2. PRELIMINARIES
Laplace Transforms, Moment Generating
Functions and Characteristic Functions
2.1 Definitions
2.2 Theorems on Laplace Transforms
2.3 Operations on Laplace Transforms
2.4 Limit Theorems
2.5 Dirac Delta Function
2.6 Appendix: Complex Numbers
2.7 Appendix: Notes on Partial Fractions
31
Laplace Transforms, Moment Generating
Functions and Characteristic Functions
2.1. Definitions:
Let ϕ(t) be defined on real line.
a. Moment Generating Function (MGF)
Z ∞
M GF =
eθt ϕ(t)dt
−∞
which exists if
Z
∞
−∞
|eθt ϕ(t)dt| < ∞
32
b. Characteristic Function (CF)
Z ∞
CF =
eiyt ϕ(t)dt
−∞
which exists if
Z
∞
−∞
|eiyt ϕ(t)|dt < ∞
Since eiyt = cos yt + i sin yt,
Z ∞
|ϕ(t)|dt < ∞
CF exists if
|eiyt | = 1
−∞
33
(i2 = −1).
c. Laplace Transform
Let ϕ(t) be defined on [0, ∞) and assume
Z ∞
e−x0 t ϕ(t)dt < ∞
0
for some value of x0 ≥ 0.
Characteristic function of e−x0 t ϕ(t) is
Z ∞
e−iyt e−x0 t ϕ(t)dt (sign of iyt could be + or -).
CF =
0
Z ∞
e−(x0 +iy)t ϕ(t)dt
CF =
0
s
CF
=
=
x0 + iy
ϕ∗ (s) =
Z
∞
0
e−st ϕ(t)dt for R(s) ≥ x0
ϕ∗ (s) is Laplace Transform of ϕ(t).
34
ϕ∗ (s) =
Z
∞
e−st ϕ(t)dt
0
Since ϕ(t) defined on [0, ∞), for ε ≥ 0
Z
Z ∞
e−st ϕ(t)dt =
ϕ∗ (s) = lim
ε→0
ε
∞
e−st ϕ(t)dt.
0+
Often a script L is used to denote a LaPlace transform; i.e.
L {ϕ(t)} = ϕ∗ (s)
Suppose f (t) is a pdf on [0, ∞),
Z ∞
f ∗ (s) =
e−st f (t)dt , R(s) ≥ 0
0
Laplace transform is appropriate for non-negative r.v.
35
If the cdf is F (t) =
Z
t
f (x)dx =
0
f ∗ (s) =
Z
Z
t
dF (x)
0
∞
e−st dF (t)
0
is Laplace-Stieltjes Transform.
Note: E(e−sT ) = f ∗ (s)
f ∗ (0) = 1, 0 ≤ f ∗ (s) ≤ 1
36
2.2 Theorems on Laplace Transforms (LT)
a. Uniqueness Theorem.
Distinct probability distributions have distinct Laplace Transforms
b. Continuity Theorem
For n = 1, 2, . . . , let {Fn (t)} be a sequence of cdf 0 s such that Fn → F .
∗
Define {fn∗ (s)} as the
sequence
of
LT
such
that
L
{f
(t)}
=
f
(s)
n
n
Z
∞
e−st dF (t).
and define f ∗ (s) =
0
Then fn∗ (s) → f ∗ (s) and conversely if fn∗ (s) → f ∗ (s), then
Fn (t) → F (t)
37
c. Convolution Theorem
If T1 , T2 are independent, non-negative r.v. with p.d.f f1 (t), f2 (t) then
the pdf of T = T1 + T2 is
Z ∞
f1 (τ )f2 (t − τ )dτ
f (t) =
0
