#### Transcript Ch8 - Department of Engineering and Physics

```ENGR 4323/5323
Digital and Analog Communication
Ch 8
Fundamentals of Probability Theory
Engineering and Physics
University of Central Oklahoma
Dr. Mohamed Bingabr
Chapter Outline
• Concept of Probability
• Random Variables
• Statistical Averages (MEANS)
• Correlation
• Linear Mean Square Estimation
• Sum of Random Variables
• Central Limit Theorem
2
Deterministic and Random Signals
Deterministic Signals: Signals that can be determined by
mathematical equation or graph. It is possible to predict the
future values with 100% certainty.
Random Process Signals: Unpredictable message signals and
noise waveform. These type of signal are information-bearing
signals and they play key roles in communications.
3
Concept of Probability
Experiment: In probability theory an experiment is a process
whose outcome cannot be fully predicted. (Throwing a die)
Sample space: A set that contain all possible outcomes of an
experiment. {1, 2, 3, 4, 5, 6}
Sample point (element): an outcome of an experiment. {3}
Event: A subset of the sample space that share some common
characteristics. {2, 4, 6} even number
Complement of event A (Ac): Event containing all points not in
A. {1, 3, 5}
4
Concept of Probability
Null event (ø): Event that has no sample point.
Union of events A and B (A U B): The event that contains all
points A and B.
Intersection (joint) of events A and B (A ∩ B, AB): The event
that contain all points common to event A and B.
Mutually Exclusive: Events A and B are mutually exclusive if A
occur then B can not occur.
Relative frequency and Probability: If event A is of interest and
an experiment is conducted N times then the relative
frequency of A occurrence (probability) is
𝑁(𝐴)
𝑓 𝐴 = lim
= 𝑃(𝐴)
𝑁→∞ 𝑁
5
Concept of Probability
S
A
B
6
Concept of Probability
Joint Probability:
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)
S
A
B
If A and B are mutually exclusive A ∩ B = ø then
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃(𝐵)
Conditional Probability: the probability of one event is
influenced by the outcome of another event.
𝑃(𝐴 ∩ 𝐵)
𝑃 𝐴/𝐵 =
𝑃(𝐵)
Independent Events: The occurrence of one event is not
influenced by the occurrence of the other event.
𝑃 𝐴∩𝐵 =𝑃 𝐴 𝑃 𝐵
𝑃 𝐴/𝐵 = 𝑃(𝐴)
7
Bernoulli Trials
Bernoulli trial is an experiment where there are two possible
outcomes, success or failure. If the probability of success is p
then the probability of failure is (1-p).
𝑝(𝑘 success in a specific order in n trials) = 𝑝𝑘 (1 − 𝑝)𝑛−𝑘
𝑛
𝑛!
Number of way to arrange k success in n trials =
= !
𝑘 (𝑛−𝑘)!
𝑘
𝑛 𝑘
𝑝(𝑘 success in n trials) =
𝑝 (1 − 𝑝)𝑛−𝑘
𝑘
𝑛!
𝑘 (1 − 𝑝)𝑛−𝑘
= !
𝑝
𝑘 (𝑛 − 𝑘)!
8
Example 1
A binary symmetric channel (BSC) has an error probability Pe =
0.001 (i.e., the probability of receiving 0 when 1 is transmitted, or
vice versa). Note that the channel behavior is symmetrical with
respect to 0 and 1. A sequence of 8 binary digits is transmitted
over this channel. Determine the probability of receiving exactly
2 digits in error.
9
Example 2
In binary communication, one of the techniques used to increase
the reliability of a channel is to repeat a message several times.
For example, we can send each message (0 or 1) three times.
Hence, the transmitted digits are 000 (for message 0) or 111 (for
message 1). Because of channel noise, we may receive any one
of the eight possible combinations of three binary digits. The
decision as to which message is transmitted is made by the
majority rule. If Pe is the error probability of one digit, and P(ϵ) is
the probability of making a wrong decision in this scheme. Find
P(ϵ) in term of Pe. If Pe = 0.01 then what is P(ϵ) ?
