measures the spread of the distribution about its center of mass. These quantities are also
known, respectively, as the mean (moment) of X and the second moment of X about the mean. Other moments
give added information. For example, the third moment about the mean E[(X-μX)3]E[(X-μX)3] gives
information about the skew, or asymetry, of the distribution about the mean. We investigate
further along these lines by examining the expectation of certain functions of X. Each of these
functions involves a parameter, in a manner that completely determines the distribution.
For reasons noted below, we refer to these as transforms. We consider three of
the most useful of these.
We define each of three transforms, determine some key properties, and use them to study
various probability distributions associated with random variables. In the section on integral transforms, we
show their relationship to well known integral transforms. These have been studied extensively
and used in many other applications,
which makes it possible to utilize the considerable literature on these transforms.
Definition. The moment generating functionMX for random variable X (i.e., for
its distribution) is the function
M
X
(
s
)
=
E
[
e
s
X
]
(
s
is
a
real
or
complex
parameter)
M
X
(
s
)
=
E
[
e
s
X
]
(
s
is
a
real
or
complex
parameter)
(2)
The characteristic functionφX for random variable X is
φ
X
(
u
)
=
E
[
e
i
u
X
]
(
i
2
=
-
1
,
u
is
a
real
parameter)
φ
X
(
u
)
=
E
[
e
i
u
X
]
(
i
2
=
-
1
,
u
is
a
real
parameter)
(3)
The generating functiongX(s)gX(s) for a nonnegative, integer-valued random variable X is
g
X
(
s
)
=
E
[
s
X
]
=
∑
k
s
k
P
(
X
=
k
)
g
X
(
s
)
=
E
[
s
X
]
=
∑
k
s
k
P
(
X
=
k
)
(4)
The generating function E[sX]E[sX] has meaning for more general random variables,
but its usefulness is greatest for nonnegative, integer-valued variables, and we limit our
consideration to that case.
The defining expressions display similarities which show useful relationships. We note
two which are particularly useful.
M
X
(
s
)
=
E
[
e
s
X
]
=
E
[
(
e
s
)
X
]
=
g
X
(
e
s
)
and
φ
X
(
u
)
=
E
[
e
i
u
X
]
=
M
X
(
i
u
)
M
X
(
s
)
=
E
[
e
s
X
]
=
E
[
(
e
s
)
X
]
=
g
X
(
e
s
)
and
φ
X
(
u
)
=
E
[
e
i
u
X
]
=
M
X
(
i
u
)
(5)
Because of the latter relationship, we ordinarily use the moment generating function instead
of the characteristic function to avoid writing the complex unit i. When desirable,
we convert easily by the change of variable.
The integral transform character of these entities implies that there is essentially a
one-to-one relationship between the transform and the distribution.
Moments
The name and some of the importance of the moment generating function arise from the
fact that the derivatives of MX evaluateed at s=0s=0 are the moments about the origin. Specifically
M
X
(
k
)
(
0
)
=
E
[
X
k
]
,
provided
the
k
th
moment
exists
M
X
(
k
)
(
0
)
=
E
[
X
k
]
,
provided
the
k
th
moment
exists
(6)
Since expectation is an integral and because of the regularity of the integrand,
we may differentiate inside the integral with respect to the parameter.
M
X
'
(
s
)
=
d
d
s
E
[
e
s
X
]
=
E
d
d
s
e
s
X
=
E
[
X
e
s
X
]
M
X
'
(
s
)
=
d
d
s
E
[
e
s
X
]
=
E
d
d
s
e
s
X
=
E
[
X
e
s
X
]
(7)
Upon setting s=0s=0, we have MX'(0)=E[X]MX'(0)=E[X]. Repeated differentiation gives
the general result. The corresponding result for the characteristic function is
φ(k)(0)=ikE[Xk]φ(k)(0)=ikE[Xk].
The density function is fX(t)=λe-λtfX(t)=λe-λt for t≥0t≥0.