and L {f (t)} = f1∗ (s)f2∗ (s)
In general if {Ti } i = 1, 2, . . . , n are independent non-negative r.v.,
then the Laplace transform of the pdf of T = T1 + · · · + Tn is
n
Y
fi∗ (s)
L {f (t)} =
i=1
38
d. Moment Generating Property
Suppose all moments exist.
∗
f (s) =
Z
∞
e
−st
f (t)dt =
0
=
=
Z
∞
0
∞
X
n=0
∞
X
n=0
(
∞
X
(−1)
n
n=0
n
s
(−1)n
n!
Z
s n tn
n!
)
f (t)dt
∞
tn f (t)dt
0
sn
mn where mn = E(T n )
(−1)
n!
39
n
e. Inversion Theorem
Knowledge of f ∗ (s). The inversion formula is written
Z c+i∞
1
est f ∗ (s)ds
f (t) =
2πi c−i∞
where the integration is in the complex plane and c is an appropriate
constant. (It is beyond the scope of this course to discuss the inversion
formula in detail).
Notation: L {f (t)} = f ∗ (s)
f (t) = L−1 {f ∗ (s)}
where L−1 { } is referred to as the “Inverse Laplace Transform”.
40
Example: Exponential Distribution
f (t)
=
E(T )
=
f ∗ (s)
=
λe−λt for t ≥ 0
1/λ, V (t) = 1/λ2 .
Z ∞
Z
e−st λe−λt dt = λ
0
=
f ∗ (s)
=
=
λ
λ+s
∞
X
(−1)n
e−(s+λ)t dt
0
∞
(λ + s)e
−(λ+s)t
0
1
λ
=
λ+s
1+
n=0
⇒
Z
∞
n
s
n!
=
s
λ
n=0
n!
λn
mn = E(T n ) =
41
∞
X
n!
λn
(−1)n
λ
dt =
λ+s
s n
λ
L{λe
−λt
λ
,
}=
λ+s
L
−1
λ
} = λe−λt
{
λ+s
L{e−λt } = (λ + s)−1 , L−1 {(λ + s)−1 } = e−λt
Note:
Z c+i∞
1
est f ∗ (s)ds = f (t)
2πi c−i∞
Z c+i∞
λ
1
st
ds = λe−λt
e
2πi c−i∞
λ+s
Z c+i∞
1
est (λ + s)−1 ds = e−λt
2πi c−i∞
L−1 {(λ + s)−1 }
42
= e−λt
L−1 {(λ + s)−1 } = e−λt
Differentiating w.r. to λ
L−1 {(λ + s)−2 } = te−λt
again
L−1 {2(λ + s)−3 } = t2 e−λt
..
.
(n − 1) times
⇒
and
L−1 {(n − 1)!(λ + s)−n } = tn−1 e−λt
L
L
tn−1 e−λt
Γ(n)
= (λ + s)−n , Γ(n) = (n − 1)!
n−1 −λt
e
λ(λt)
Γ(n)
43
= λn /(λ + s)n
λ(λt)n−1 e−λt
is gamma distribution.
Note:
Γ(n)
If {Ti } i = 1, 2, . . . , n are independent non-negative identically
distributed r.v. following an exponential distribution with parameter λ and
T = T1 + . . . + T n
n
n
Y
λ
f ∗ (s) =
fi∗ (s) =
λ+s
1
44
Example: Suppose T1 , T2 are independent non-negative random variables
following exponential distributions with parameters λ1 , λ2 (λ1 6= λ2 ).
What is distribution of T = T1 + T2 ?
λ1
λ2
f ∗ (s) = f1∗ (s)f2∗ (s) =
.
λ1 + s
λ2 + s
f ∗ (s) can be written as a partial fraction; i.e.
∗
f (s)
where
A1
=
A2
A1 (λ2 + s) + A2 (λ1 + s)
