Download The Normal Distribution

2 - Some Basic Probability Distributions
This second chapter contains a short summary of the basic properties of some of the
probability distributions that we will be concerned with, i.e. the normal, gamma,
chi-square and exponential distributions.
2.1
The Normal Distribution
A random variable X is said to have a normal distribution with mean  and standard
deviation  if its probability density function has the form
(1)
f(t) = f(t; ,) =
1
2
2
e-(x-) /(2 )
2 
It is assumed  > 0. In this case we shall say X is a normal random variable with mean 
and standard deviation  and write X ~ N(, ).
This section collects some basic properties of normal random variables, all of which are
well known; see Hogg and Tanis [6].
Proposition 1. If m > 0 and  > 0 then

(2)
 1 e-(x-)2/(22) dx = 1
 2 
-
which confirms that f(t) defined by (1) is a valid density function.
Proof. If one makes the change of variables z = (x - )/( 2) then one has



-

1
2
2
e-(x-) /(2 ) dx =
2 
 1 e-z2 dz. It is well known that this latter integral is one. 
 
-
2.1 - 1
Proposition 2. If X is a normal random variable with mean  and standard deviation 
then the mean of X is 
1
2
2
e-x /(2 ).
2 
Proof. The density function of X -  is
This is an even function so its
mean is zero. So the mean of X is . 
Proposition 3. If X is a normal random variable with mean  and standard deviation 
then the variance of X is 2.
Proof. X and X -  have the same variance, so it suffices to show that the variance of
X -  is 2. It was noted in the proof of the previous proposition that the density function
of Y = X -  is fY(y) =
1
2
2
e-y /(2 ). The variance of Z = Y/ is the variance of Y/2, so
2 
it suffices to show that the variance of Z is one. The density function of Z is


1 -z2/2
1
1
2
2 -z2/2

fZ(z) = fY(z) =
e . One has var(Z) =  z2 
e-z /2 dz =
dz.
ze
2 - 

2
  2
-
2/2
Integrate by parts letting u = z and dv = ze-z
var(Z) = -
1
2

2
ze-z /2 |-
+
1
-z2/2

e
2 - 
2
dz. One has du = dz and v = - e-z /2. So
dz = 0 – 0 + 1 = 1. 
2.1 - 2