Normal Random Variables
Lecture XII
Univariate Normal Distribution.
• Definition 5.2.1. The normal density is
given by
2

1
1 x  m  
f x  
exp 
    x  , s  0
2
2 s
 2 s

• When X has the above density, we write
symbolically X~N(m,s2).
Mean and Variance of Normal
Distribution
• Theorem 5.2.1. Let X be distributed
N(m,s2). Then E[X]= m and V[X]= s2
EX   


z
 1 x  m 
x
exp


2
2
2 s
 2 s
2
1
xm
s
 x  zs  m
dx  sdz

 dx

1
 1 2

EX   
zs  m  exp  z  dz

2 s
 2 

1
 1 2
2

zs exp  z  dz

2 s
 2 

1
 1 2
 m
exp  z  dz

2
 2 





1
 1 2
 1 2
2
zs exp  z  dz  C  z exp  z  dz

2 s
 2 
 2 


 1 2

 C  exp  z   0

2

 

2

1
1 x  m  
2

V X  
x  m  exp 
dx

2

2 s 
 2 s


1
 1 2
2


zs  m  m  exp  z sdz

2 s 
 2 
1  2 2
 1 2

z s exp  z  dz

2 
 2 

 1 2
u   z dv   z exp  z 
 2 
 1 2
du  1 v  exp  z 
 2 




 1 2
 1 2
 1 2

 z z exp  2 z  dz   z exp  2 z    exp  2 z  dz


• The first term of the integration by parts is
clearly zero while the second is defined by
polar integral. Thus,
V X   0  s
2



1
 1 2
2
exp  z  dz  s
2
 2 
Normality of a Linear
Transformation
• Theorem 5.2.2. Let X be distributed N(m,s2)
and let Y=a+bX. Then we have
Y~N(a+bm,b2s2).
• This theorem can be demonstrated using
Theorem 3.6.1 (the theorem on changes in
variables):
d
gy  f  y
dy

1

1
 x   a  bx  
1
y 
y a
b
d  y  1

dy
b
1
2

 y a
 

 m  

1
1 b

 
gy 
exp 
 2

s2
2 s b




 1  y  a  bm 2 

exp 

2 2
s b
2 s b
 2

1
• Note that probabilities can be derived for
any normal based on the standard normal
integral. Specifically, in order to find the
probability that X~N(10,4) lies between 4
and 8 (P[4<X<8]) implies:
P4  X  8  PX  8  PX  4
• Transforming each boundary to standard
normal space:
x1  10  6
z1 

 3
2
2
x2  10  2
z2 

 1
2
2
• Thus, the equivalent boundary becomes
P 3  z  1  Pz  1  Pz  3
where z is a standard normal variable.
These values can be found in a standard
normal table as P[z<-1]=.1587 and
P[z<-3]=.0013.
• As a first step, I want to graph two
distributions with the same mean and
variance.
• I want to start with a binomial distribution with
probability .5 and 20 draws.
• This distribution has a mean of 10 and a
variance of 5.
0.4500
0.4000
0.3500
0.3000
0.2500
0.2000
0.1500
0.1000
0.0500
0.0000
0
5
10
15
Binomial
Normal
20
25
0.0900
0.0800
0.0700
0.0600
0.0500
0.0400
0.0300
0.0200
0.0100
0.0000
0
-0.0100
20
40
60
80
100
120