A1
+
=
(λ1 + s) (λ2 + s)
(λ1 + s)(λ2 + s)
=
λ1 λ2
, A2 = −A1
λ2 − λ 1
45
Since L−1 {(λ + s)−1 } = e−λt
f (t) =
L−1 {f ∗ (s)} =
=
=
A1 L−1 {(λ1 + s)−1 } + A2 L−1 {(λ2 + s)−1 }
A1 e−λ1 t + A2 e−λ2 t
λ1 λ2 −λ1 t
−λ2 t
−e
e
λ2 − λ 1
Suppose T1 , T2 , T3 are non-negative independent r.v. so that T1 is
exponential with parameter λ1 and T2 , T3 are exponential each with
parameter λ2 . Find pdf of T = T1 + T2 + T3 .
2
λ2
B1
B2
A1
λ1
∗
+
+
=
f (s) =
λ1 + s
λ2 + s
λ1 + s λ2 + s (λ2 + s)2
f (t) = A1 e−λ1 t + B1 e−λ2 t + B2 te−λ2 t
46
Homework:
Show A1 = (λ1 − λ2 )−2 λ1 λ22
−1
−2
B1 = − (λ1 − λ2 ) + (λ1 − λ2 )
λ1 λ22
B2 = (λ1 − λ2 )−1 λ1 λ22
In general if fi∗ (s) = (λi /λi + s) i = 1, 2, . . . , n and
Qn
∗
f (s) = i=1 fi∗ (s), then f(t) is called the Erlangian distribution.
(Erlang was a Danish telephone engineer who used this distribution to
model telephone calls).
47
2.3 Operations on Laplace Transforms
L{e−λt }
L{tn−1 e−λt }
=
(λ + s)−1
=
Γ(n)/(λ + s)n
Setting λ = 0 in above
L{1}
= 1/s
L{tn−1 }
= Γ(n)/sn
48
Homework: Prove the following relationships
Z t
ϕ∗ (s)
ϕ(x)dx =
1. L
s
0
2. L {ϕ0 (t)} = sϕ∗ (s) − ϕ(0+ )
3. L {ϕ(r) (t)} = sr ϕ∗ (s) − sr−1 ϕ(0+ ) − sr−2 ϕ0 (0+ )
− . . . − ϕ(0) (0+ )
Prove (1) and (2) using integration by parts
Prove (3) by successive use of (2).
49
Suppose f (t) is a pdf for a non-negative random variable and
Z ∞
Z t
f (x)dx, Q(t) =
f (x)dx
F (t) =
0
Since L {
Rt
0
t
ϕ∗ (s)
ϕ(x)dx} = s
∗
f
(s)
L {F (t)} = F ∗ (s) = s
Since Q(t) = 1 − F (t)
∗
1
−
f
(s)
1
L {Q(t)} = Q∗ (s) = s − F ∗ (s) =
s
50
2.4 Limit Theorems
lim ϕ(t) = lim sϕ∗ (s)
t→0+
s→∞
Proof:
L {ϕ0 (t)} = sϕ∗ (s) − ϕ(0+ )
0∗
lim ϕ (s) = lim [sϕ∗ (s) − ϕ(0+ )]
s→∞
But lim
s→∞
Z
s→∞
∞
e−st ϕ0 (t)dt = 0
0
..˙ lim [sϕ∗ (s) − ϕ(0+ )] = 0
s→∞
51
lim ϕ(t) = lim sϕ∗ (s)
t→∞
s→0
Proof:
0
lim L {ϕ (t)} = lim
s→0
s→0
= lim
Z
t→∞
∞
e
−st
0
ϕ (t)dt =
0
Z
Z
∞
ϕ0 (t)dt
0
t
0
ϕ0 (x)dx = lim [ϕ(t) − ϕ(0+ )].
t→∞
Since L {ϕ0 (t)} = sϕ∗ (s) − ϕ(0+ )
lim [sϕ∗ (s) − ϕ(0+ )] = lim [ϕ(t) − ϕ(0+ )]
t→∞
s→0
⇒ lim sϕ∗ (s) = lim ϕ(t)
t→∞
s→0
52
2.5. Dirac Delta Function
Sometimes it is useful to use the Dirac delta function. It is a “strange”
function and has the property.

