Chapter 2 Transformations and Expectations

Discrete Random Variables Bernoulli Trials Discrete Calculus Geometric Interpretation of Expectation

Chapter 2Transformations and Expectations

Expected Values

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Outline

Discrete Random VariablesDefinitionProperties

Bernoulli Trials

Discrete Calculus

Geometric Interpretation of Expectation

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Discrete Random VariablesLet x1, x2, · · · xn be observations, the empirical mean,

x =1

n(x1 + x2 · · ·+ xn).

So, for the observations, 0, 1, 3, 2, 4, 1, 2, 4, 1, 1, 2, 0, 3, x = 24/13.

We could also organize these observations taking advantage of the distributive propertyof the real numbers, compute x as follows

x n(x) xn(x)0 2 01 4 42 3 63 2 64 2 8

13 24

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Discrete Random VariablesFor h(x) = x2, we can perform a similar computation:

x h(x) = x2 n(x) h(x)n(x)

0 0 2 01 1 4 42 4 3 123 9 2 184 16 2 32

13 66

Then,

h(x) =1

n

∑

x

h(x)n(x)=∑

x

h(x)p(x) =66

13.

where

p(x) =n(x)

nfor the proportion of observations equal to x .

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Definition

In analogy for the mean from a set of observations, we define, for a finite sample spaceΩ = ω1, ω2, . . . , ωN and a probability P on Ω, we can define the expectation or theexpected value of a random variable X by an analogous average,g(X ) by an analogousaverage,

EX =N∑

j=1

X (ωj)Pωj.

Eg(X ) =N∑

j=1

g(X (ωj))Pωj.

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Definition

Roll one die. Then Ω = 1, 2, 3, 4, 5, 6. Let X be the value on the die. So, X (ω) = ω.If the die is fair, then the probability model has Pω = 1/6 for each outcome ω. Anexample of an unfair dice would be the probability with P1 = P2 = P3 = 1/12and P4 = P5 = P6 = 1/4.

ω X (ω) Pω X (ω)Pω Pω X (ω)Pω1 1 1/6 1/6 1/12 1/122 2 1/6 2/6 1/12 2/123 3 1/6 3/6 1/12 3/124 4 1/6 4/6 3/12 12/125 5 1/6 5/6 3/12 15/126 6 1/6 6/6 3/12 18/12

sum 1 EX = 21/6 = 7/2 1 EX = 51/12 = 17/4

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PropertiesTwo properties of expectation are immediate from the formula

EX =∑

ω

X (ω)Pω.

1. If X (ω) ≥ 0 for every outcome ω ∈ Ω, then every term in the sum is nonnegativeand consequently their sum EX ≥ 0.

2. Let X1 and X2 be two random variables and c1, c2 be two real numbers, then byusing g(x1, x2) = c1x1 + c2x2 and the distributive property, we find out that

E [c1X1 + c2X2] = c1EX1 + c2EX2.

Taking these two properties, we say that the operation of taking an expectation

X 7→ EX

is a positive linear functional.7 / 21


Discrete Random Variables

Example. Toss a coin three times and X denote the number of heads.

A B C D E F G

ω X (ω) x Pω PX = x X (ω)Pω xPX = xHHH 3 3 PHHH PX = 3 X (HHH)PHHH 3PX = 3HHT 2 PHHT X (HHT )PHHTHTH 2 2 PHTH PX = 2 X (HTH)PHTH 2PX = 2THH 2 PTHH X (THH)PTHHHTT 1 PHTT X (HHT )PHHTTTH 1 1 PTHT PX = 1 X (HTH)PHTH 1PX = 1THT 1 PTTH X (THH)PTHHTTT 0 0 PTTT PX = 0 X (TTT )PTTT 0PX = 0

EX = 0 · PX = 0+ 1 · PX = 1+ 2 · PX = 2+ 3 · PX = 3.

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Discrete Random VariablesTo find a similar formula for Eg(X ) for a discrete random variable X ,

A B C D E F G

ω X (ω) x Pω PX = x g(X (ω))Pω g(x)PX = x...

......

......

......

ωk X (ωk) Pωk g(X (ωk))Pωkωk+1 X (ωk+1) xi Pωk+1 PX = xi g(X (ωk+1))Pωk+1 g(xi )PX = xi

......

......