M
X
(
s
)
=
E
[
e
s
X
]
=
∫
0
∞
λ
e
-
(
λ
-
s
)
t
d
t
=
λ
λ
-
s
M
X
(
s
)
=
E
[
e
s
X
]
=
∫
0
∞
λ
e
-
(
λ
-
s
)
t
d
t
=
λ
λ
-
s
(8)
M
X
'
(
s
)
=
λ
(
λ
-
s
)
2
M
X
'
'
(
s
)
=
2
λ
(
λ
-
s
)
3
M
X
'
(
s
)
=
λ
(
λ
-
s
)
2
M
X
'
'
(
s
)
=
2
λ
(
λ
-
s
)
3
(9)
E
[
X
]
=
M
X
'
(
0
)
=
λ
λ
2
=
1
λ
E
[
X
2
]
=
M
X
'
'
(
0
)
=
2
λ
λ
3
=
2
λ
2
E
[
X
]
=
M
X
'
(
0
)
=
λ
λ
2
=
1
λ
E
[
X
2
]
=
M
X
'
'
(
0
)
=
2
λ
λ
3
=
2
λ
2
(10)
From this we obtain Var [X]=2/λ2-1/λ2=1/λ2 Var [X]=2/λ2-1/λ2=1/λ2.
The generating function does not lend itself readily to computing moments, except that
g
X
'
(
s
)
=
∑
k
=
1
∞
k
s
k
-
1
P
(
X
=
k
)
so
that
g
X
'
(
1
)
=
∑
k
=
1
∞
k
P
(
X
=
k
)
=
E
[
X
]
g
X
'
(
s
)
=
∑
k
=
1
∞
k
s
k
-
1
P
(
X
=
k
)
so
that
g
X
'
(
1
)
=
∑
k
=
1
∞
k
P
(
X
=
k
)
=
E
[
X
]
(11)
For higher order moments, we may convert the generating function to the moment generating function
by replacing s with es, then work with MX and its derivatives.
P(X=k)=e-μμkk!,k≥0P(X=k)=e-μμkk!,k≥0, so that
g
X
(
s
)
=
e
-
μ
∑
k
=
0
∞
s
k
μ
k
k
!
=
e
-
μ
∑
k
=
0
∞
(
s
μ
)
k
k
!
=
e
-
μ
e
μ
s
=
e
μ
(
s
-
1
)
g
X
(
s
)
=
e
-
μ
∑
k
=
0
∞
s
k
μ
k
k
!
=
e
-
μ
∑
k
=
0
∞
(
s
μ
)
k
k
!
=
e
-
μ
e
μ
s
=
e
μ
(
s
-
1
)
(12)
We convert to MX by replacing s with es to get MX(s)=eμ(es-1)MX(s)=eμ(es-1). Then
M
X
'
(
s
)
=
e
μ
(
e
s
-
1
)
μ
e
s
M
X
'
'
(
s
)
=
e
μ
(
e
s
-
1
)
[
μ
2
e
2
s
+
μ
e
s
]
M
X
'
(
s
)
=
e
μ
(
e
s
-
1
)
μ
e
s
M
X
'
'
(
s
)
=
e
μ
(
e
s
-
1
)
[
μ
2
e
2
s
+
μ
e
s
]
(13)
so that
E
[
X
]
=
M
X
'
(
0
)
=
μ
,
E
[
X
2
]
=
M
X
'
'
(
0
)
=
μ
2
+
μ
,
and
Var
[
X
]
=
μ
2
+
μ
-
μ
2
=
μ
E
[
X
]
=
M
X
'
(
0
)
=
μ
,
E
[
X
2
]
=
M
X
'
'
(
0
)
=
μ
2
+
μ
,
and
Var
[
X
]
=
μ
2
+
μ
-
μ
2
=
μ
(14)
These results agree, of course, with those found by direct computation with the distribution.
Operational properties
We refer to the following as operational properties.
- (T1): If Z=aX+bZ=aX+b, then
MZ(s)=ebsMX(as),φZ(u)=eiubφX(au),gZ(s)=sbgX(sa)MZ(s)=ebsMX(as),φZ(u)=eiubφX(au),gZ(s)=sbgX(sa)(15)
For the moment generating function, this pattern follows from
E[e(aX+b)s]=sbsE[e(as)X]E[e(aX+b)s]=sbsE[e(as)X](16)
Similar arguments hold for the other two.
