# probability and random processes - ?· probability and random processes main aims i understand...

Post on 30-Aug-2018

213 views

Embed Size (px)

TRANSCRIPT

Probability and Random Processes

Probability and Random Processes

Jinho ChoiGIST

February 2017

1 / 99

Probability and Random Processes

What Albert Einstein said:

I As I have said so many times, God doesnt play dice with theworld.

I Two things are infinite: the universe and human stupidity.

So, he does believe in infinity,but not in randomness.

2 / 99

Probability and Random Processes

Main aims

I Understand fundamental issues of probability theory (pdf,mean, and variance)

I Understand joint and conditional pdf, independence, andcorrelation

I Learn properties of Gaussian random variables and be able toderive Chernoff bound

I Understand random processes with key definitions

I Be able to compute the mean and variance of samples fromrandom processes

3 / 99

Probability and Random Processes

Probability Theory

Probability Theory

The three axioms with a sample space, , a family of sets, F , forallowable events, and measure Pr():

I The first axiom: The probability of an event, E, is anon-negative real number:

Pr(E) 0.

I The second axiom:Pr() = 1.

I The third axiom: For any countable sequence of mutuallyexclusive events E1, E2, . . .,

Pr(E1 E2 ) =

i=1

Pr(Ei).

4 / 99

Probability and Random Processes

Probability Theory

Random variables: A random variable is a mapping from thesample space to the set of real numbers.

Sample space (abstract space) Real number

The main idea of random variables is to describe some randomevents by numbers.

real numbers

Event space

event X( )

X: random variable:

X() (,)

5 / 99

Probability and Random Processes

Probability Theory

Example of random variables: A game of dice

I Before you draw a dice, the number of dots is unknown. This number can be considered as a random variable.I Once a dice is drawn, we have a particular number, which

would be one of {1, . . . , 6}. This is called a realization.

realizations of {2, 3, 4}6 / 99

Probability and Random Processes

Probability Theory

Cumulative distribution function (CDF):

FX(x) = Pr(X x) ,

where X is the random variable (r.v.) and x is a real number. Bydefinition, the CDF is a nondecreasing function.Example: a dice

Cummulative distribution function (CDF):

FX(x) = Pr(X x),where X is the random variable (r.v.) and x is a real number.

Probability density function (PDF):

fX(x) =d

dxFX(x)

Example: a dice

1 4 5 62 3

2/6

3/6

4/6

5/6

1

1/6

FX(x) = Pr(X x)

x

3

7 / 99

Probability and Random Processes

Probability Theory

There are different types of r.v.s such as:

I continuous r.v.: X has a continuous value

I discrete r.v.: X has a discrete value

Examples:

I the phase of a sinusoid, : sin(ct+ ) continuous r.v.

I the number of dots of a dice discrete r.v.

8 / 99

Probability and Random Processes

Probability Theory

A discrete r.v. with binomial distribution

I Consider a random experiement that has two possibleoutcomes. For example, the outcome of this experiment canbe expressed (1 to denote success; 0 to denote failure) as

Y =

{1, with probability p;0, with probability 1 p.

I Consider a sum of n outcomes from independent experiments:

X =

n

j=1

Yj = {0, 1 . . . , n}.

I Then, X is the binomial random variable with parameters nand p, X B(n, p):

Pr(X = j) =

(n

j

)pj(1 p)nj , j = 0, . . . , n.

9 / 99

Probability and Random Processes

Probability Theory

Continuous r.v., X, has the probability density function (pdf) as

fX(x) =d

dxFX(x) .

I As FX(x) is nondecreasing, fX(x) 0.

I In general,

t

fX(x)dx = FX(t) .

I Since limx FX(x) = 1, fX(x)dx = 1.

For a discrete r.v., the pdf becomes the probability massfunction (pmf) which is actually probability.

I Example of dice:

fX(X = k) Pr(X = k) =1

6, k = 1, 2, . . . , 6.

10 / 99

Probability and Random Processes

Probability Theory

Mean and VarianceFor a r.v. X, the mean of X (or g(X), where g(x) is a function ofx) is given by

I a continuous r.v.:

E[X] =xfX(x)dx or E[g(X)] =

g(x)fX(x)dx

I a discrete r.v.:

E[X] =

k

xk Pr(xk) or E[g(X)] =

k

g(xk) Pr(xk)

The variance is given by

V ar(X) = E[(X E[X])2]

The mean and variance are used to characterize a random variable.11 / 99

Probability and Random Processes

Probability Theory

Mean of X B(n, p)

E[X] =n

j=0

j

(n

j

)pj(1 p)nj

=

n

j=1

jn!

j!(n j)!pj(1 p)nj

=

n

j=1

n(n 1)!(j 1)!(n j)!

ppj1(1 p)nj

= np

n

j=1

(n 1)!(j 1)!(n j)!

pj1(1 p)nj

= np(p+ (1 p))n1 = np.