10
Multiplication Rule for Conditional
Probability
𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 𝑃 𝐵/𝐴
𝑃 𝐴1 𝐴2 … 𝐴𝑛 = 𝑃 𝐴1 . 𝑃 𝐴2 /𝐴1 . 𝑃 𝐴3 /𝐴1 𝐴2 … 𝑃 𝐴𝑛 /𝐴1 𝐴2 … 𝐴𝑛−1
Example
Suppose a box of diodes consist of Ng good diodes and Nb bad
diodes. If five diodes are randomly selected, one at a time,
without replacement, determine the probability of obtaining the
sequence of diodes in the order of good, bad, good, good, bad.
11
The Total Probability Theorem
Let n disjoint events A1, …, An from a partition of the sample
spaces S such that
𝑛
𝐴𝑖 = 𝑆
and 𝐴𝑖 ∩ 𝐴𝑗 = ∅,
if 𝑖 ≠ 𝑗
𝑖=1
Then the probability of an event
B can be written as
𝑛
𝑃(𝐵) =
𝑃(𝐵/𝐴𝑖 )𝑃(𝐴𝑖 )
𝑖=1
This theorem simplifies the analysis of the more complex events
of interest, B, by identifying all different causes Ai.
12
Example
The decoding of a data packet may be in error because of N
distinct error patterns E1, E2, …, En it encounters. These error
patterns are mutually exclusive, each with probability P(Ei) = pi.
When the error pattern Ei occurs, the data packet would be
incorrectly decoded with probability qi. Find the probability that
the data packet is incorrectly decoded.
13
Baye’s Theorem
Baye’s theorem determines the likelihood of a particular cause
of an event among many disjoint possible causes.
Theorem
Let n disjoint events A1, …, An form a partition of the sample
space S. Let B be an event with P(B) >0. Then for j=1, …, n,
𝑃 𝐵/𝐴𝑗 𝑃(𝐴𝑗 )
𝑃(𝐴𝑗 /𝐵) =
=
𝑃(𝐵)
𝑃 𝐵/𝐴𝑗 𝑃(𝐴𝑗 )
𝑛
𝑖=1 𝑃(𝐵/𝐴𝑖 )𝑃(𝐴𝑖 )
14
Example
A communication system always encounter one of three
possible interference waveforms: F1, F2, or F3. The probability of
each interference is 0.8, 0.16, and 0.04, respectively. The
communication system fails with probability 0.01, 0.1, and 0.4
when it encounters F1, F2, and F3, respectively. Given that the
system has failed, find the probability that the failure is a result
of F1, F2, or F3, respectively.
15
Random Variable
A discrete random variable has numerical values that resulted
from mapping sample points (outcomes of experiment) to these
numbers.
The outcomes of tossing a coin are {H, T} we can assign 1 for
head and -1 for tail. The random variable X = {1, -1}
𝑃𝑥 𝑥𝑖 = 1
𝑖
16
Random Variable
For two independent random variables X and Y (tossing two
coins):
𝑃𝑥𝑦 𝑥𝑖 , 𝑦𝑗 = 𝑃𝑥 𝑥𝑖 𝑃𝑦 𝑦𝑗
𝑃𝑥𝑦 𝑥𝑖 , 𝑦𝑗 = 1
𝑖
𝑗
17
Conditional Probabilities
If x and y are two RVs, then the conditional probability of
x = xi given y = yj is denoted by Px|y(xi|yj)
𝑃x|y 𝑥𝑖 𝑦𝑗 =
𝑖
𝑃y|x 𝑦𝑗 𝑥𝑖 = 1
𝑗
𝑃xy 𝑥𝑖 , 𝑦𝑗 = 1
𝑖
𝑗
𝑃xy (𝑥𝑖 , 𝑦𝑗 ) =
𝑖
𝑃x|y 𝑥𝑖 𝑦𝑗 𝑃y 𝑦𝑗 = 𝑃y (𝑦𝑗 )
𝑖
𝑃xy (𝑥𝑖 , 𝑦𝑗 ) =
𝑗
𝑃y|x 𝑦𝑗 𝑥𝑖 𝑃x 𝑥𝑖 = 𝑃x (𝑥𝑖 )
𝑗
18
Conditional Probabilities
If x and y are two RVs, then the conditional probability of
x = xi given y = yj is denoted by Px|y(xi|yj)
𝑃y 𝑦𝑗 =
𝑃y|x 𝑦𝑗 𝑥𝑖 𝑃x 𝑥𝑖
𝑖
𝑃x 𝑥𝑖 =
𝑃x|y 𝑥𝑖 𝑦𝑗 𝑃y 𝑦𝑗
𝑗
19
Example
A binary symmetric channel (BSC) error probability is Pe. The
probability of transmission 1 is Q, and that of transmitting 0 is 1Q. Determine the probability of receiving 1 and 0 at the receiver.
