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Chapter 4: Part a – The Expectation and Variance of Distributions
We will be discussing
 The Algebra of Expectation
 The Algebra of Variance
 The Normal Distribution
(These topics are needed for Chapter 5)
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Slide 4a.1
Distributions
The Expectation of a Discrete Random Variable
Assume we have a random variable, ai.
If ai can take on only certain values: 1, 2, ···, J, then we have
as the definition of Expectation of ai, or E(ai)
J
E(a i )   Pr(a i  j)  j
j
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Slide 4a.2
Distributions
The Expectation of a Continuous Variable
Imagine we have a scalar, ai, that can take on
any possible values with probability f(ai), i. e.
f(ai)
ai
By definition, the expectation of that scalar is

E(a i )   f (a i ) a i da i

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Slide 4a.3
Distributions
Three Rules for E(·)
The Expectation of a Constant is That Constant
E(c) = c
The Expectation of a Sum is the Sum of the Expectations
E(a + b) = E(a) + E(b)
In the Expectation of a Linear Combination,
a Constant Matrix Can Pass Through E(·)
E(Da) = DE(a)
E(a′F) = E(a′)F
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Slide 4a.4
Distributions
The Variance of a Random Variable
The Variance of the Random Vector a is Given by
V(a)  E[a  E(a)][a  E(a)] 
  a1 
 a1 
a1  
  
a 
a  
a



V  2   E   2   E  2  


  
 


  

  a n 
a n 
a n  

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a
1
a 2  a n   Ea 1
a2



 a n 



Slide 4a.5
Distributions
The Variance of a Mean Centered Vector
If E(a) = 0, i. e. a is mean centered, we have just
V(a) = E(aa′)
  a1 
 
 a
E   2  a 1
 
 a n 
a2



 an  


NB – just because E(a) = 0 doesn’t mean that E(aa′) = 0!
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Slide 4a.6
Distributions
The Variance Matrix
For mean centered a, we have V(a) = E(aa′)
Cov(a 1 , a 2 )
 V (a 1 )
Cov(a , a )
V (a 2 )
2
1
V(a)  




Cov(a n , a 1 ) Cov(a n , a 2 )
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 Cov(a 1 , a n ) 
 Cov(a 2 , a n )






V (a n ) 
Slide 4a.7
Distributions
Two Rules for V(·)
Adding a Constant Vector Does Not Change the Variance
V(a + c) = V(a)
The Variance of a Linear Combination is a Quadratic Form
V(Da) = DV(a)D′
Hint: The Above Theorem Will Figure Many Many Times in What Is to Come!
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Slide 4a.8
Distributions
The Normal Density Function
Consider a random scalar x. Under the normal distribution,
the probability that x takes on the value xa is given by the
equation
1
  ( x a  ) 2 
Pr(x )  Pr(x  x a ) 
exp 
.
2 2
2


1.0
Pr(x)
0
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xa

x
Slide 4a.9
Distributions
The Standard Normal Density
If  = 0 and 2 = 1, so that z = (x - ) / 
the expression simplifies to
1
 z a2 
Pr(z)  Pr(z  z a ) 
exp 

2
 2 
  (z a )
Note very common notation
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Slide 4a.10
Distributions
The Normal Distribution Function
According to the normal distribution function,
the probability that the random scalar x is less
than or equal to some value xb is
1 x
  ( x a  ) 2 
Pr[x  x b ] 
 exp
 dx
2 2
2  

b
1.0
Pr(x)
0
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xb
x
Slide 4a.11
Distributions
Other Notational Conventions
For our scalar x that is distributed according to the normal
distribution function, we say x ~ N(, 2).
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Slide 4a.12
Distributions
The Standardized Normal Distribution
If we set z = (x - ) /  then
zb
Pr(z  z b )   (z) dz

 (z b )
Again, note the notational convention and that
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(z)
 (z) .
z
Slide 4a.13
Distributions
The Normal Ogive
zb
Pr(z  z b )   (z) dz  (z b )

1.0
(z)
0
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zb
Slide 4a.14
Distributions