 0 for t 6= 0
δ(t) =
 ∞ t=0
Define
Let h > 0

 1 if t > 0
U (t) =
 0 t≤0


0



U (t + h) − U (t)
1
=
δ(t, h) =

h
h


 0
53
if t > 0
if − h < t ≤ 0
if t ≤ −h
Define

for t 6= 0
U (t + h) − U (t)  0
=
δ(t) = lim δ(t, h) = lim
 ∞ for t = 0
h→0
h→0
h
Consider
Z ∞
0
=
Z
ϕ(x)δ(t − x)dx
∞
0
= lim
h→0
U (t − x + h) − U (t − x)
ϕ(x) lim
dx
h→0
h
1
h
Z
∞
0
ϕ(x)U (t − x + h)dx −

 1
Since U (t − x + h) =
 0
Z
t−x+h>0
otherwise
54
∞
0
ϕ(x)U (t − x)dx
Z
∞
ϕ(x)δ(t − x)dx
0
)
(Z
Z t
t+h
1
ϕ(x)dx −
ϕ(x)dx
= lim
h→0 h
0
0
= lim
h→0
1
h
(Z
t+h
ϕ(x)dx
t
)
= lim
h→0
ϕ(t + θh)h
h
= ϕ(t)
(by mean value theorem)
for some value of θ 0 < θ < 1.
Z ∞
ϕ(x)δ(t − x)dx = ϕ(t)
..˙
0
Similarly ϕ(t) =
Z
∞
0
ϕ(t − x)δ(x)dx
55
ϕ(t) =
Z
∞
0
ϕ(x)δ(t − x)dx =
Z
∞
0
ϕ(t − x)δ(x)dx
Special Cases
ϕ(x) = 1 for all x
1=
Z
∞
0
δ(t − x)dx =
⇒
Z
Z
∞
δ(x)dx for all t
0
∞
δ(x)dx = 1
0
Suppose ϕ(x) = e−sx . Then
Z
If t = 0,
Z
∞
0
e−sx δ(x − t)dx = e−st
∞
0
e−sx δ(x)dx = L{δ(x)} = 1
56
Example 1 Consider a non-negative random variable T such that
pn = P {T = tn }. Define
∞
X
f (t) =
n=1
Z
∞
f (t)dt =
0
∞
X
pn
n=1
F (tn ) = P {T ≤ tn } =
∞
X
j=1
Z
pn δ(t − tn )
∞
0
δ(t − tn )dt =
pj U (tn+1 − tj ) =
Z
∞
X
pn = 1
1
tn
f (t)dt =
0
X
pj
j≤n
Note: If tj < tn+1 ⇒ tj ≤ tn
f ∗ (s) =
Z
∞
e−st f (t)dt =
0
∞
X
n=1
pn
Z
∞
0
57
e−st δ(t − tn )dt =
∞
X
n=1
pn e−stn
Example 2
Consider a random variable T such that p = P {T = 0} but for T > 0
Z t2
q(x)dx
P {t1 < T ≤ t2 } =
t1
⇒
f (t)
f ∗ (s)
= pδ(t) + (1 − p)q(t)
= p + (1 − p)q ∗ (s).
Another way of formulating this problem is to define the random variable
T by

