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Lecture Notes ApSc 3115/6115:

Engineering Analysis III

Chapter 7: Expectation and Variance

Version: 5/30/2014

Text Book:

A Modern Introduction to Probability and Statistics,

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7 Expectation and Variance

7.1 Expected Valuesá

______________________________________

• , Random variables are complicated objects containing a lot of information on the experiments that are modeled by them.

• Typically, random variables are summarized by two numbers:

: also called , gives the center

The expected value the expectation or mean

- in the sense of average value - of the distribution of the random variable.

The variance: a measure of spread of the distribution of the random variable.

Example Expected Value: An oil company needs 10 drill bits in an exploration project. Suppose that it is known that drill bits will last # $, , or % hours with

probabilities . !Þ" !Þ(, , and !Þ# How long can we expect the exploration to

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7 Expectation and Variance

7.1 Expected Valuesá

______________________________________

One drill bit lasts on average: !Þ" ‚ #  !Þ( ‚ $  !Þ# ‚ % œ $Þ"hours 10 drill bits Ê Exploration can continue (on average) for hours$"

But could be as short as "! ‚ # œ #!hours or as long as "! ‚ % œ %!hours!

Mathematical Fact: For large 8, drill bits last around 8 8 ‚ $Þ" hours.

Definition: The expectation of a discrete random variable \ taking the values + ß + ß á" # and with probability mass function is the number: À

IÒ\Ó œ + T Ð\ œ + Ñ œ + :Ð+ Ñ

3 3

3 3 3 3

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7 Expectation and Variance

7.1 Expected Valuesá

______________________________________

Quick exercise 7.1: Let be the discrete random variable that takes the values ,\ " # % ), , , and , each with probability "' "Î&. Compute the expectation of .\

Solution QE7.1: IÒ\Ó œ " † &"  # † "&  % † &"  ) † "&  "' † "& œ $"& œ 'Þ#Þ

Additional Interpetation:

Expected Value is the center of gravity or the balancing point

of the probability distribution. For the random variable

associated with the drill bit, this is illustrated in Figure 7.1.

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7 Expectation and Variance

7.1 Expected Valuesá

______________________________________

How to define the expected value of a continuous random variable?

Suppose is a RV on \ Ò!ß "Ó and we want to estimate/calculate IÒ\Ó.

Step 1: Approximate by \ ]8 with outcomes !ß ß á ß8" 8"8 ß " and probabilities

T Ð] œ 5Ñ œ T Ð5  " Ÿ \ Ÿ 5Ñ Ê IÒ] Ó œ 5T Ð] œ 5ÑÞ 8 8 8 8 8 8 8 8 5œ! 8  Step 2: For large , we know 8 T Ð5"8 Ÿ \ Ÿ Ñ ¸ 0 Ð Ñ Ê85 81 85

IÒ] Ó œ 5T Ð] œ 5Ñ ¸ 5 0 Ð Ñ Þ5 8 8 8 8 8 8 8 5œ! 5œ! 8 8   1

Step 3: Now set IÒ\Ó œ

8 Ä ∞lim   5œ! 8 ! " 5 5 80 Ð Ñ8 8 œ B0 ÐBÑ.BÞ 1

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7 Expectation and Variance

7.1 Expected Valuesá

______________________________________

Definition: The expectation expected value or mean of a continuousß random variable \ is the number À

IÒ\Ó œ  B0 ÐBÑ.B

∞ ∞

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7 Expectation and Variance

7.1 Expected Valuesá

______________________________________

Quick exercise 7.2: Compute the expectation of a random variable that isY

uniformly distributed over Ò#ß &Ó.

Solution QE 7.2: 0 Ð?Ñ œ ß ? − Ò#ß &Ó"$ and elsewhere. Hence,!

IÒY Ó œ ? † .? œ" " "† ? œ " ? œ #&  % œ #" œ $ Þ" $ $ # ' ' ' ' #      # & # # # # & &

which is the balancing point! Y µ Y Ð ß Ñ Ê IÒY Ó œα " α "# . • The expected value of a random variable may not exist!

M œ  B0 ÐBÑ.B œ  B0 ÐBÑ.B   B0 ÐBÑ.B œ M  M M  !ß M  !Þ

∞ ∞ !

∞ ! ∞

 ,  

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7 Expectation and Variance

7.1 Expected Valuesá

______________________________________

Example: The Cauchy distribution 0 ÐBÑ œ 1Ð"B Ñ" # ß  ∞  B  ∞Þ

M œ B † " .B œ " Ð"  B Ñ œ  ∞ Ð"  B Ñ # M œ B † " .B œ " Ð"  B Ñ œ  ∞ Ð"  B Ñ #  # ! ∞ # ∞ !  # ∞ ! # ! ∞   1 1 1 1     ln ln

The expected value may be of infinite value! When M is finite, but isM

infinite, the expected value is infinite.

Example: A distribution that has an infinite expectation the Pareto is

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7 Expectation and Variance

7.2 Three Examplesá

______________________________________

\ ´ The number of weeks until success µ K/9Ð:Ñß where : œ "!%.

Definition: The expectation of a geometric distribution.

