MATH 256 Probability and Random Processes

MATH 256 Probability and Random Processes

Yrd. Doç. Dr. Didem Kivanc [email protected]

[email protected]

14/10/2011 Lecture 3

OKAN UNIVERSITYFACULTY OF ENGINEERING AND ARCHITECTURE

04 Random Variables

Fall 2011

What is a random variable• Random Variables

– A random variable X is not a “variable” like algebra– A random variable X is a function:

• From a set of outcomes of a random event (the sample space S of an experiment)

• To the set of real numbers• Realizations of a random variable are called random variates.

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1

-1

Set of outcomesof a coin toss

Random Variable

ℝ

heads

tails

X

Example• Experiment: throw 3 coins• Sample Space: S = {(H,H,H), (H,H,T), (H,T,H), (T,H,H), (H, T, T),

(T,H,T), (T,T,H),(T,T,T)}• Y is a random variable, giving the number of heads that

landed:

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(H,H,H)

(H,H,T) (H,T,H)(T,H,H)

(H,T,T)(T,H,T) (T,T,H)

(T,T,T)

3210

10

83

183

281

38

P Y

P Y

P Y

P Y

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Three balls are to be randomly selected without replacement from an urn containing 20 balls numbered 1 through 20. If we bet that at least one of the balls that are drawn has a number as large as or larger than 17, what is the probability that we win the bet?

Let X be the largest of the three numbers drawn.

1

23,4,..., 20

203

i

P X i i

17 17 18 19 200.508

P X P X P X P X P X

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Independent trials consisting of the flipping of a coin having probability p of coming up heads are continually performed until either a head occurs or a total of n flips is made. If we let X denote the number of times the coin is flipped, then X is a random variable taking on one of the values 1, 2, 3, . . . , n with respective probabilities:

2

2

1

1

2 1

3 1

1 1

1

n

n

P X p

P X p p

P X p p

P X n p p

P X n p p

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Three balls are randomly chosen from an urn containing 3 white, 3 red, and 5 black balls. Suppose that we win $1 for each white ball selected and lose $1 for each red ball selected. If we let X denote our total winnings from the experiment, then X is a random variable taking on the possible values 0, 1, 2, 3 with respective probabilities

Suppose every ball has a number. Then your balls are:W1, W2, W3, R1, R2, R3, B1, B2, B3, B4, B5Or for convenience I will number them from 1 to 11. So there are ways to choose three balls from this set.11

3

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The list of possible values for X is {-3,-2,-1,0,1,2,3}To get -3, we must choose RRR. To get -2, we must choose 2 R and one BTo get -1, we must choose 2 R and one W or one R and two B.To get 0, we must choose one R, one W and one B or BBBTo get +1, we must choose 2 W and one R or one W and two BTo get +2, we must choose 2W and one BTo get +3, we must choose WWW.

So:

33 1

3 311 1653

P X P X

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3 52 1 15

2 211 1653

P X P X

3 3 3 52 1 1 2 39

1 111 1653

P X P X

5 3 3 53 1 1 1 55

011 1653

P X

The cumulative distribution function• For a random variable X, the function F defined by

• is called the cumulative distribution function, or, the distribution function, of X.

• Thus, the distribution function specifies, for all real values x, the probability that the random variable is less than or equal to x.

• F(x) is a nondecreasing function of x, that is, • If a < b then F(a) < F(b).

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F x P X x x

For the previous example:

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13 3

16515

2 216539

1 1165

550

165

P X P X

P X P X

P X P X

P X

For the previous example:

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3 01

2165

1 15 161

165 165 1651 15 39 55

0165 165 165 165

1 15 39 55 1101

165 165 165 165 1651 15 39 55 39 149

2165 165 165 165 165 165

1 15 39 55 39 15 1643

165 165 165 165 165 165 1651 15 39 55

4165 165 165 16

F

F

F

F

F

F

F

F

39 15 1 165

15 165 165 165 165

Probability Mass Function

• Is defined for a discrete variable X.

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p a P X a

0 1,2,...( ) 0

ip x ip a

p x x

for for all other values of

1

1ii

p x

• Suppose that

• Then since x must be one of the values xi,

Example of probability mass function

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0 0 1 4

1 1 1 2

2 2 1 4

p P X

p P X

p P X

Example• The probability mass function of a random variable X is given

by i=0,1,2,… where λ is some positive value.

• Find (a) P{X = 0} and (b) P{X > 2}.