......

......

......

...

Eg(X ) = g(x1)PX = x1+ · · ·+ g(xn)PX = xn= g(x1)fX (x1) + · · ·+ g(xn)fX (xn)

Eg(X ) =∑

x

g(x)fX (x)

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Discrete Random Variables

A similar formula holds if we have a vector of random variables X = (X1,X2, . . . ,Xn),

fX (x1, x2, . . . , xn) = PX1 = x1,X2 = x2, . . . ,Xn = xn,the joint probability mass function and g a real-valued function ofx = (x1, x2, . . . , xn).

In the two dimensional case, this takes the form

Eg(X1,X2) =∑

x1

∑

x2

g(x1, x2)fX1,X2(x1, x2).

We will return to this identity in computing the covariance of two random variables.

Exercise. Find EX 2 for both the fair dice and the unfair dice.

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Bernoulli Trials

Bernoulli trials are the simplest and among the most common models for anexperimental procedure. Each trial has two possible outcomes, variously called,

heads-tails, yes-no, up-down, left-right, win-lose, female-male, green-blue, dominant-recessive,success-failure.

Random variables X1,X2, . . . ,Xn are called a sequence of Bernoulli trials provided that:

• Each Xi takes on two values, namely, 0 and 1. We call the value 1 a success andthe value 0 a failure.

• Each trial has the same probability for success, i.e., PXi = 1 = p for each i .

• The outcomes on each of the trials is independent.

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Bernoulli Trials

For each trial i , the expected value

EXi = 0 · PXi = 0+ 1 · PXi = 1 = 0 · (1− p) + 1 · p = p

is the same as the success probability.

Let Sn = X1 + X2 + · · ·+ Xn be the total number of successes in n Bernoulli trials.

Using the linearity of expectation, we see that

ESn = E [X1 + X2 + · · ·+ Xn] = p + p + · · ·+ p = np,

the expected number of successes in n Bernoulli trials is np.

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Bernoulli TrialsTo determine the probability mass function for Sn. Beginning with a concrete example,let n = 8, and the outcome

failure, success, failure, failure, success, failure, failure, success

Using the independence of the trials, we can compute the probability of this outcome:

(1− p)× p × (1− p)× (1− p)× p × (1− p)× (1− p)× p = p3(1− p)5.

• Any of the(83

)sequences of 8 Bernoulli trials having 3 successes also has

probability p3(1− p)5.• Each of the outcomes are mutually exclusive.• Their union is the event S8 = 3.• Consequently, by the axioms of probability,

PS8 = 3 =

(8

3

)p3(1− p)5.

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Bernoulli TrialsReplace 8 by n and 3 by x to see that any sequence of n Bernoulli trials having xsuccesses has probability

px(1− p)n−x .

In addition, we know that we have (n

x

)

mutually exclusive sequences of n Bernoulli trials that have x successes. Thus, we havethe mass function

fSn(x) = PSn = x =

(n

x

)px(1− p)n−x , x = 0, 1, . . . , n.

The fact that the sumn∑

x=0

fSn(x) =n∑

x=0

(n

x

)px(1− p)n−x = (p + (1− p))n = 1n = 1

follows from the binomial theorem. Thus, Sn is called a binomial random variable 14 / 21


Bernoulli TrialsExercise. Let S4 be the number of successes in 4 Bernoulli trials with p = 1/3.

1. Find the mass function fS4(x), x=0,1,2,3,4.

2. Use fS4 to compute ES4.

3. Sketch the cumulative distribution function FS4 for S4.

Answer. fS4(x) =(4x

) (13

)x (23

)(4−x)

x fS4(x) xfS4(x)

0 16/81 01 32/81 32/812 24/81 48/813 8/81 24/814 1/81 4/81

108/81=4/3

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Discrete CalculusLet h be a function on whose domain and range are integers. The (positive) differenceoperator

∆+h(x) = h(x + 1)− h(x).

If we take, h(x) = (x)k = x(x − 1) · · · (x − k + 1), then

∆+(x)k = (x + 1)x · · · (x − k + 2)− x(x − 1) · · · (x − k + 1)

= ((x + 1)− (x − k + 1))(x(x − 1) · · · (x − k + 2)) = k(x)k−1.

Thus, the falling powers and the difference operator plays a role similar to the powerfunction and the derivative,

d

dxxk = kxk−1.

