Chapter 8 Probability and Random variables

Chapter 8 Probability and Random variables

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F. G. Stremler, Introduction to Communication Systems 3/e

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• Probability • All possible outcomes (A1 to AN) are included

• Joint probability

• Conditional probability

NNAP A

N lim)(

1)(1

N

iiAP

NNABP AB

N lim)(

)()(

/)|(

APABP

NNNN

NNABP

A

AB

A

AB )()(

/)|(

BPABP

NNNN

NNBAP

B

AB

B

AB

)()|()()|()( BPBAPAPABPABP

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• Bayes’ theorem

• Random 2/52 playing cards. After looking at the first card, P(2nd is heart)=? if 1st is or isn’t heart

A: a heart on the 1st; B: a heart on the 2nd; C: no heart on the 1st

P(B|A) = 12/51; P(B|C) = 13|51• Probability of two mutually exclusive events

P(A+B)=P(A)+P(B)• If the events are not mutually exclusive

P(A+B)=P(A)+P(B)-P(AB)

)()|()()|(

APBAPBPABP

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Random variables• A real valued random variable is a real-value

function defined on the events of the probability system.

• Cumulative distribution function (CDF) of x is

• Properties of F(a)• Nondecreasing, • 0≤F(a)≤1,

)(lim)()(nanaxPaF x

n

1)(0)(

FF

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Probability density function (PDF)

xadaadFxf |)()(

Properties of PDF

.0)( xf

1)()(

Fdxxf

aXPdxaf x )(

)()( axPaF

Tutorial Q.2 • Consider the experiment that consists in the rolling of two

honest dice. The random variable X is assigned to the sum of the numbers showing up to the two dice. Determine and plot the cumulative distribution function (CDF) and the probability distribution function (pdf) of X.

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Discrete and continuous distributions• Discrete: random variable has M discrete values

CDF or F(a) was discontinuous as a increase.Digital communicationsPDF

CDF

events discretely ofnumber theisM

)()()(1

M

iii xxxPxf

MLax

xPaF

L

L

ii

, that suchinteger largest theis L

)()(1

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• Continuous distributions: if a random variable is allowed to take on any value in some interval.

CDF and PDF would be continuous functions.

Analogue communications, noise.• Expected value of a discretely distributed random

variable

)()()]([1

i

M

ii xPxhxhy

Normalized average power

P = i

yi2 p(yi)

exampleA discrete random signal, y(t), can take one of the

four predefined voltage levels, y1 = 0.5 V, y2 = 0.4 V, y3 = 0.2 V, and y4 = 0.1 V. Assume that these levels occur with probabilities, p(y1) = 0.2, p(y2) = 0.3, p(y3) = 0.1, and p(y4) = 0.4.

Calculate the average power delivered by y(t) into a 100Ω resistive load.

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The normalized average power, P, is given by:

P = i

yi2 p(yi)

i.e. (0.5)2 0.2 + (0.4)2 0.3 + (0.2)2 0.1 + (0.1)2 0.4 = 0.106/100 W = 1.06 mW

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Important distributions• Binomial : discrete

• Poisson: discrete

• Uniform: continuous, a random variable that is equally likely to take on any value within a given range.

• Gaussian (normal): continuous• Normalised Gaussian pdf having zero mean and

unit variance

• Sinusoidal: continuous

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2/2

21)( xexp

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