The Distribution of a Linear Combination of Random Variables Bernd - - PowerPoint PPT Presentation

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The Distribution of a Linear Combination

• f Random Variables

Bernd Schr¨

• der

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Introduction

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 3

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Introduction

This section mostly summarizes earlier results in a slightly more abstract light.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Definition. Let X1,...,Xn be random variables and let

a1,...,an be numbers.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Definition. Let X1,...,Xn be random variables and let

a1,...,an be numbers. Then the random variable Y =

n

∑

k=1

akXk

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Definition. Let X1,...,Xn be random variables and let

a1,...,an be numbers. Then the random variable Y =

n

∑

k=1

akXk = a1X1 +···+anXn

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Definition. Let X1,...,Xn be random variables and let

a1,...,an be numbers. Then the random variable Y =

n

∑

k=1

akXk = a1X1 +···+anXn is called a linear combination of X1,...,Xn.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Definition. Let X1,...,Xn be random variables and let

a1,...,an be numbers. Then the random variable Y =

n

∑

k=1

akXk = a1X1 +···+anXn is called a linear combination of X1,...,Xn. Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Definition. Let X1,...,Xn be random variables and let

a1,...,an be numbers. Then the random variable Y =

n

∑

k=1

akXk = a1X1 +···+anXn is called a linear combination of X1,...,Xn.

• Example. X =

n

∑

k=1

1 nXk is a linear combination of X1,...,Xk.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Proposition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

• 1. E
• n

∑

k=1

akXk

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

• 1. E
• n

∑

k=1

akXk

• =

n

∑

k=1

akE(Xk)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

• 1. E
• n

∑

k=1

akXk

• =

n

∑

k=1

akE(Xk) =

n

∑

k=1

akµk.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

• 1. E
• n

∑

k=1

akXk

• =

n

∑

k=1

akE(Xk) =

n

∑

k=1

akµk.

• 2. V
• n

∑

k=1

akXk

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

• 1. E
• n

∑

k=1

akXk

• =

n

∑

k=1

akE(Xk) =

n

∑

k=1

akµk.

• 2. V
• n

∑

k=1

akXk

• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

• 1. E
• n

∑

k=1

akXk

• =

n

∑

k=1

akE(Xk) =

n

∑

k=1

akµk.

• 2. V
• n

∑

k=1

akXk

• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk).

• 3. If X1,...,Xn are independent

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

• 1. E
• n

∑

k=1

akXk

• =

n

∑

k=1

akE(Xk) =

n

∑

k=1

akµk.

• 2. V
• n

∑

k=1

akXk

• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk).

• 3. If X1,...,Xn are independent, then

V

• n

∑

k=1

akXk

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

• 1. E
• n

∑

k=1

akXk

• =

n

∑

k=1

akE(Xk) =

n

∑

k=1

akµk.

• 2. V
• n

∑

k=1

akXk

• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk).

• 3. If X1,...,Xn are independent, then

V

• n

∑

k=1

akXk

• =

n

∑

k=1

a2

kV (Xk)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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• Proposition. Let X1,...,Xn be random variables with means

µ1,...,µn and standard deviations σ1,...,σn.

• 1. E
• n

∑

k=1

akXk

• =

n

∑

k=1

akE(Xk) =

n

∑

k=1

akµk.

• 2. V
• n

∑

k=1

akXk

• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk).

• 3. If X1,...,Xn are independent, then

V

• n

∑

k=1

akXk

• =

n

∑

k=1

a2

kV (Xk) = n

∑

k=1

a2

kσ2 k .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Proof of Part 1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y) and it is easy to see that E(aX) = aE(X).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y) and it is easy to see that E(aX) = aE(X). (Remember that expected values are sums or integrals

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y) and it is easy to see that E(aX) = aE(X). (Remember that expected values are sums or integrals and constants can be factored out of sums and integrals.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y) and it is easy to see that E(aX) = aE(X). (Remember that expected values are sums or integrals and constants can be factored out of sums and integrals.) E(a1X1 +···an−1Xn−1 +anXn)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y) and it is easy to see that E(aX) = aE(X). (Remember that expected values are sums or integrals and constants can be factored out of sums and integrals.) E(a1X1 +···an−1Xn−1 +anXn) = E

• (a1X1 +···an−1Xn−1)+anXn
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 29

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y) and it is easy to see that E(aX) = aE(X). (Remember that expected values are sums or integrals and constants can be factored out of sums and integrals.) E(a1X1 +···an−1Xn−1 +anXn) = E

• (a1X1 +···an−1Xn−1)+anXn
• =

E(a1X1 +···an−1Xn−1)+E(anXn)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 30

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y) and it is easy to see that E(aX) = aE(X). (Remember that expected values are sums or integrals and constants can be factored out of sums and integrals.) E(a1X1 +···an−1Xn−1 +anXn) = E