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( ) !ip i c i

0

0 0

( ) 1

( ) 1!

ii

i i

p i

p i c cei

c e

0 !

ix

i

xe

i

The cumulative distribution function• The cumulative distribution function F can be expressed in

terms of p(a) by

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all

( ) ( )x a

F a p x

• If X is a discrete random variable whose possible values are

x1, x2, x3, … where x1< x2 < x3 < … then the distribution function F of X is a step function.

Example

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1 1 1 11 2 3 4

4 2 8 8p p p p

• then the distribution function F of X is

0 11 4 1 23 4 2 37 8 3 41 4

aa

F a aa

a

• For example, suppose the probability mass function (pmf) of X is

Expectation of a random variable• If X is a discrete random variable having a probability mass

function p(x) then the expectation or the expected value of X denoted by E[X] is defined by

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: ( ) 0

( )x p x

E X xp x

• In other words,

• Take every possible value for X• Multiply it by the probability of getting that value• Add the result.

Examples of expectation• For example, suppose you have a fair coin. You flip the coin,

and define a random variable X such that – If the coin lands heads, X = 1

– If the coin lands tails, X = 2

• Then the probability mass function of X is given by

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11 2

2p p 1 2 if 1 or 2,

0 otherwise.x x

p x Or we can write

1 11 2 1.5

2 2E X

Examples of expectation• Next, suppose you throw a fair die. You flip the die, and define

a random variable Y such that – If the die lands a number less than or equal to 5, then Y = 0

– If the die lands a number greater than 5, then Y = 1

• Then the probability mass function of Y is given by

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5 6 if 0,

Pr 1 6 if 1,0 otherwise.

yp y Y y y

5 1 10 1

6 6 6E X

Frequency interpretation of probabilities• The law of large numbers – we will see in chapter 8 – assumes

that if we have an experiment (e.g. tossing a coin) and we perform it an infinite number of times, then the proportion of time that any event E occurs will be P(E).

• [Recall here than event means a subset of the sample space, or a set of outcomes for the experiment]

• So for instance suppose X is a random variable which will be equal to x1 with probability p(x1), x2 with probability p(x2), …, xn with probability p(xn).

• By the frequency interpretation, if we keep playing this game, then the proportion of time that we win xi will be p(xi).

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Frequency interpretation of probabilities• Or we can say that when we play the game N times, where N

is a very big number, we will win xi about Np(xi) times.

• Then the average winnings per game will be:

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1 1 2 2

1 2 2

1

No. of times I won No. of times I won ... No. of times I won

No. of times I played...

n n

nn n

n ni

x x x x x x

Np x x Np x x Np xx p x E X

N

Example 3a• Question:

– Find E[X] where X is the outcome when we roll a fair die. • Solution:

– Since

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11 2 3 4 5 6

6p p p p p p

1 1 2 2 3 3 4 4 5 5 6 6

1 1 6 71 2 3 4 5 6 2.5

6 6 2

E X p p p p p p

Example 3b• Question:

– We say that I is an indicator variable for an event A if

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1 0 cE I p A p A p A

1 if occurs0 if occursc

AI

A

– What is E[I] ?

Example 3d• A school class of 120 students is driven in 3 buses to a

symphonic performance. There are 36 students in one of the busses, 40 in another, and 44 in the third bus. When the busses arrive, one of the 120 students is randomly chosen. Let X denote the number of students on the bus of that randomly chosen student, and find E[X].

• Solution:

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36 Pr Student is on 1st bus 40 Pr Student is on 2nd bus

44 Pr Student is on 3rd bus

E X

Pr Student is on 1st bus 36 120 Pr Student is on 2nd bus 40 120

Pr Student is on 2nd bus 44 120

36 36 120 40 40 120 44 44 120 40.27E X

Example 3d• Same problem as before, but assume that the bus is chosen

randomly instead of the student, and find E[X].• Solution:

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36 Pr 1st bus is chosen 40 Pr 2nd bus is chosen

44 Pr 3rd bus is chosen

E X

36 1 3 40 1 3 44 1 3 40.00E X

1Pr 1st bus is chosen Pr 2nd bus is chosen Pr 3rd bus is chosen 3

Expectation of a function of a random variable

• To find E[g(x)], that is, the expectation of g(X)• Two step process:

– find the pmf of g(x)– find E[g(x)]

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Let X denote a random variable that takes on any of the values –1, 0, and 1 with respective probabilities

{ 1} 0.2 { 0} 0.5 { 1} 0.3P X P X P X Compute 2E X

SolutionLet Y = X 2.