- (T2): If the pair {X,Y}{X,Y} is independent, then
MX+Y(s)=MX(s)MY(s),φX+Y(u)=φX(u)φY(u),gX+Y(s)=gX(s)gY(s)MX+Y(s)=MX(s)MY(s),φX+Y(u)=φX(u)φY(u),gX+Y(s)=gX(s)gY(s)(17)
For the moment generating function, esXesX and esYesY form an independent pair for
each value of the parameter s. By the product rule for expectation
E[es(X+Y)]=E[esXesY]=E[esX]E[esY]E[es(X+Y)]=E[esXesY]=E[esX]E[esY](18)
Similar arguments are used for the other two transforms.
A partial converse for (T2) is as follows:
- (T3): If MX+Y(s)=MX(s)MY(s)MX+Y(s)=MX(s)MY(s), then the pair {X,Y}{X,Y} is uncorrelated.
To show this, we obtain two expressions for E[(X+Y)2]E[(X+Y)2], one by direct expansion and
use of linearity, and the other by taking the second derivative of the moment generating
function.
E[(X+Y)2]=E[X2]+E[Y2]+2E[XY]E[(X+Y)2]=E[X2]+E[Y2]+2E[XY](19)
MX+Y''(s)=[MX(s)MY(s)]''=MX''(s)MY(s)+MX(s)MY''(s)+2MX'(s)MY'(s)MX+Y''(s)=[MX(s)MY(s)]''=MX''(s)MY(s)+MX(s)MY''(s)+2MX'(s)MY'(s)(20)
On setting s=0s=0 and using the fact that MX(0)=MY(0)=1MX(0)=MY(0)=1, we have
E[(X+Y)2]=E[X2]+E[Y2]+2E[X]E[Y]E[(X+Y)2]=E[X2]+E[Y2]+2E[X]E[Y](21)
which implies the equality E[XY]=E[X]E[Y]E[XY]=E[X]E[Y].
Note that we have not shown that being uncorrelated implies the product rule.
We utilize these properties in determining the moment generating and generating functions
for several of our common distributions.
Some discrete distributions
-
Indicator function
X
=
I
E
P
(
E
)
=
p
X
=
I
E
P
(
E
)
=
p
g
X
(
s
)
=
s
0
q
+
s
1
p
=
q
+
p
s
M
X
(
s
)
=
g
X
(
e
s
)
=
q
+
p
e
s
g
X
(
s
)
=
s
0
q
+
s
1
p
=
q
+
p
s
M
X
(
s
)
=
g
X
(
e
s
)
=
q
+
p
e
s
(22)
- Simple random variableX=∑i=1ntiIAiX=∑i=1ntiIAi (primitive form) P(Ai)=piP(Ai)=pi
MX(s)=∑i=1nestipiMX(s)=∑i=1nestipi(23)
- Binomial(n,p)(n,p). X=∑i=1nIEiwith{IEi:1≤i≤n}iidP(Ei)=pX=∑i=1nIEiwith{IEi:1≤i≤n}iidP(Ei)=p
We use the product rule for sums of independent random variables and the generating
function for the indicator function.