 0 with probability p
T =
 t with pdf q(t) for t > 0
This is an example of a random variable having both a discrete and
continuous part.
58
2.6 Appendix
ELEMENTS OF COMPLEX NUMBERS
∧
∧
r(x, y)
r
r
0
θ
>
0
>
Rectangular
Polar Coordinates
Coordinates
r = Modulus
θ = Amplitude
x = r cos θ,
y = r sin θ
59
Complex no. representation:
p
z = x + iy, r = x2 + y 2
θ = tan−1
|z| = absolute value= r
x
y
i: to be determined
z = r(cos θ + i sin θ) = reiθ
Suppose zj = aj + ibj
Addition/Subtraction: z1 ± z2 ⇒ (a1 ± a2 ) + i(b1 ± b2 )
60
Multiplication:
zj = rj eiθj , z1 z2 = r1 r2 ei(θ1 +θ2 ) z1 z2 ⇒ r = r1 r2 ,
θ = θ 1 + θ2
z1 z2 = r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 ) = x + iy
x = r1 r2 cos(θ1 + θ2 ) = r1 r2 (cos θ1 cos θ2 − sin θ1 sin θ2 ) = a1 a2 − b1 b2
y = r1 r2 sin(θ1 + θ2 ) = r1 r2 (sin θ1 cos θ2 + cos θ1 sin θ2 ) = b1 a2 + a1 b2
⇒ z1 z2 = (a1 + ib1 )(a2 + ib2 ) = a1 a2 + i[a1 b2 + a2 b1 ] + i2 [b1 b2 ]
If i2 = −1 Re(z1 z2 ) = a1 a2 − b1 b2
i, i2 = −1,
⇒ i4n+1 = i,
i3 = −i,
i4 = 1
i4n+2 = −1,
i4n+3 = −i,
i4n = 1
z n = [r(cos θ + i sin θ)]n = rn (cos nθ + i sin nθ)
61
Exponential function
eθ
=
∞
X
θn
0
eiθ
eiθ
e−iθ
n!
i3 θ 3
i2 θ 2
+
+ ...
= 1 + iθ +
2
3!
θ4
θ6
θ3
θ5
θ2
+
−
+ . . . ) + i(θ −
+
...)
= (1 −
2
4!
6!
3!
5!
= cos θ + i sin θ,
= cos θ − i sin θ, cos(−θ) = cos θ; sin −θ = − sin θ
eiθ − e−iθ
eiθ + e−iθ
, sin θ =
cos θ =
2
2i
cos 0 = 1, sin 0 = 0
Since eiθ = cos θ + i sin θ; z = reiθ
If z = ex+iy = ex eiy :
ex : modulus, ey : amplitude
z1 z2 = ex1 +iy1 · ex2 +iy2 = ex1 +x2 +i(y1 +y2 )
62
MOMENT GENERATING AND
CHARACTERISTIC FUNCTIONS
tY
R∞
MGF: ψ(t) = E(e ) = −∞ ety dF (y)
R∞
ψ(t) exists if −∞ | ety | dF (y) ≤ ∞
Sometimes mgf does not exist.
Importance of M GF
Uniqueness Theorem: If two random variables have the same mgf , they
have the same cdf except possibly at a countable number of points having
0 probability.
63
Continuity Theorem: Let {Xn } and X have mgf {ψn (t)} and ψ(t) with
cdf 0 s Fn (x) and F (x). Then a necessary and sufficient condition for
lim Fn (X) = F (X) is that for every t, limn→∞ ψn (t) = ψ(t), where
n→∞
ψ(t) is continuous at t = 0.
Inversion Formula: Knowledge of ψ(t) enables the pdf or frequency
function to be calculated.
Convolution Theorem: If Yi are independent with mgf ψi (t), then the
n
n
X
Y
Yi is ψS (t) =
ψi (t)
mgf of S =
1
If ψi (t) = ψ(t)
1
⇒ ψS (t) = ψ(t)n
64
MOMENT GENERATING FUNCTIONS
DO NOT ALWAYS EXIST!
For that reason one ordinarily uses a characteristic function of a
distribution rather than the mgf .
function of a random variable y is
Def. The characteristic
Z
∞
ϕ(t) = E(eity ) =
−∞
ity |ϕ(t)| = E(e ) ≤
ity as e = 1.
Z
eity dF (y) for −∞ < t < ∞
∞
−∞
ity e dF (y) =
⇒ Characteristic Functions always exist.
65
Z
∞
dF (y) = 1
−∞
Relation between mgf and cf .
ϕ(t) = ψ(it)
ϕ(t) =
X (it)n
n!
ϕ(r) (0) = ir mr
mn
The characteristic function is a Fourier Transform; i.e. for any function
g(y)
F.T.(q(y)) =
Z
∞
eity q(y)dy
−∞
66
if q(y) is pdf ⇒ c.f.
RELATION TO LaPLACE TRANSFORMS
Let Y be a non-negative r.v. with pdf f (y); i.e. P {Y ≥ 0} = 1. Then it is
common to use Laplace Transforms instead of characteristic functions.
Def.: s = a + ib (a > 0). Then the Laplace Transform of f (y) is
Z ∞
e−sy f (y)dy
f ∗ (s) =
0
Since e−sy f (y) = e−iby e−ay f (y)
The Laplace Transform is the equivalent of taking the Fourier Transform
of e−ay f (y)
Ex. Exponential ϕ(t) = (1 −
Z ∞
f ∗ (s) =
e−sy λe−λy dy =
0
it −1
λ)
s
λ
= (1 + )−1
λ+s
λ
67
MOMENT GENERATING FUNCTIONS AND
CHARACTERISTIC FUNCTIONS
mgf
cf
Bernoulli
(pet + q)
(peit + q)
Binomial
(pet + q)n
(peit + q)n
e
Poisson
Geometric
pet /(1 − qet )
etb −eta
t(b−a)
Uniform over (a, b)
Normal
Exponential
λ(et−1 )
peit /(1 − qeit )
eitb − eita
it(b − a)
itm− 12 σ 2 t2
tm+ 12 σ 2 t2
e
= (1 − λt )−1
(1 −
e
λ
λ−t
e
λ(eit −1)
68
it −1
)
λ
INVERSION THEOREM
Integer valued R.V.
Let Y = 0, ±1, ±2, . . . with prob. f (j) = P (Y = j)
cf ϕ(t) = E(eitY ) =
∞
X
eitj f (j)
−∞
Inversion Formula:
1
f (k) =
2π
Z
π
e−ikt ϕ(t)dt
−π
69
1
f (k) =
2π
Z
Z