\ µ K/9Ð:Ñ Ê IÒ\Ó œ 5 ‚ T Ð\ œ 5Ñ œ 5Ð"  :Ñ : œ " :   5œ" 5œ" ∞ ∞ 5"

The geometric distribution If you buy a lottery ticket every week and you have a chance of in " "!ß !!!of winning the jackpot, what is the expected

number of weeks you have to buy tickets before you get the jackpot?

Answer: "!ß !!! weeks (almost two centuries ;-) !!!).

•  

5œ" 5œ"

∞ ∞

5" " 5" "

: Ð"BÑ

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7 Expectation and Variance

7.2 Three Examplesá

______________________________________

  5œ" 5œ" ∞ ∞ 5" 5" # # 5:Ð"  :Ñ œ : 5Ð"  :Ñ œ : " œ : œ " Ò"  Ð"  :ÑÓ : :

The exponential distribution: Recall the chemical reactor example in chapter 5. X ´ Residence time in min. µ IB:Ð!Þ#&Ñ Ê ÒX Ó œ %E minutes.

Definition: The expectation of an exponential distribution.

\ µ IB:Ð Ñ Ê IÒ\Ó œ-  ! ∞  B B /- .B œ "

Definition: The expectation of a normal distribution.

\ µ R Ð ß. 5#Ñ Ê IÒ\Ó œ B " / Ð Ñ .B œ .

"#

B. #

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7 Expectation and Variance

7.3 The change-of-variable formulaá

______________________________________

Example: Suppose a construction company builds square buildings with width and depth \ß \ µ Y Ò!ß "!Ó. Suppose we have for the price of a buildingT À

T œ G ‚ \ ß# where is the price per square meter a constant)G Ð Þ

Annual revenue is proportional to, the average building size IÒ] Ó, where ] œ \#. Thus, we first have to determine the distribution of ] œ \#,

\ − Ò!ß "!Ó Ê ] œ \ − Ò!ß "!!ÓÞ# J ÐCÑ œ T Ð] Ÿ CÑ œ T Ð\ Ÿ CÑ œ T Ð\ Ÿ CÑ œ Cß \ µ Y Ò!ß "!ÓÞ "! #   recall 0 ÐCÑ œ . J ÐCÑ œ . C œ " " œ " "

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7 Expectation and Variance

7.3 The change-of-variable formulaá

______________________________________

IÒ\ Ó œ IÒ] Ó œ C " " .C œ " C.C œ #! C #! # ! ! "!! "!!   

#! $

" #

C

$#

œ $$ 7

"

$

"!! ! #

Conclusion: Annual Revenue ¸

$$ ‚"$ (# buildings per year) (price per 7 ÑÞ#

Observation: \ µ Y Ò!ß "!Óß IÒ\Ó œ & IÒ\ Ó Á IÒ\Ó † IÒ\Ó œ #&7 Þ # # • Alternative Method to evaluate IÒ\ Ó À# Realize that buildings with area B#

get build with the same frequency as buildings with width Thus, recallingBÞ \ µ Y Ð!ß "!Ñ one has: IÒ\ Ó œ B 0 ÐBÑ.B œ B " .B œ "! # "! # "! # \  

"! $

" "

B

$

"!

œ $$ 7

"

$

#

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7 Expectation and Variance

7.3 The change-of-variable formulaá

______________________________________

Definition: The change of variable formula. Let be a rv and \ 1 À ‘ Ä Þ‘

\ + ß á ß + Ê IÒ1Ð\ÑÓ œ 1Ð+ ÑT Ð\ œ + Ñ œ 1Ð+ Ñ:Ð+ Ñ \ \ µ 0 Ð † Ñ Ê IÒ1Ð\ÑÓ œ 1ÐBÑ0 ÐBÑ.B discrete on continuous, " 8 3 3 3 3 3 3 ∞ ∞   

Quick exercise 7.3: Let \ µ F/<Ð:ÑÞCompute IÒ# ÓÞ\

Solution QE 7.3:

T Ð\ œ !Ñ œ "  :ß T Ð\ œ "Ñ œ :Þ Thus:

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7 Expectation and Variance

7.3 The change-of-variable formulaá

______________________________________

• Suppose 1ÐBÑ œ +B  ,ß +ß , − ‘ and is a continuous RV\ Þ IÒ1Ð\ÑÓ œ IÒ+\  ,Ó œ œ     ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ 1ÐBÑ0 ÐBÑ.B Ð+B  ,Ñ0 ÐBÑ.B œ + B0 ÐBÑ.B  , 0 ÐBÑ.B œ +IÒ\Ó  ,

Same applies when is a discrete RV!\

• AnExpected value of linear transformation of random variable : \

operation that occurs very often in practice is a change of units, e.g., changing from Fahrenheit to Celsius, from minutes to hours, etc.

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7 Expectation and Variance

7.4 Varianceá

______________________________________

• Suppose you are offered an opportunity for an investment at the cost of $&!!

whose expected return is $500. Seems an OK opportunity.

What if we have for payoff $ %, $5 % ? What if we have for payoff $ %, $1000 % ?