20
Example
Over a certain binary communication channel, the symbol 0 is
transmitted with probability 0.4 and 1 is transmitted with
probability 0.6. It is given that P(ϵ|0) = 10-6 and P(ϵ|1) = 10-4,
where P(ϵ|xi) is the probability of detecting the error given that xi
is transmitted. Determine P(ϵ), the error probability of the
channel.
21
Cumulative Distribution Function (CDF)
A CDF, Fx(x), of an RV x is the probability that x takes a value
less than or equal to x.
Property of CDF
1) Fx(x) 0
2) Fx() = 1
3) Fx (-)=0
4) Fx(x) is a nondecreasing function.
22
Example
In an experiment, a trial consists of four successive tosses of a
coin. If we define an RV x as the number of heads appearing in
a trial, determine Px(x) and Fx(x).
23
Continuous Random Variable
The random variable has continuous value.
px(x) is the probability density function (pdf) that describes the
relative frequency of occurrence of different values of x.
Properties of the probability density function:
∞
𝑝x 𝑥 ≥ 0
−∞
𝑝x 𝑥 𝑑𝑥 = 1
Cumulative distribution function: 𝐹x 𝑥 =
𝑑𝐹x 𝑥
𝑝x 𝑥 =
𝑑𝑥
𝑃 𝑥1 < 𝑥 ≤ 𝑥2 =
𝑥2
𝑥1
𝑥
𝑝x 𝑢 𝑑𝑢 = 1
−∞
𝑝x 𝑥 𝑑𝑥 = 𝐹x (𝑥2 ) − 𝐹x (𝑥1 )
24
Continuous Random Variable
25
The Gaussian (Normal) Random Variable
Standard Gaussian RV (µ = 0, σ = 1)
1 −𝑥 2 /2
𝑝x 𝑥 =
𝑒
2𝜋
𝐹x 𝑥 =
Q 𝑦 =
𝑥
1
2𝜋
1
2𝜋
𝑒 −𝑥
2 /2
𝑑𝑥
−∞
∞
2 /2
−𝑥
𝑒
𝑑𝑥
𝑦
𝑄 𝑥 = 1 − 𝐹x (𝑥)
𝐹x 𝑥 = 𝑃 x ≤ 𝑥 = 1 − 𝑄(𝑥)
𝑃 x > 𝑥 = 𝑄(𝑥)
26
27
28
The Gaussian (Normal) Random Variable
General Gaussian RV (µ , σ)
𝑝x 𝑥 =
𝐹x 𝑥 =
1
𝜎 2𝜋
𝑒
−(𝑥−𝑚)2 /2𝜎 2
1
𝑥
𝜎 2𝜋
−∞
2 /2𝜎 2
−(𝑥−𝑚)
𝑒
𝑑𝑥
𝑥−𝑚
𝐹x 𝑥 = 𝑃 x ≤ 𝑥 = 1 − 𝑄
𝜎
𝑥−𝑚
𝑃 x>𝑥 =𝑄
𝜎
29
Example
Over a certain binary channel, message m = 0 and 1 are
transmitted with equal probability by using a positive and
negative pulse, respectively. The received pulse corresponding
to 1 is p(t), shown in the figure, and the received pulse
corresponding to 0 is –p(t). Let the peak amplitude of p(t) be Ap
at t = Tp. The channel noise n(t) has a normal distribution with
zero mean and 𝜎𝑛 standard deviation. Because of the channel
noise, the received pulse will be
𝑟 𝑡 = ±𝑝 𝑡 + 𝑛(𝑡)
What is the probability of error Pe.