gX(s)=∏i=1n(q+ps)=(q+ps)nMX(s)=(q+pes)ngX(s)=∏i=1n(q+ps)=(q+ps)nMX(s)=(q+pes)n(24)
- Geometric(p)(p). P(X=k)=pqk∀k≥0P(X=k)=pqk∀k≥0E[X]=q/pE[X]=q/p
We use the formula for the geometric series to get
gX(s)=∑k=0∞pqksk=p∑k=0∞(qs)k=p1-qsMX(s)=p1-qesgX(s)=∑k=0∞pqksk=p∑k=0∞(qs)k=p1-qsMX(s)=p1-qes(25)
- Negative binomial(m,p)(m,p)
If Ym is the number of the trial in a Bernoulli sequence on which the mth success occurs, and
Xm=Ym-mXm=Ym-m is the number of failures before the mth success, then
P(Xm=k)=P(Ym-m=k)=C(-m,k)(-q)kpmP(Xm=k)=P(Ym-m=k)=C(-m,k)(-q)kpm(26)
whereC(-m,k)=-m(-m-1)(-m-2)⋯(-m-k+1)k!whereC(-m,k)=-m(-m-1)(-m-2)⋯(-m-k+1)k!(27)
The power series expansion about t=0t=0 shows that
(1+t)-m=1+C(-m,1)t+C(-m,2)t2+⋯for-1<t<1(1+t)-m=1+C(-m,1)t+C(-m,2)t2+⋯for-1<t<1(28)
Hence
MXm(s)=pm∑k=0∞C(-m,k)(-q)kesk=p1-qesmMXm(s)=pm∑k=0∞C(-m,k)(-q)kesk=p1-qesm(29)
Comparison with the moment generating function for the geometric distribution shows that
Xm=Ym-mXm=Ym-m has the same distribution as the sum of m iid random variables, each
geometric (p)(p). This suggests that the sequence is characterized by independent, successive
waiting times to success. This also shows that the expectation and variance of Xm are
m times the expectation and variance for the geometric. Thus
E[Xm]=mq/pand Var [Xm]=mq/p2E[Xm]=mq/pand Var [Xm]=mq/p2(30)
- Poisson(μ)(μ)P(X=k)=e-μμkk!∀k≥0P(X=k)=e-μμkk!∀k≥0
In Example 2, above, we establish gX(s)=eμ(s-1)gX(s)=eμ(s-1) and MX(s)=eμ(es-1)MX(s)=eμ(es-1).
If {X,Y}{X,Y} is an independent pair, with X∼X∼ Poisson (λ)(λ) and Y∼Y∼ Poisson (μ)(μ),
then Z=X+Y∼Z=X+Y∼ Poisson (λ+μ)(λ+μ). Follows from (T1) and product of exponentials.
Some absolutely continuous distributions
- Uniform on (a,b)(a,b)fX(t)=1b-aa<t<bfX(t)=1b-aa<t<b
MX(s)=∫estfX(t)dt=1b-a∫abestdt=esb-esas(b-a)MX(s)=∫estfX(t)dt=1b-a∫abestdt=esb-esas(b-a)(31)
- Symmetric triangular(-c,,c)(-c,,c)
fX(t)=I[-c,0)(t)c+tc2+I[0,c](t)c-tc2fX(t)=I[-c,0)(t)c+tc2+I[0,c](t)c-tc2(32)
MX(s)=1c2∫-c0(c+t)estdt+1c2∫0c(c-t)estdt=ecs+e-cs-2c2s2MX(s)=1c2∫-c0(c+t)estdt+1c2∫0c(c-t)estdt=ecs+e-cs-2c2s2(33)
=ecs-1cs·1-e-cscs=MY(s)MZ(-s)=MY(s)M-Z(s)=ecs-1cs·1-e-cscs=MY(s)MZ(-s)=MY(s)M-Z(s)(34)
where MY is the moment generating function for Y∼Y∼ uniform (0,c)(0,c) and similarly
for MZ. Thus, X has the same distribution as the difference of two independent random
variables, each uniform on (0,c)(0,c).
- Exponential(λ)(λ)fX(t)=λe-λt,t≥0fX(t)=λe-λt,t≥0
In example 1, above, we show that MX(s)=λλ-sMX(s)=λλ-s.
- Gamma(α,λ)(α,λ)fX(t)=1Γ(α)λαtα-1e-λtt≥0fX(t)=1Γ(α)λαtα-1e-λtt≥0
MX(s)=λαΓ(α)∫0∞tα-1e-(λ-s)tdt=λλ-sαMX(s)=λαΓ(α)∫0∞tα-1e-(λ-s)tdt=λλ-sα(35)
For α=nα=n, a positive integer,
MX(s)=λλ-snMX(s)=λλ-sn(36)
which shows that in this case X has the distribution of the sum of n independent
random variables each exponential (λ)(λ).