{ 1} { 1} { 1} 0.5{ 0} { 0} 0.5

P Y P X P XP Y P X

2 1(0.5) 0(0.5) 0.5E X E Y

Then the probability mass function of Y is given by

0.5 if 0 or 10 otherwise.

y yp y

Statistics vs. Probability• You may have noticed that the concept of “expectation”

seems a lot like the concept of “average”.• So why do we use this fancy new word “expectation”? Why

not just call it “average”?• We find the average of a list of numbers. The numbers are

already known. • We find the expectation of a random variable. We may have

only one such random variable. We may only toss the coin or die once.

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Statistics vs. Probability• For instance, let us define a random variable X using the result of a coin toss:

let X = 1 if the coin lands heads, X = 0 if the coin lands tails.• If we perform this experiment K times, we will get a list of values for X. We can

find the average value for K by adding all the values for X, and dividing by K.

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1

1 K

ii

XK

• Is this coin fair? We don’t know, but we can find out.

Number of times the coin lands heads0 Pr 0

Number of times the coin lands tails1 Pr 1

p XK

p XK

Statistics vs. Probability• What we did on the previous slide was statistics: we analyzed

the data to draw some conclusions about the process or mechanism (i.e. the coin) that generated that data.

• Probability is how we draw conclusions about the future. • So suppose I did the experiments on the previous slide

yesterday. Today I will come into the class and toss the coin exactly once.

• Then I can use the statistics from yesterday to help find out what I can expect the result of the coin toss to be today:

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1

0

0 0 1 1i

E X ip i p p

Statistics vs. Probability• Okay, so I got 0.5.• What does this mean? X can never equal 0.5. • Expectation makes more sense with continuous random variables, e.g. when

you measure a voltage on a voltmeter. • With the coin toss you can think of it this way: • Suppose someone wants you to guess X. But you will pay a lot of money if

you’re wrong, and the money you pay is proportional to how wrong you are. • If you guess g, and the result was actually a, then you have to pay• What should you guess? • You must minimize • If you guess g=E[X], then this penalty is minimized.

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2 21 1 0 0g p g p

2100 g a

Statistics: how to find the pmf of a random voltage from measurements

• Suppose you are going to measure a voltage. • You know that the voltage is really about 5V. • But you have an old voltmeter that doesn’t measure very

well. • The voltmeter is digital and has 1 decimal place. So you can

only read voltages 0.00, 0.1, …, 4.7, 4.8, 4.9, 5.0, 5.1, …, 9.9. • You start measuring the voltage. You get the following

measurements: 4.7, 5.0, 4.9, 5.0, 5.3, 4.9, 4.8, 5.2, …• From these measurements you can construct a probability

mass function graph as follows.

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Pmf drawn from results of experiment

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10

4.64.5 4.7 4.8 4.9 5.0 5.1 5.2 5.3 5.4 5.5

1

Measurements: 4.7, 5.0, 4.9, 5.0, 5.3, 4.9, 4.8, 5.2,5.0, 4.5, 4.8, 5.1, 5.0, 5.1, 4.9, 5.3, 5.1, 5.2, 5.1, 5.4

23

4

5

6

7 8

9

11

12

13

14

15

1618

17

18

19

20

And to show this with animation

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10

4.64.5 4.7 4.8 4.9 5.0 5.1 5.2 5.3 5.4 5.5

1

Measurements: 4.7, 5.0, 4.9, 5.0, 5.3, 4.9, 4.8, 5.2,5.0, 4.5, 4.8, 5.1, 5.0, 5.1, 4.9, 5.3, 5.1, 5.2, 5.1, 5.4

23

4

5

6

7 8

9

11

12

13

14

15

1618

17

18

19

20

pmf derived mathematically• Based on the frequency interpretation, we can define the pmf

as follows:

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14.5 204.6 0

14.7 2024.8 2034.9 2055.0 20

p

p

p

p

p

p

45.1 2025.2 2025.3 2015.4 20

5.5 0

p

p

p

p

p

• Now I can predict the future based on this pmf. • Probability does not bother with data. Statistics is all about

data.

Statistics vs. Probability• Are these the correct probabilities? I don’t know. Even if we ran the experiment millions of times, we would be wrong, probably a little wrong, maybe even very

wrong. It is always possible to throw 1000 heads in a row even with a fair die, although it is very unlikely that this will happen. • In any case, when studying probability we are not concerned with whether the pmf is correct for this experiment, because we do not care about experiments or data. • Statisticians, or the people who designed this experiment must take care to design it well, so they can give us a good statistical model. • All we know is the statistical model (that is the pmf) and we derive, mathematically, predictions about the future based on this pmf.

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