The normal (or Gaussian) pdf with zero mean is given by:
fX
x=12πσ2e−x22σ2
fX
x
1
2
σ
2
x
2
2
σ
2
(6)
What is the
nnth order central
moment for the Gaussian?
Since the mean is zero, the nnth
order central moment is given by
EXn=∫xn
fX
xdx=12πσ2∫xne−x22σ2dx
X
n
x
x
n
fX
x
1
2
σ
2
x
x
n
x
2
2
σ
2
(7) is a function of
x2
x
2
and therefore is symmetric about zero. So all the
odd-order moments will integrate to zero (including the
lst-order moment, giving zero mean). The even-order moments
are then given by:
EXn=22πσ2∫0∞xne−x22σ2dx
X
n
2
2
σ
2
x
0
x
n
x
2
2
σ
2
(8)
where
nn is even. The integral is
calculated by substituting
u=x22σ2
u
x
2
2
σ
2
to give:
∫0∞xne−x22σ2dx=122σ2n+12∫0∞un−12e−udu=122σ2n+12Γn+12
x
0
x
n
x
2
2
σ
2
1
2
2
σ
2
n
1
2
u
0
u
n
1
2
u
1
2
2
σ
2
n
1
2
Γ
n
1
2
(9)
Here
Γz
Γ
z
is the Gamma function, which is defined as an
integral for all real
z>0
z
0
and is rather like the factorial function but
generalized to allow non-integer arguments. Values of the
Gamma function can be found in mathematical tables. It is
defined as follows:
Γz=∫0∞uz−1e−udu
Γ
z
u
0
u
z
1
u
(10)
and has the important (factorial-like) property that
∀z,z≠0:Γz+1=zΓz
z
z
0
Γ
z
1
z
Γ
z
(11)
∀z,z∈Z∧(z>0):Γz+1=z!
z
z
z
0
Γ
z
1
z
(12)
The following results hold for the Gamma function (see below
for a way to evaluate
Γ12
Γ
1
2
etc.):
and hence
Hence
EXn={0 if n=odd1π2σ2n2Γn+12 if n=even
X
n
0
n
odd
1
2
σ
2
n
2
Γ
n
1
2
n
even
(17)-
Valid pdf,
n=0
n
0
:
EX0=1πΓ12=1
X
0
1
Γ
1
2
1
(18)
The normalization factor
12πσ2
1
2
σ
2
in the expression for the pdf of a unit
variance Gaussian (e.g.
Equation 6) arises directly from the above result.
-
Mean,
n=1
n
1
:
so the mean is zero.
-
Variance,
n=2
n
2
:
EX−X-2=EX2=1π(2σ2)Γ32=1π(2σ2)π2=σ2
X
X
2
X
2
1
2
σ
2
Γ
3
2
1
2
σ
2
2
σ
2
(20) -
Skewness,
n=3
n
3
:
so the skewness is zero.
-
Kurtosis,
n=4
n
4
:
EX−X-4=EX4=1π2σ22Γ52=1π2σ223π4=3σ4
X
X
4
X
4
1
2
σ
2
2
Γ
5
2
1
2
σ
2
2
3
4
3
σ
4
(22)
κ=EX−X-4σ4−3=3−3=0
κ
X
X
4
σ
4
3
3
3
0
(23)