- Normal(μ,σ2)(μ,σ2).
- The standardized normal, Z∼N(0,1)Z∼N(0,1)
MZ(s)=12π∫-∞∞este-t2/2dtMZ(s)=12π∫-∞∞este-t2/2dt(37)
Now st-t22=s22-12(t-s)2st-t22=s22-12(t-s)2 so that
MZ(s)=es2/212π∫-∞∞e-(t-s)2/2dt=es2/2MZ(s)=es2/212π∫-∞∞e-(t-s)2/2dt=es2/2(38)
since the integrand (including the constant 1/2π1/2π) is the density for N(s,1)N(s,1).
- X=σZ+μX=σZ+μ implies by property (T1)
MX(s)=esμeσ2s2/2=expσ2s22+sμMX(s)=esμeσ2s2/2=expσ2s22+sμ(39)
Suppose {X,Y}{X,Y} is an independent pair with X∼N(μX,σX2)X∼N(μX,σX2) and
Y∼N(μY,σY2)Y∼N(μY,σY2). Let Z=aX+bY+cZ=aX+bY+c. Then Z is normal, for
by properties of expectation and variance
μ
Z
=
a
μ
X
+
b
μ
Y
+
c
and
σ
Z
2
=
a
2
σ
X
2
+
b
2
σ
Y
2
μ
Z
=
a
μ
X
+
b
μ
Y
+
c
and
σ
Z
2
=
a
2
σ
X
2
+
b
2
σ
Y
2
(40)
and by the operational properties for the moment generating function
M
Z
(
s
)
=
e
s
c
M
X
(
a
s
)
M
Y
(
b
s
)
=
exp
(
a
2
σ
X
2
+
b
2
σ
Y
2
)
s
2
2
+
s
(
a
μ
X
+
b
μ
Y
+
c
)
M
Z
(
s
)
=
e
s
c
M
X
(
a
s
)
M
Y
(
b
s
)
=
exp
(
a
2
σ
X
2
+
b
2
σ
Y
2
)
s
2
2
+
s
(
a
μ
X
+
b
μ
Y
+
c
)
(41)
=
exp
σ
Z
2
s
2
2
+
s
μ
Z
=
exp
σ
Z
2
s
2
2
+
s
μ
Z
(42)
The form of MZ shows that Z is normally distributed.
Moment generating function and simple random variables
Suppose X=∑i=1ntiIAiX=∑i=1ntiIAi in canonical form. That is,
Ai is the event {X=ti}{X=ti} for each of the distinct values in the range of X,
with pi=P(Ai)=P(X=ti)pi=P(Ai)=P(X=ti). Then the moment generating function for X is
M
X
(
s
)
=
∑
i
=
1
n
p
i
e
s
t
i
M
X
(
s
)
=
∑
i
=
1
n
p
i
e
s
t
i
(43)
The moment generating function MX is thus related directly and simply to the distribution
for random variable X.
Consider the problem of determining the sum of an independent pair {X,Y}{X,Y} of
simple random variables. The moment generating function for the sum is the product of
the moment generating functions. Now if Y=∑j=1mujIBjY=∑j=1mujIBj, with P(Y=uj)=πjP(Y=uj)=πj, we have
M
X
(
s
)
M
Y
(
s
)
=
∑
i
=
1
n
p
i
e
s
t
i
∑
j
=
1
m
π
j
e
s
u
j
=
∑
i
,
j
p
i
π
j
e
s
(
t
i
+
u
j
)
M
X
(
s
)
M
Y
(
s
)
=
∑
i
=
1
n
p
i
e
s
t
i
∑
j
=
1
m
π
j
e
s
u
j
=
∑
i
,
j
p
i
π
j
e
s
(
t
i
+
u
j
)
(44)
The various values are sums ti+ujti+uj of pairs (ti,uj)(ti,uj) of values. Each
of these sums has probability piπjpiπj for the values corresponding to ti,ujti,uj.