12 / 99

Probability and Random Processes

Probability Theory

Geometric random variable:

I pmf: Pr(X = k) = (1 p)k1p, k 1Note that

k=1

(1 p)k1p = p

k=0

(1 p)k = p 11 (1 p)

= 1

I Example: The number of independent flips of a coin untilhead first apprears.

I Mean: Letting q = 1 p, it can be shown that

E[X] =

k=1

k(1 p)k1p = p ddq

k=0

qk

= pd

dq

1

1 q= p

1

(1 q)2=

1

p

13 / 99

Probability and Random Processes

Probability Theory

A continuous r.v. with uniform distributionLet us consider a uniform r.v. X that has the pdf as

fX(x) =

{1A , 0 x A;0, otherwise

The mean is

E[X] = A

0x

1

Adx =

1

A

x2

2

A

x=0

=A

2.

The variance is

E[(X E[X])2] = A

0

(x A

2

)2 1Adx

=

A/2

A/2z2

1

Adz

=1

A 1

3z3A/2

A/2=A2

1214 / 99

Probability and Random Processes

Probability Theory

More examples on expectationQ) Let X be a r.v. and a and c are constants. Show that

E[aX] = aE[X] and E[X + c] = E[X] + c

A)

E[aX] =

(a x)fX(x)dx = a

xfX(x)dx = aE[X]

E[X + c] =

(x+ c)fX(x)dx =

xfX(x)dx+

cfX(x)dx = E[X] + c

Q) Show that E[X2] = V ar(X) + (E[X])2A)

E[(X E[X])2] = E[X2 2(E[X])X + (E[X])2]= E[X2] 2(E[X])E[X] + (E[X])2= E[X2] (E[X])2

15 / 99

Probability and Random Processes

Probability Theory

Q) Show that V ar(aX + c) = a2V ar(X)

Q) Suppose that X is a r.v. with mean 1 and variance 3. FindE[3X2 + 2X].A)

E[3X2 + 2X] = 3E[X2] + 2E[X]= 3(V ar(X) + E2[X]) + 2E[X]

= 3 (3 + 12) + 2 1 = 14.

16 / 99

Probability and Random Processes

Probability Theory

Jensens inequality:For a convex function, g(x),

E[g(X)] g(E[X]) .

A convex function satisfiesg(x1) + (1 )g(x2) g(x1 + (1 )x2), [0, 1]

17 / 99

Probability and Random Processes

Probability Theory

Gaussian or normal random variable:

fX(x) = N (, 2) =12

exp

((x )

2

22

),

where the mean

E[X] =

xfX(x)dx =

and the variance

E[(X E[X])2] =

(x )2fX(x)dx = 2

18 / 99

Probability and Random Processes

Probability Theory

Normal or Gaussian pdfs:

19 / 99

Probability and Random Processes

Probability Theory

Normal or Gaussian cdfs:

20 / 99

Probability and Random Processes

Probability Theory

Q-function: a tail of the normal pdf.

Pr(X x) = Q(x) 4=

x

12

exp

( t

2

2

)dt

21 / 99

Probability and Random Processes

Probability Theory

Conditional probabilityThe conditional probability of an event A given B is denoted andgiven by

Pr(A |B) = Pr(A,B)Pr(B)

Q) Find the probability that the face of one dot occurs assumingodd dots are observed:A) We have

Pr(odd) =1

2and

Pr(1, odd) = Pr(1) =1

6.

Hence, it follows

Pr(1| odd) = Pr(1, odd)Pr(odd)

=1

3.

22 / 99

Probability and Random Processes

Probability Theory

Multiple random variables: The joint pdf is written as

fXY (x, y) =2

xyFXY (x, y) =

2

xyPr(X x, Y y)

Conditional pdf:

fX|Y (x|y) =

{fXY (x,y)fY (y)

, if fY (y) 6= 0.0, otherwise.

Marginalization:

fX(x) =

fXY (x, y)dy or fY (y) =

fXY (x, y)dx

23 / 99

Probability and Random Processes

Probability Theory

Expectation with two r.v.s:For continuous r.v.s, double-integral should take place:

E[g(X,Y )] =

g(x, y)fXY (x, y)dxdy (continuous)

E[g(X,Y )] =

x

y

g(x, y) Pr(X = x, Y = y) (discrete).

The conditional expectation is defined by

E[X|Y ] =xfX|Y (x|Y )dx (continuous)

E[X|Y ] =

x

xPr(X = x |Y ) (discrete).

Note that E[X|Y ] = g(Y ) is a function of Y , which is a randomvariable.

24 / 99

Probability and Random Processes

Probability Theory

Q) Show thatE[XY ] = E[Y E[X|Y ]]

A)

E[XY ] =

xyfXY (x, y)dxdy

=

xyfX|Y (x|y)fY (y)dxdy

=

y

(xfX|Y (x|y)dx

)fY (y)dy

=

yE[X|y]fY (y)dy = E[Y E[X|Y ]].

25 / 99

Probability and Random Processes

Recommended