30
Example (cont.)
𝑃𝑒 =
𝑃(𝜖, 𝑚𝑖 )
𝑖
𝑃𝑒 =
𝑃(𝑚𝑖 )𝑃(𝜖|𝑚𝑖 )
𝑖
𝑃𝑒 = 𝑃 0 𝑃 𝜖 0 + 𝑃(1)𝑃(𝜖|1)
𝑃 𝜖 0 = 𝑃 𝑛 > 𝐴𝑃
𝑃 𝜖 1 = 𝑃 𝑛 < −𝐴𝑃
𝐴𝑝
𝑃𝑒 = 𝑄
𝜎𝑛
𝐴𝑝
=𝑄
𝜎𝑛
𝐴𝑝
=𝑄
𝜎𝑛
31
Joint Distribution
For two RVs x and y, the CDF Fxy(x,y)
𝐹xy 𝑥, 𝑦 = 𝑃(x ≤ 𝑥 and y ≤ 𝑦)
𝑝xy
𝜕2
𝑥, 𝑦 =
𝐹xy (𝑥, 𝑦)
𝜕𝑥𝜕𝑦
𝑥2 𝑦2
𝑃 𝑥1 < x ≤ 𝑥2 , 𝑦1 < y ≤ 𝑦2 =
𝑝xy 𝑥, 𝑦 𝑑𝑥𝑑𝑦
𝑥1 𝑦1
∞
𝑝x (𝑥) =
𝑝xy 𝑥, 𝑦 𝑑𝑦
−∞
∞
𝑝y (𝑦) =
𝑝xy 𝑥, 𝑦 𝑑𝑥
−∞
32
Conditional Densities
For two RVs x and y, the CDF Fxy(x,y)
𝐹xy 𝑥, 𝑦 = 𝑃(x ≤ 𝑥 𝑎𝑛𝑑 y ≤ 𝑦)
𝑝x|y
𝑝xy 𝑥, 𝑦
𝑥|𝑦 =
𝑝y 𝑦
𝑝y|x
𝑝xy 𝑥, 𝑦
𝑦|𝑥 =
𝑝x 𝑥
Bayes’ rule
𝑝x|y 𝑥|𝑦 𝑝y 𝑦 = 𝑝y|x 𝑦|𝑥 𝑝x 𝑥
Independent Random Variables
𝑝x|y 𝑥|𝑦 = 𝑝x 𝑥
𝑝x|y 𝑥|𝑦 = 𝑝x 𝑥
𝑝xy 𝑥, 𝑦 = 𝑝x 𝑥 𝑝y 𝑦
33
Rayleigh Density Example
Derive the Rayleigh probability density function (pdf).
𝑟 −𝑟 2/2𝜎2
𝑒
𝑝𝑟 𝑟 = 𝜎 2
0
𝑟≥0
𝑟<0
34
Statistical Averages (MEANS)
The average value or expected value of RV x
𝑛
x = 𝐸[𝑥] =
𝑥𝑖 𝑃x (𝑥𝑖 )
𝑖=1
∞
x = 𝐸[𝑥] =
𝑥𝑝x 𝑥 𝑑𝑥
−∞
Mean of a function g(x) of a random variable x
𝑛
𝑔(𝑥) =
𝑔(𝑥𝑖 )𝑃x (𝑥𝑖 )
𝑖=1
∞
𝑔(𝑥) =
−∞
𝑔(𝑥)𝑝x 𝑥 𝑑𝑥
The random variable x can be
the alphabetic letters and the
function could be the PCM
35
Example
Example:
The output voltage of sinusoid generator is A cos(ωt). This
output is sampled randomly. The sampled output is an RV x,
which can take on any value in the range (-A, A). Determine the
mean value x and the mean square value x 2 of the sample
output.