Since more than one pair sum may have the same value, we need to sort the values,
consolidate like values and add the probabilties for like values to achieve the
distribution for the sum. We have an m-function mgsum for
achieving this directly. It produces the pair-products for the probabilities and the pair-sums for
the values, then performs a csort operation. Although not directly dependent upon the
moment generating function analysis, it produces the same result as that produced by multiplying
moment generating functions.
Suppose the pair {X,Y}{X,Y} is independent with distributions
X
=
[
1
3
5
7
]
Y
=
[
2
3
4
]
P
X
=
[
0
.
2
0
.
4
0
.
3
0
.
1
]
P
Y
=
[
0
.
3
0
.
5
0
.
2
]
X
=
[
1
3
5
7
]
Y
=
[
2
3
4
]
P
X
=
[
0
.
2
0
.
4
0
.
3
0
.
1
]
P
Y
=
[
0
.
3
0
.
5
0
.
2
]
(45)
Determine the distribution for Z=X+YZ=X+Y.
X = [1 3 5 7];
Y = 2:4;
PX = 0.1*[2 4 3 1];
PY = 0.1*[3 5 2];
[Z,PZ] = mgsum(X,Y,PX,PY);
disp([Z;PZ]')
3.0000 0.0600
4.0000 0.1000
5.0000 0.1600
6.0000 0.2000
7.0000 0.1700
8.0000 0.1500
9.0000 0.0900
10.0000 0.0500
11.0000 0.0200
This could, of course, have been achieved by using icalc and csort, which has the advantage
that other functions of X and Y may be handled. Also, since the random
variables are nonnegative, integer-valued, the MATLAB convolution function may be
used (see Example 7). By repeated use of the function mgsum, we
may obtain the distribution for the sum of more than two simple random variables. The
m-functions mgsum3 and mgsum4 utilize this strategy.
The techniques for simple random variables may be used with the simple approximations to
absolutely continuous random variables.
The moment generating functions for the uniform and the symmetric triangular show that
the latter appears naturally as the difference of two uniformly distributed random
variables. We consider X and Y iid, uniform on [0,1].
tappr
Enter matrix [a b] of x-range endpoints [0 1]
Enter number of x approximation points 200
Enter density as a function of t t<=1
Use row matrices X and PX as in the simple case
[Z,PZ] = mgsum(X,-X,PX,PX);
plot(Z,PZ/d) % Divide by d to recover f(t)
% plotting details --- see Figure 1
The generating function
The form of the generating function for a nonnegative, integer-valued random variable
exhibits a number of important properties.
X
=
∑
k
=
0
∞
k
I
A
i
(canonical
form)
p
k
=
P
(
A
k
)
=
P
(
X
=
k
)
g
X
(
s
)
=
∑
k
=
0
∞
s
k
p
k
X
=
∑
k
=
0
∞
k
I
A
i
(canonical
form)
p
k
=
P
(
A
k
)
=
P
(
X
=
k
)
g
X
(
s
)
=
∑
k
=
0
∞
s
k
p
k
(46)
- As a power series in s with nonnegative coefficients whose partial sums converge to one,
the series converges at least for |s|≤1|s|≤1.
- The coefficients of the power series display the distribution: for value k the
probability pk=P(X=k)pk=P(X=k) is the coefficient of sk.
- The power series expansion about the origin of an analytic function is unique. If the
generating function is known in closed form, the unique power series expansion
about the origin determines the distribution. If the power series converges to a known closed
form, that form characterizes the distribution,
- For a simple random variable (i.e., pk=0pk=0 for k>nk>n), gX is a polynomial.
In Example 2, above, we establish the generating function for X∼X∼ Poisson (μ)(μ) from
the distribution. Suppose, however, we simply encounter the generating function
g
X
(
s
)
=
e
m
(
s
-
1
)
=
e
-
m
e
m
s
g
X
(
s
)
=
e
m
(
s
-
1
)
=
e
-
m
e
m
s
(47)
From the known power series for the exponential, we get
g
X
(
s
)
=
e
-
m
∑
k
=
0
∞
(
m
s
)
k
k
!