• (a1X1 +···an−1Xn−1)+anXn
• =

E(a1X1 +···an−1Xn−1)+E(anXn) = ···

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 31

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y) and it is easy to see that E(aX) = aE(X). (Remember that expected values are sums or integrals and constants can be factored out of sums and integrals.) E(a1X1 +···an−1Xn−1 +anXn) = E

• (a1X1 +···an−1Xn−1)+anXn
• =

E(a1X1 +···an−1Xn−1)+E(anXn) = ··· = E(a1X1)+···E(an−1Xn−1)+E(anXn)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 32

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Proof of Part 1. We know that E(X +Y) = E(X)+E(Y) and it is easy to see that E(aX) = aE(X). (Remember that expected values are sums or integrals and constants can be factored out of sums and integrals.) E(a1X1 +···an−1Xn−1 +anXn) = E

• (a1X1 +···an−1Xn−1)+anXn
• =

E(a1X1 +···an−1Xn−1)+E(anXn) = ··· = E(a1X1)+···E(an−1Xn−1)+E(anXn) = a1E(X1)+···an−1E(Xn−1)+anE(Xn)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

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Proof of Part 2.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 34

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Proof of Part 2.

V

• n

∑

k=1

akXk

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 35

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Proof of Part 2.

V

• n

∑

k=1

akXk

• = E

 

• n

∑

k=1

akXk −E

• n

∑

j=1

ajXj 2 

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 36

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Proof of Part 2.

V

• n

∑

k=1

akXk

• = E

 

• n

∑

k=1

akXk −E

• n

∑

j=1

ajXj 2  = E  

• n

∑

k=1

akXk −

n

∑

j=1

ajE(Xj) 2 

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 37

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Proof of Part 2.

V

• n

∑

k=1

akXk

• = E

 

• n

∑

k=1

akXk −E

• n

∑

j=1

ajXj 2  = E  

• n

∑

k=1

akXk −

n

∑

j=1

ajE(Xj) 2  = E  

• n

∑

k=1

akXk 2 −2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)+

• n

∑

j=1

ajE(Xj) 2 

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 38

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Proof of Part 2.

V

• n

∑

k=1

akXk

• = E

 

• n

∑

k=1

akXk −E

• n

∑

j=1

ajXj 2  = E  

• n

∑

k=1

akXk −

n

∑

j=1

ajE(Xj) 2  = E  

• n

∑

k=1

akXk 2 −2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)+

• n

∑

j=1

ajE(Xj) 2  = E  

• n

∑

k=1

akXk 2 

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 39

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Proof of Part 2.

V

• n

∑

k=1

akXk

• = E

 

• n

∑

k=1

akXk −E

• n

∑

j=1

ajXj 2  = E  

• n

∑

k=1

akXk −

n

∑

j=1

ajE(Xj) 2  = E  

• n

∑

k=1

akXk 2 −2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)+

• n

∑

j=1

ajE(Xj) 2  = E  

• n

∑

k=1

akXk 2  −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 40

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Proof of Part 2.

V

• n

∑

k=1

akXk

• = E

 

• n

∑

k=1

akXk −E

• n

∑

j=1

ajXj 2  = E  

• n

∑

k=1

akXk −

n

∑

j=1

ajE(Xj) 2  = E  

• n

∑

k=1

akXk 2 −2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)+

• n

∑

j=1

ajE(Xj) 2  = E  

• n

∑

k=1

akXk 2  −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• +E

 

• n

∑

j=1

ajE(Xj) 2 

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 41

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Proof of Part 2 (cont.).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 42

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Proof of Part 2 (cont.).

= E  

• n

∑

k=1

akXk 2 −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• +E

 

• n

∑

j=1

ajE(Xj) 2 

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 43

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Proof of Part 2 (cont.).

= E  

• n

∑

k=1

akXk 2 −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• +E

 

• n

∑

j=1

ajE(Xj) 2  = E

• n

∑

j=1

ajXj

n

∑

k=1

akXk

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 44

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Proof of Part 2 (cont.).

= E  

• n

∑

k=1

akXk 2 −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• +E

 

• n

∑

j=1

ajE(Xj) 2  = E

• n

∑

j=1

ajXj

n

∑

k=1

akXk

• −2

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 45

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Proof of Part 2 (cont.).

= E  

• n

∑

k=1

akXk 2 −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• +E

 

• n

∑

j=1

ajE(Xj) 2  = E

• n

∑

j=1

ajXj

n

∑

k=1

akXk

• −2

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj) +

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 46

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Proof of Part 2 (cont.).