36
Statistical Averages (MEANS)
Mean of the Sum
x+y=x+y
Mean of the product
∞
𝑔1 (𝑥)𝑔2 (𝑦) =
∞
𝑔1 (𝑥)𝑔2 (𝑦)𝑝xy 𝑥, 𝑦 𝑑𝑥𝑑𝑦
−∞ −∞
If RVs x and y are independent, then
xy = x y
∞
𝑔1 (𝑥)𝑔2 (𝑥) =
−∞
∞
𝑔1 (𝑥)𝑝x 𝑥 𝑑𝑥
𝑔2 (𝑦)𝑝y 𝑦 𝑑𝑦
−∞
37
Moments
The nth moment of an RV x
x𝑛 =
∞
𝑥 𝑛 𝑝x 𝑥 𝑑𝑥
−∞
The nth central moment of an RV x
(x − x)𝑛 =
∞
−∞
(𝑥 − x)𝑛 𝑝x 𝑥 𝑑𝑥
The variance 𝜎x2 and standard deviation 𝜎x
𝜎x2 = (x − x)2 = x 2 − x 2
38
Example
Find the mean, variance, and the Mean Square of the Uniform
Quantization Error in PCM.
39
Example
Find the variance and the Mean Square Error Caused by
Channel Noise in PCM.
40
Variance of a Sum of Independent RVs
z=x+y
𝜎z2 = 𝜎x2 + 𝜎y2
Example
Find the total mean square error in PCM
m
Quantization
𝑞 =𝑚−𝑚
m
Channel
m
𝜖 =m−m
41
Chebyshev’s Inequality
𝑃 |x| ≤ 𝑘𝜎x
1
≥1− 2
𝑘
𝑃 |x − x| ≤ 𝑘𝜎x
1
≥1− 2
𝑘
The standard deviation σ of an RV x is a measure of the width
of its PDF. The standard deviation in communication is also
used to estimate the bandwidth of a signal spectrum.
42
Correlation
The covariance 𝜎xy is a measure of the nature of dependence
between the RVs x and y.
𝜎xy = (x − x)(y − y)
𝜎xy = xy − xy
Correlation coefficient is
a normalized covariance.
𝜎xy
𝜌xy =
𝜎x 𝜎y
−1 ≤ 𝜌xy ≤ 1
Independent variable are uncorrelated, the converse is not
necessarily true.
43
Linear Mean Square Estimation
When two random variables x and y are related (dependent),
then it is possible to estimate the value of y from a knowledge
of the value of x.
Minimum square error is one possible criterion for the estimation
of y.
𝜖2 = y − y 2
y = 𝑎x
𝜕𝜖 2
= 2𝑎x 2 − 2xy = 0
𝜕𝑎
𝑅xy
𝑎= 2=
𝑅xx
x
xy
The optimum estimation is to choose a to make 𝜖 2 = 0.
𝜖 2 = y − 𝑎x
2
= y − 𝑎x y − 𝑎. 𝜖x = y − 𝑎x y
𝜖 2 = 𝑅yy − 𝑎𝑅xy
44
Using n Random Variable for Estimation
Using n random variables x1, x2,…,xn to estimate a random
variable x0.
x0 = 𝑎1 x1 + 𝑎2 x2 + ⋯ + 𝑎𝑛 x𝑛
𝜖 2 = x0 − 𝑎1 x1 + 𝑎2 x2 + ⋯ + 𝑎𝑛 x𝑛
2
𝜕𝜖 2
= −2 x0 − 𝑎1 x1 + 𝑎2 x2 + ⋯ + 𝑎𝑛 x𝑛 x𝑖 = 0
𝜕𝑎𝑖
𝑅0𝑖 = 𝑎1 𝑅𝑖1 + 𝑎2 𝑅𝑖2 + ⋯ + 𝑎𝑛 𝑅𝑖𝑛
𝑅11 𝑅12
𝑎1
𝑎2
𝑅
𝑅22
= 21
⋮
⋯
⋯
𝑎𝑛
𝑅𝑛1 𝑅𝑛2
… 𝑅1𝑛
… 𝑅2𝑛
⋯
⋯
⋯ 𝑅𝑛𝑛
−1
where
𝑅𝑖𝑗 = x𝑖 x𝑗
𝑅01
𝑅02
⋮
𝑅0𝑛
𝜖 2 = 𝑅00 − 𝑎1 𝑅01 + 𝑎2 𝑅02 + ⋯ + 𝑎𝑛 𝑅0𝑛
45
Example
In differential pulse code modulation (DPCM), instead of
transmitting sample values directly, we estimate (predict) the
value of each sample from the knowledge of previous n samples.