=
e
-
m
∑
k
=
0
∞
s
k
m
k
k
!
g
X
(
s
)
=
e
-
m
∑
k
=
0
∞
(
m
s
)
k
k
!
=
e
-
m
∑
k
=
0
∞
s
k
m
k
k
!
(48)
We conclude that
P
(
X
=
k
)
=
e
-
m
m
k
k
!
,
0
≤
k
P
(
X
=
k
)
=
e
-
m
m
k
k
!
,
0
≤
k
(49)
which is the Poisson distribution with parameter μ=mμ=m.
For simple, nonnegative, integer-valued random variables, the generating
functions are polynomials. Because of the product rule (T2), the
problem of determining the distribution for the sum of independent
random variables may be handled by the process of multiplying polynomials.
This may be done quickly and easily with the MATLAB convolution function.
Suppose the pair {X,Y}{X,Y} is independent, with
g
X
(
s
)
=
1
10
(
2
+
3
s
+
3
s
2
+
2
s
5
)
g
Y
(
s
)
=
1
10
(
2
s
+
4
s
2
+
4
s
3
)
g
X
(
s
)
=
1
10
(
2
+
3
s
+
3
s
2
+
2
s
5
)
g
Y
(
s
)
=
1
10
(
2
s
+
4
s
2
+
4
s
3
)
(50)
In the MATLAB function convolution, all powers of s must be accounted for by
including zeros for the missing powers.
gx = 0.1*[2 3 3 0 0 2]; % Zeros for missing powers 3, 4
gy = 0.1*[0 2 4 4]; % Zero for missing power 0
gz = conv(gx,gy);
a = [' Z PZ'];
b = [0:8;gz]';
disp(a)
Z PZ % Distribution for Z = X + Y
disp(b)
0 0
1.0000 0.0400
2.0000 0.1400
3.0000 0.2600
4.0000 0.2400
5.0000 0.1200
6.0000 0.0400
7.0000 0.0800
8.0000 0.0800
If mgsum were used, it would not be necessary to be concerned about missing powers and
the corresponding zero coefficients.
We consider briefly the relationship of the moment generating function and the
characteristic function with well known integral transforms (hence the name of this chapter).
Moment generating function and the Laplace transform
When we examine the integral forms of the moment generating function, we see that they
represent forms of the Laplace transform, widely used in engineering and applied mathematics.
Suppose FX is a probability distribution function with FX(-∞)=0FX(-∞)=0. The
bilateral Laplace transform for FX is given by
∫
-
∞
∞
e
-
s
t
F
X
(
t
)
d
t
∫
-
∞
∞
e
-
s
t
F
X
(
t
)
d
t
(51)
The Laplace-Stieltjes transform for FX is
∫
-
∞
∞
e
-
s
t
F
X
(
d
t
)
∫
-
∞
∞
e
-
s
t
F
X
(
d
t
)
(52)
Thus, if MX is the moment generating function for X, then MX(-s)MX(-s) is the
Laplace-Stieltjes transform for X (or, equivalently, for FX).
The theory of Laplace-Stieltjes transforms shows that under conditions sufficiently
general to include all practical distribution functions
M
X
(
-
s
)
=
∫
-
∞
∞
e
-
s
t
F
X
(
d
t
)
=
s
∫
-
∞
∞
e
-
s
t
F
X
(
t
)
d
t
M
X
(
-
s
)
=
∫
-
∞
∞
e
-
s
t
F
X
(
d
t
)
=
s
∫
-
∞
∞
e
-
s
t
F
X
(
t
)
d
t
(53)
Hence
1
s
M
X
(
-
s
)
=
∫
-
∞
∞
e
-
s
t
F
X
(
t
)
d
t
1
s
M
X
(
-
s
)
=
∫
-
∞
∞
e
-
s
t
F
X
(
t
)
d
t
(54)
The right hand expression is the bilateral Laplace transform of FX. We may use tables
of Laplace transforms to recover FX when MX is known. This is particularly useful
when the random variable X is nonnegative, so that FX(t)=0FX(t)=0 for t<0t<0.