= E  

• n

∑

k=1

akXk 2 −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• +E

 

• n

∑

j=1

ajE(Xj) 2  = E

• n

∑

j=1

ajXj

n

∑

k=1

akXk

• −2

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj) +

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj) =

n

∑

j=1 n

∑

k=1

ajakE[XjXk]

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 47

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Proof of Part 2 (cont.).

= E  

• n

∑

k=1

akXk 2 −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• +E

 

• n

∑

j=1

ajE(Xj) 2  = E

• n

∑

j=1

ajXj

n

∑

k=1

akXk

• −2

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj) +

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj) =

n

∑

j=1 n

∑

k=1

ajakE[XjXk] −

n

∑

j=1 n

∑

k=1

ajakE(Xj)E[Xk]

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 48

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Proof of Part 2 (cont.).

= E  

• n

∑

k=1

akXk 2 −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• +E

 

• n

∑

j=1

ajE(Xj) 2  = E

• n

∑

j=1

ajXj

n

∑

k=1

akXk

• −2

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj) +

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj) =

n

∑

j=1 n

∑

k=1

ajakE[XjXk] −

n

∑

j=1 n

∑

k=1

ajakE(Xj)E[Xk] =

n

∑

j=1 n

∑

k=1

ajak

• E(XjXk)−E(Xj)E(Xk)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 49

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Proof of Part 2 (cont.).

= E  

• n

∑

k=1

akXk 2 −E

• 2

n

∑

k=1

akXk

n

∑

j=1

ajE(Xj)

• +E

 

• n

∑

j=1

ajE(Xj) 2  = E

• n

∑

j=1

ajXj

n

∑

k=1

akXk

• −2

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj) +

n

∑

k=1

akE[Xk]

n

∑

j=1

ajE(Xj) =

n

∑

j=1 n

∑

k=1

ajakE[XjXk] −

n

∑

j=1 n

∑

k=1

ajakE(Xj)E[Xk] =

n

∑

j=1 n

∑

k=1

ajak

• E(XjXk)−E(Xj)E(Xk)
• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 50

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Proof of Part 3.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 51

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Proof of Part 3. Recall that the covariance of independent random variables is zero.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 52

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Proof of Part 3. Recall that the covariance of independent random variables is zero. V

• n

∑

k=1

akXk

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 53

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Proof of Part 3. Recall that the covariance of independent random variables is zero. V

• n

∑

k=1

akXk

• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 54

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Proof of Part 3. Recall that the covariance of independent random variables is zero. V

• n

∑

k=1

akXk

• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk) =

n

∑

k=1

akakCov(Xk,Xk)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 55

logo1

Proof of Part 3. Recall that the covariance of independent random variables is zero. V

• n

∑

k=1

akXk

• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk) =

n

∑

k=1

akakCov(Xk,Xk) =

n

∑

k=1

a2

kV (Xk)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 56

logo1

Proof of Part 3. Recall that the covariance of independent random variables is zero. V

• n

∑

k=1

akXk

• =

n

∑

j=1 n

∑

k=1

ajakCov(Xj,Xk) =

n

∑

k=1

akakCov(Xk,Xk) =

n

∑

k=1

a2

kV (Xk)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 57

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Corollary.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 58

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• Corollary. Let X1 and X2 be random variables.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 59

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• Corollary. Let X1 and X2 be random variables. Then

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 60

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• Corollary. Let X1 and X2 be random variables. Then

E(X1 −X2) = E(X1)−E(X2)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 61

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• Corollary. Let X1 and X2 be random variables. Then

E(X1 −X2) = E(X1)−E(X2) and

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 62

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• Corollary. Let X1 and X2 be random variables. Then

E(X1 −X2) = E(X1)−E(X2) and V(X1 −X2) = V(X1)−2Cov(X1,X2)+V(X2).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 63

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• Corollary. Let X1 and X2 be random variables. Then

E(X1 −X2) = E(X1)−E(X2) and V(X1 −X2) = V(X1)−2Cov(X1,X2)+V(X2). Moreover, if X1 and X2 are independent we have

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 64

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• Corollary. Let X1 and X2 be random variables. Then

E(X1 −X2) = E(X1)−E(X2) and V(X1 −X2) = V(X1)−2Cov(X1,X2)+V(X2). Moreover, if X1 and X2 are independent we have V(X1 −X2) = V(X1)+V(X2).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 65

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Proposition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 66

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• Proposition. Let X1,...,Xn be independent normally

distributed random variables

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 67

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• Proposition. Let X1,...,Xn be independent normally

distributed random variables with no assumption made on the means and variances.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables

SLIDE 68

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• Proposition. Let X1,...,Xn be independent normally

distributed random variables with no assumption made on the means and variances. Then any linear combination of these random variables is again normally distributed.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science The Distribution of a Linear Combination of Random Variables