] À T Ð] œ %&!Ñ œ &! T Ð] œ &!Ñ œ &! ] À T Ð] œ !Ñ œ &! T Ð] œ Ñ œ &!

" " "

# # #

Clearly, the spread (around the mean) makes you feel different. Usually this is measured by the expected squared deviation from the mean.

Definition: The variance Z +<Ð\Ñof a random variable is the number\ À

Z +<Ð\Ñ œ IÒÐ\  IÒ\ÓÑ Ó œ IÒ\ Ó  ÐIÒ\ÓÑ# # #

Note that: Z +<Ð\Ñ   !Þ

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7 Expectation and Variance

7.4 Varianceá

______________________________________

Quick exercise 7.4:Calculate the mean and variance for and ]" ]#

Payoff $ %, $5 % ?

Payoff $ %, $1000 % ?

] À T Ð] œ %&!Ñ œ &! T Ð] œ &!Ñ œ &! ] À T Ð] œ !Ñ œ &! T Ð] œ Ñ œ &!

" " "

# # #

Solution QE 7.4:

IÒ] Ó œ %&! † T Ð] œ %&!Ñ  &&! † T Ð] œ &!Ñ œ %&!  &&! œ &!! # IÒ] Ó œ ! † T Ð] œ !Ñ  "!!! † T Ð] œ "!!!Ñ œ !  "!!! œ &!! # " " " # " " $ $5 $ $ $ $

Z +<Ð] Ñ œ Ð%&!  &!!Ñ † T Ð] œ %&!Ñ  Ð&&!  &!!Ñ † T Ð] œ &!Ñ œ #&!!  #&!! œ #&!! Ê W>ÞH/@Ð\Ñ œ &!Þ

# Z +<Ð] Ñ œ Ð!  &!!Ñ † T Ð] œ !Ñ  Ð"!!!  &!!Ñ † T Ð] œ "! " # " # " # # " # " $ $5 $ $ $ !!Ñ

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7 Expectation and Variance

7.4 Varianceá

______________________________________

Z +<Ð\Ñ œ IÒÐ\  IÒ\ÓÑ Ó œ IÒ\  #IÒ\Ó † \  ÐIÒ\ÓÑ Ó œ IÒ\ Ó  IÒ#IÒ\Ó † \Ó  IÒÐIÒ\ÓÑ Ó

IÒ\Ó † IÒ\Ó  ÐIÒ\ÓÑ œ IÒ\ Ó  ÐIÒ\ÓÑ

# # #

# #

# # #

œ IÒ\ Ó  ##

An alternative expression for the variance: For any random variable \ À

Z +<Ð\Ñ œ IÒ\ Ó  ÐIÒ\ÓÑ# #

Variance of a normal distribution:

Definition: The expectation of a normal distribution.

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7 Expectation and Variance

7.4 Varianceá

______________________________________

Recall Example : \ ´ Time a drill bit lastsÞ

T Ð\ œ #Ñ œ !Þ"ß T Ð\ œ $Ñ œ !Þ(ß T Ð\ œ %Ñ  !Þ"Þ IÒ\Ó œ !Þ" ‚ #  !Þ( ‚ $  !Þ# ‚ % œ $Þ" hours

Method 1: Z +<Ð\Ñ œ IÒ\  ÐIÒ\ÓÑ Ó#

Z +<Ð\Ñ œ Ð#  $Þ"Ñ † !Þ"  Ð$  $Þ"Ñ † !Þ(  Ð%  $Þ"Ñ † !Þ# œ Ð  "Þ"Ñ † !Þ"  Ð  !Þ"Ñ † !Þ(  Ð!Þ*Ñ † !Þ# œ "Þ#" † !Þ"  !Þ!" † !Þ(  !Þ)" † !Þ# œ œ !Þ"#"  !Þ!!(  !Þ"'# œ ! # # # # # # Þ#*

Method 2: Z +<Ð\Ñ œ IÒ\ Ó  ÐIÒ\ÓÑ# #

Z +<Ð\Ñ œ Ð#Ñ † !Þ"  Ð$Ñ † !Þ(  Ð%Ñ † !Þ#  Ð$Þ"Ñ œ % † !Þ"  * † !Þ(  "' † !Þ#  *Þ'"

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7 Expectation and Variance

7.4 Varianceá

______________________________________

Expectation and Variance under a change of units: For any random random variable and any real numbers and \ + , À

IÒ+\  ,Ó œ +IÒ\Ó  , Z +<Ð+\  ,Ñ œ + Z +<Ð\Ñ#

Without calculation, why is Z +<Ð\Ñ not affected above by ?,

Z +<Ð+\  ,Ñ œ IÒ Ð+\  ,Ñ  IÒ+\  ,Ó Ó œ IÒ +\  ,  +IÒ\Ó  , Ó œ IÒ +\  ,  +IÒ\Ó  , Ó

IÒ +\  +IÒ\Ó Ó œ IÒ+ \  IÒ\Ó Ó + IÒ \  IÒ\Ó Ó œ + Z +<Ð\Ñ Ð Ñ Ð Ñ Ð Ñ Ð Ñ Ð Ñ Ð Ñ # # # # # # # # # Ð Ñ œ œ

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