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Important Examples of Expectation

Module by: Nick Kingsbury

Summary: This module gives some important examples of expectation.

We get Moments of a pdf by setting gX=Xn g X X n in this previous equation,
nth order moment EXn=∫xnfXxdx X n x x n fX x (1)
  • n=1 n 1 : 1st order moment, Ex x = Mean value
  • n=2 n 2 : 2nd order moment, Ex2 x 2 = Mean-squared value (Power or energy)
  • n>2 n 2 : Higher order moments, Exn x n , give more detail about fXx fX x .

Central Moments

Central moments are moments about the centre or mean of a distribution,
nth order central moment EX-X¯n=∫x-X¯nfXxdx X X n x x X n fX x (2)
Some important parameters from central moments of a pdf are:
  • Variance, n=2 n 2 :
    σ2=EX-X¯2=∫x-X¯2fXxdx=∫x2fXxdx-2X¯∫xfXxdx+X¯2∫fXxdx=EX2-2X¯2+X¯2=EX2-X¯2 σ 2 X X 2 x x X 2 fX x x x 2 fX x 2 X x x fX x X 2 x fX x X 2 2 X 2 X 2 X 2 X 2 (3)
  • Standard deviation, σ=variance σ variance .
  • Skewness, n=3 n 3 :
    γ=EX-X¯3σ3 γ X X 3 σ 3 (4)
    γ=0 γ 0 if the pdf of XX is symmetric about X¯ X , and becomes more positive if the tail of the distribution is heavier when X>X¯ X X .
  • Kurtosis (or excess), n=4 n 4 :
    κ=EX-X¯4σ4-3 κ X X 4 σ 4 3 (5)
    κ=0 κ 0 for a Gaussian pdf and becomes more positive for distributions with heavier tails.
Note: Skewness and kurtosis are normalized by dividing the central moments by appropriate powers of σσ to make them dimensionless. Kurtosis is usually offset by -3-3 to make it zero for Gaussian pdfs.

Example: Central Moments of a Normal Distribution

The normal (or Gaussian) pdf with zero mean is given by:
fXx=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=∫xnfXxdx=12πσ2∫xnⅇ-x22σ2dx X n x x n fX x 1 2 σ 2 x x n x 2 2 σ 2 (7)
fXx fX x 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.):
Γ12=π Γ 1 2 (13)
Γ1=1 Γ 1 1 (14)
and hence
Γ32=π2 Γ 3 2 2 (15)
Γ2=1 Γ 2 1 (16)
Hence
EXn=0ifn=odd1π2σ2n2Γn+12ifn=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)
    as required for a valid pdf.
    Note: 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 :
    EX=0 X 0 (19)
    so the mean is zero.
  • Variance, n=2 n 2 :
    EX-X¯2=EX2=1π2σ2Γ32=1π2σ2π2=σ2 X X 2 X 2 1 2 σ 2 Γ 3 2 1 2 σ 2 2 σ 2 (20)
    Therefore standard deviation = variance=σ variance σ .
  • Skewness, n=3 n 3 :
    EX3=0 X 3 0 (21)
    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)
    Hence
    κ=EX-X¯4σ4-3=3-3=0 κ X X 4 σ 4 3 3 3 0 (23)

Evaluation of the Gamma Function

From the definition of ΓΓ and substituting u=x2 u x 2 :
Γ12=∫0∞u-12ⅇ-udu=∫0∞x-1ⅇ-x22xdx=2∫0∞ⅇ-x2dx=∫-∞∞ⅇ-x2dx Γ 1 2 u 0 u 1 2 u x 0 x -1 x 2 2 x 2 x 0 x 2 x x 2 (24)
Using the following squaring trick to convert this to a 2-D integral in polar coordinates:
Γ212=∫-∞∞ⅇ-x2dx∫-∞∞ⅇ-y2dy=∫-∞∞∫-∞∞ⅇ-x2+y2dxdy=∫-ππ∫0∞ⅇ-r2rdrdθ=2π-12ⅇ-r2|0∞=π Γ 1 2 2 x x 2 y y 2 y x x 2 y 2 θ r 0 r 2 r 2 0 1 2 r 2 (25)
and so (ignoring the negative square root):
Γ12=π≈1.7725 Γ 1 2 1.7725 (26)
Hence, using Γz+1=zΓz Γ z 1 z Γ z :
Γ32527292…=12π34π158π10516π… Γ 3 2 5 2 7 2 9 2 … 1 2 3 4 15 8 105 16 … (27)
The case for z=1 z 1 is straightforward:
Γ1=∫0∞u0ⅇ-udu=-ⅇ-u|0∞=1 Γ 1 u 0 u 0 u 0 u 1 (28)
so
Γ2345…=12624… Γ 2 3 4 5 … 1 2 6 24 … (29)

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