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Discrete CalculusTo find the analog to the integral, note that

h(n + 1)− h(0) = (h(n + 1)− h(n)) + (h(n)− h(n − 1)) + · · ·+ · · · (h(2)− h(1)) + (h(1)− h(0))

=n∑

x=0

∆+h(x).

For example,

n∑

x=1

(x)k =1

k + 1

n∑

x=1

∆+(x)k+1 =1

k + 1(n + 1)k+1.

Exercise. Use the ideas above to find the sum to a geometric series.

∆+rx = r x+1 − crn = (r − 1)rn.

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Discrete CalculusExample. Let X be a discrete uniform random variable on S = 1, 2, . . . n, then

EX =n∑

x=1

x1

n=

1

n

n∑

x=1

(x)1 =1

n· (n + 1)2

2=

1

n· (n + 1)n

2=

n + 1

2.

Using the difference operator, EX (X − 1) is usually easier to determine for integervalued random variables than EX 2. In this example,

EX (X − 1) =n∑

j=1

x(x − 1)1

n=

1

n

n∑

j=2

(x)2

=1

3n

n∑

j=2

∆(x)3 =1

3n(n + 1)3 =

1

3n(n + 1)n(n − 1) =

n2 − 1

3

We can find EX 2 = EX (X − 1) + EX by using the linearity of expectation.18 / 21


Geometric Interpretation of Expectation

Exercise 3. Use the ideas above to find the sum to a geometric series.

Example 4. Let X be a discrete uniform random variable on S = 1, 2, . . . n, then

EX =

nX

x=1

x1

n=

1

n

nX

x=1

(x)1 =1

n· (n + 1)2

2=

1

n· (n + 1)n

2=

n + 1

2.

Using the di↵erence operator, EX(X 1) is usually easier to determine for integer valued random vari-ables than EX2. In this example,

EX(X 1) =nX

j=1

x(x 1)1

n=

1

n

nX

j=2

(x)2

=1

3n

nX

j=2

(x)3 =1

3n(n + 1)3 =

1

3n(n + 1)n(n 1) =

n2 1

3

We can find EX2 = EX(X 1) + EX by using the linearity of expectation.

1.2 Geometric Interpretation of Expectation

-

6

x1 x2 x3 x4

x1P X = x1

x2P X = x2

x3P X = x3

x4P X = x4

...

1(x1 x0)P X > x0

(x2 x1)P X > x1(x3 x2)P X > x2

(x4 x3)P X > x3 · · ·

Figure 1: Graphical illustration of EX, the expected value of X, as the area above the cumulative distributionfunction and below the line y = 1 computed two ways.

We can realize the computation of expectation for a nonnegative random variable

EX = x1PX = x1 + x2PX = x2 + x3PX = x3 + x4PX = x4 + · · ·

4

Figure: Graphical illustration of EX , the expected value of X , as the area above the cumulativedistribution function and below the horizontal line at 1 computed two ways.

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Geometric Interpretation of ExpectationWe can realize the computation of expectation for a nonnegative random variable

EX = x1PX = x1+ x2PX = x2+ x3PX = x3+ x4PX = x4+ · · ·.As illustrated in the Figure, each term in this sum can be seen as a horizontalrectangle of width xj and height PX = xj.Summation by parts, the analog in calculus to integration by parts, suggests that wecan also compute this area by looking at the vertical rectangles. The j-th rectangle haswidth xj+1 − xj and height PX > xj. Thus,

EX =∞∑

j=0

(xj+1 − xj)PX > xj.

if X take values in the nonnegative integers, then xj = j and

EX =∞∑

j=0

PX > j.20 / 21


Geometric Interpretation of ExpectationExample. For a geometric random variable based on the number of tails before thefirst heads from successive flips of a biased coin, we have that X > j = X ≥ j + 1precisely when the first j + 1 coin tosses results in tails. Thus,

PX > j = (1− p)j+1

and

EX =∞∑

j=0

PX > j =∞∑

j=0

(1− p)j+1 =1− p

1− (1− p)=

1− p

p.

Exercise. For a Z+ valued random variable, choose xj = (j)k to see that

E (X )k = k∞∑

j=0

(j)k−1PX > j.

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Chapter 2 Transformations and Expectations