Download The Gaussian Distribution

The Gaussian Distribution
(F.G.Stremler, intro to communication systems 3/e, pp482--484)
A widely known probability density function is the Gaussian or "normal" density
function. It arises when a large number of independent factors (within some fairly
broad restrictions) contribute additively to an end result. This result, known as a
"central-limit theorem," states that the sum of N independent random variables
approaches a gaussian distribution as N becomes large. This result is not dependent
upon the distribution of each random variable (within broad restrictions) as long as
the contribution of each is small compared to the sum. This theorem is not easy to
prove and interpret correctly, however, and we shall not pursue the topic further
except to observe that electrical noise often results from summations of a large
number of random effects and therefore tends to be gaussian-distributed.
The gaussian pdf is continuous and is defined by
2
2
1
e ( x m ) / 2 ,
2 
where m, σ2 are the mean and the variance respectively. For convenience in tabulating
numerical values, we define a normalized gaussian pdf having zero mean and unit
variance,
1 z2 / 2
p( z ) 
e
,
2
p( x ) 
The cumulative distribution function corresponding to the gaussian pdf is
m  k
2
2
1
e ( x m ) / 2 dx,

2 
to normalize, let z=(x-m)/σ, so that,
k
1 z2 / 2
P{ X  (m  k )}  
e
dz,

2
P{ X  (m  k )}  
This integral cannot be evaluated in closed form and requires numerical evaluation.
The procedure is to first normalize and then refer to a table of tabulated values.
For convenice in referring to tabulated value, we define the function Q(x):
1  z2 / 2
e
dz,
2 x
Q(x) can be approximated by
2
1
1
Q( x ) 
[1  2 ]e  x / 2
x
x 2
or
1
1  x2 / 2
Q( x )  [
]
e
(1  a ) x  a x 2  b 2
where a  1 and b  2 .
Q( x) 

As we observe in Fig. 8. 10, Q(x) represents the area beneath a plot of the
normalized gaussian pdf over ( x ,  ) ; the remainder of the area beneath this pdf is
[ 1 - Q(x)] Therefore our answer to the above example can be expressed as
P{ X  (m  k )}  
k

1 z2 / 2
e
dz  1  Q(k )
2
For example,

2
2
1
1 z2 / 2
e  x / 2 dx  
e
dz  Q(3)
3
3
2 
2
In the examination, you can stop at Q(3).
P{ X  3 }  

(0.0013)