The estimation error 𝜖 k, the difference between the actual value
and the estimated value of the kth sample, is quantized and
transmitted. Because the estimation error 𝜖k is smaller than the
sample value mk, for the same number of quantization levels, the
SNR is increased. The SNR improvement is equal to 𝑚2 /𝜖 2 ,
where 𝑚2 and 𝜖 2 are the mean square values of the speech
signal and the estimation error 𝜖, respectively.
Find the optimum linear second-order
predictor and the corresponding SNR
improvement.
46
Sum of Random Variables
z=x+y
How does the pdf of z relate to the pdfs of x and y?
∞
𝐹z 𝑧 = 𝑃 z ≤ 𝑧 = 𝑃 x ≤ ∞, y ≤ 𝑧 − 𝑥 =
𝑑𝑥
−∞
𝑑𝐹𝑧 (𝑧)
𝑝z 𝑧 =
=
𝑑𝑧
𝑧−𝑥
−∞
𝑝xy 𝑥, 𝑦 𝑑𝑦
∞
𝑝xy 𝑥, 𝑧 − 𝑥 𝑑𝑥
−∞
If x and y are independent random variables
∞
𝑝z 𝑧 =
𝑝x 𝑥 𝑝y 𝑧 − 𝑥 𝑑𝑥
−∞
The PDF of z is the convolution of the PDFs of x and y.
47
Sum of Gaussian Random Variables
The sum of jointly distributed Gaussian random variables is
also a Gaussian random variable regardless of their
relationship such as independence.
y = x1 + x2
y is a Gaussian RV with
y = x1 + x2
𝜎y2 = 𝜎 2x1 + 𝜎x22
If x1 and x2 are jointly Gaussian but not necessarily independent
then
𝜎y2 = 𝜎 2x1 + 𝜎x22 + 2𝜎x1 x2
48
Sum of Gaussian Random Variables
The fact that the sum of jointly distributed Gaussian random
variables is also a Gaussian random variable, has important
practical application.
For example, if xk is a sequence of jointly Gaussian signal
samples passing through a discrete time filter with impulse
response {hi}, then the filter output y is also Gaussian
∞
𝑦=
ℎ𝑖 x𝑘−𝑖
𝑖=0
49
The Central Limit Theorem
The sum of a large number of independent RVs tends to be a
Gaussian random variable, independently of the probability
densities of the variable added.
50
The Central Limit Theorem
(for the Sample Mean)
Let x1, x2, …, xn be independent random variables from a given
distribution with mean µ and variance σ2 with 0< σ2<. Then
the sample mean
x1 + ⋯ + xn
x𝑛 =
𝑛
is a Gaussian random variable with mean equals µ and
variance equals σ2/n.
𝑥
x𝑛 − 𝜇
1 −𝑣 2/2
lim 𝑃
≤𝑥 =
𝑒
𝑑𝑣
𝑛→∞
𝜎/ 𝑛
−∞ 2𝜋
x𝑛 − 𝜇
lim 𝑃
> 𝑥 = 𝑄(𝑥)
𝑛→∞
𝜎/ 𝑛
Also 𝑛𝑖=1 x𝑖 is a Gaussian random variable with mean equals
nµ and variance equals nσ2.
51
Example
Consider the communication system that transmits a data
packet of 1024 bits. Each bit can be in error with probability of
10-2. Find the (approximate) probability that more than 30 of the
1024 bits are in error.
52
```