

The probability mass function is\({\rm{p(k) = P(X = k) = }}\left\{ {\begin{array}{*{20}{c}}{{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}{\rm{p}}}&{{\rm{k = 1,2,3,4,5,6,7,8,9}}}\\{{\rm{1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p}}}&{{\rm{k = 10}}}\end{array}} \right.\).

The expected number of times a packet is transmitted is \({\rm{\mu = 10 - 9}}\sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p}}\).

Step by step solution

01

Concept Introduction

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

The complement rule is a statistical theorem that establishes a link between the probability of an occurrence and the probability of its complement, such that if one of these probabilities is known, automatically the other is also known.

02

Probability Mass Function

It is given that\({\rm{P(}}\)successful\({\rm{) = p}}\).

A packet can be transmitted at most\({\rm{10}}\)times 鈥揬({\rm{k}} \le {\rm{10}}\).

The transmissions are independent.

Probability Mass Function 鈥�

\({\rm{X = }}\)the number of times a packet is transmitted, thus\({\rm{X}}\)is the number of trials until the first success\({\rm{Y}}\)(successful transmission) occurs up to\({\rm{10}}\)packets.

The number of trials until the first success follows a geometric distribution with probability of success\(p\).

\({\rm{Y}} \sim {\rm{Geometric(p)}}\)

Definition geometric probability 鈥�

\({\rm{P(Y = k) = }}{{\rm{q}}^{{\rm{k - 1}}}}{\rm{p = (1 - p}}{{\rm{)}}^{{\rm{k - 1}}}}{\rm{p}}\)

The definition of geometric probability is for all nonnegative integers, however for \({\rm{X}}\) it is given \({\rm{k}} \le {\rm{10}}\). Use the definition for \({\rm{k = 1, 2, 3, 4, 5, 6, 7, 8, 9}}\).

03

The Complement Rule

The complement rule is represented as 鈥�

\({\rm{P(not A) = 1 - P(A)}}\)

Addition rule for disjoint or mutually exclusive events 鈥�

\({\rm{P(A or B) = P(A) + P(B)}}\)

Use the complement rule and the addition rule 鈥�

\(\begin{array}{c}{\rm{P(X = 10) = P(Y}} \ge {\rm{10) = 1 - P(Y}} \le {\rm{9) = 1 - P(Y = 1) - P(Y = 2) - \ldots - P(Y = 9)}}\\{\rm{ = 1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {\rm{P}} {\rm{(Y = k) = 1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p}}\end{array}\)

Combining these results to obtain the probability distribution 鈥�

\(\begin{array}{l}{\rm{p(k) = P(X = k)}}\\{\rm{ = }}\left\{ {\begin{array}{*{20}{c}}{{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}{\rm{p}}}&{{\rm{k = 1,2,3,4,5,6,7,8,9}}}\\{{\rm{1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p}}}&{{\rm{k = 10}}}\end{array}} \right.\end{array}\)

04

The Expected Value

The expected value (or mean) is the sum of the product of each possibility \({\rm{x}}\) with its probability \({\rm{P(x)}}\)鈥�

\(\begin{aligned}\mu &= \sum\limits_{{\rm{10}}} {\rm{x}} {\rm{P(x)}}\\&= \sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {\rm{P(X = k)}}\\&= \sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p + 10}}\left( {{\rm{1 - }}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p}}} \right)\\&= \sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p + 10 - 10}}\sum\limits_{{\rm{k = 0}}}^{\rm{9}} {{{{\rm{(1 - p)}}}^{{\rm{k - 1}}}}} {\rm{p Distributive Property}}\\&= 10 + (1 - 10)\sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p Combine like terms}}\\&= 10 - 9\sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p}}\end{aligned}\)

Therefore, the expression is obtained as \({\rm{10 - 9}}\sum\limits_{{\rm{k = 1}}}^{\rm{9}} {\rm{k}} {{\rm{(1 - p)}}^{{\rm{k - 1}}}}{\rm{p}}\).

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