If X is absolutely continuous, then
M
X
(
-
s
)
=
∫
-
∞
∞
e
-
s
t
f
X
(
t
)
d
t
M
X
(
-
s
)
=
∫
-
∞
∞
e
-
s
t
f
X
(
t
)
d
t
(55)
In this case, MX(-s)MX(-s) is the bilateral Laplace transform of fX. For nonnegative
random variable X, we may use ordinary tables of the Laplace transform to recover fX.
Suppose nonnegative X has moment generating function
M
X
(
s
)
=
1
(
1
-
s
)
M
X
(
s
)
=
1
(
1
-
s
)
(56)
We know that this is the moment generating function for the exponential (1) distribution. Now,
1
s
M
X
(
-
s
)
=
1
s
(
1
+
s
)
=
1
s
-
1
1
+
s
1
s
M
X
(
-
s
)
=
1
s
(
1
+
s
)
=
1
s
-
1
1
+
s
(57)
From a table of Laplace transforms, we find 1/s1/s is the transform for the constant 1 (for t≥0t≥0)
and 1/(1+s)1/(1+s) is the transform for e-t,t≥0e-t,t≥0, so that FX(t)=1-e-tt≥0FX(t)=1-e-tt≥0, as expected.
Suppose the moment generating function for a nonnegative random variable is
M
X
(
s
)
=
λ
λ
-
s
α
M
X
(
s
)
=
λ
λ
-
s
α
(58)
From a table of Laplace transforms, we find that for α>0α>0,
Γ
(
α
)
(
s
-
a
)
α
is
the
Laplace
transform
of
t
α
-
1
e
a
t
t
≥
0
Γ
(
α
)
(
s
-
a
)
α
is
the
Laplace
transform
of
t
α
-
1
e
a
t
t
≥
0
(59)
If we put a=-λa=-λ, we find after some algebraic manipulations
f
X
(
t
)
=
λ
α
t
α
-
1
e
-
λ
t
Γ
(
α
)
,
t
≥
0
f
X
(
t
)
=
λ
α
t
α
-
1
e
-
λ
t
Γ
(
α
)
,
t
≥
0
(60)
Thus, X∼X∼ gamma (α,λ)(α,λ), in keeping with the determination, above, of the
moment generating function for that distribution.
The characteristic function
Since this function differs from the moment generating function by the interchange of parameter
s and iuiu, where i is the imaginary unit, i2=-1i2=-1, the integral expressions make
that change of parameter. The result is that Laplace transforms become Fourier transforms.
The theoretical and applied literature is even more extensive for the characteristic function.
Not only do we have the operational properties (T1) and (T2) and the result on moments as derivatives
at the origin, but there is an important expansion for the characteristic function.
An expansion theorem
If E[|X|n]<∞E[|X|n]<∞, then
φ
(
k
)
(
0
)
=
i
k
E
[
X
k
]
,
for
0
≤
k
≤
n
and
φ
(
u
)
=
∑
k
=
0
n
(
i
u
)
k
k
!
E
[
X
k
]
+
o
(
u
n
)
as
u
→
0
φ
(
k
)
(
0
)
=
i
k
E
[
X
k
]
,
for
0
≤
k
≤
n
and
φ
(
u
)
=
∑
k
=
0
n
(
i
u
)
k
k
!
E
[
X
k
]
+
o
(
u
n
)
as
u
→
0
(61)
We note one limit theorem which has very important consequences.
A fundamental limit theorem
Suppose {Fn:1≤n}{Fn:1≤n} is a sequence of probability distribution functions and
{φn:1≤n}{φn:1≤n} is the corresponding sequence of characteristic functions.
- If F is a distribution function such that Fn(t)→F(t)Fn(t)→F(t) at every point
of continuity for F, and φ is the characteristic function for F, then
φn(u)→φ(u)∀uφn(u)→φ(u)∀u(62)
- If φn(u)→φ(u)φn(u)→φ(u) for all u and φ is continuous at 0, then
φ is the characteristic function for distribution function F such that
Fn(t)→F(t)ateachpointofcontinuityofFFn(t)→F(t)ateachpointofcontinuityofF(63)
— □□