π
e
∞
X
1
e−ikt 
eitj f (j) dt =
2π
−π
j=−∞
π
i(j−k)t

dt
Z
=
−π
∞
X
j=−∞
π
ei(j−k)t dt
−π
π
−π
[cos(j − k)t + i sin(j − k)t]dt
π
π sin(j − k)t
cos(j − k)t
+ i
for j 6= k
−
j
−
k
(j
−
k)
−π
−π
=
sin nπ = 0 for any integer n.
cos nπ = cos(−nπ)
Z
f (j)
Z
π
ei(j−k)t
−π

 0 for j 6= k
dt =
 2π for j = k
70
INVERSION FORMULA FOR CONTINUOUS
TYPE RANDOM VARIABLES
Let
Z ∞Y have pdf f (y) and cf ϕ(t) which is integrable; i.e.
|ϕ(t)| dt < ∞
−∞
Inversion Formulae:
1
f (y) =
2π
Z
∞
eiyt ϕ(t)dt
−∞
Ex. Normal Distribution (standard normal)
Z ∞
−y 2 /2
e
ity
−t2 /2
√
dy = e
e
ϕ(t) =
2π
−∞
71
Replace t by −t
Z
2
∞
−y /2
2
−ity e
√
dy = e−t /2
e
2π
−∞
Interchange symbols t and y
Z ∞
−t2 /2
−ity e
−y 2 /2
√
dt = e
e
2π
−∞
Divide by
√
2π
1
2π
Z
2
∞
e−ity e−t
−∞
2
−y /2
e
/2
dt = √
2π
which is the inversion formular for N (0, 1).
72
UNIQUENESS THEOREM
Let X be an arbitrary r.v. and let Y be N (0, 1) where X and Y are
independent. Consider Z = X + cY (c is a constant).
ϕZ (t) = ϕX (t)e
−c2 t2 /2
Note that ϕZ (t) is integrable as |ϕX (t)| ≤ 1
Z ∞
1
e−itz ϕz (t)dt
f (z) =
2π −∞
73
P (a < Z ≤ b) =
=
=
FZ (b) − FZ (a) =
Z
b
a
1
2π
1
2π
Z
Z
∞
−∞
Z
b
f (z)dz
a
∞
e
−it(x+cy)
ϕx (t)e
−c2 t2 /2
dtdz
−∞
"Z
#
b
e
−itx
dx ϕx (t)e
a
−c2 t2 /2 −itcy
e
Let c → 0 ⇒ z → x and dz → dx
Fx (b) − Fx (a) =
1
2π
Z
∞
−∞
e
−ibt
−e
−it
−iat
ϕx (t)dt
⇒ The distribution function is determined by its c.f.
⇒ If two r.v.’s have same characteristic function they have the same
distribution function.
74
dt
Examples
Double Exponential f (x) = 12 e−|x|
−∞<x<∞
Z ∞
Z ∞
1
1
eitx e−|x| dx =
[cos tx + i sin tx]e−|x| dx
ϕ(t) =
2
2 −∞
−∞
Since sin(−tx) = − sin tx and cos(−tx) = cos(tx)
Z ∞
Z
1 ∞
(cos tx)e−|x| dx =
(cos tx)e−x dx
ϕ(t) =
2 −∞
0
as sin 0 = 0,
∞ e−x [t sin tx − cos tx]
1
=
ϕ(t) = 2
1
+
t
1 + t2
0
cos 0 = 1
Double Exponential is sometimes called Laplace’s Distribution.
75
Ex. What is c.f. of the Cauchy distribution which has
1
− ∞ < y < ∞?
pdf f (y) = π(1+y
2)
Note that by the inversion formula for the double exponential
Z ∞
1
1
1 |y|
−ity
e =
e
dt
2
2π −∞
1 + t2
Hence for the Cauchy distribution
Z ∞
eity
ϕ(t) =
−∞
1
−|t|
dy
=
e
π(1 + y 2 )
76
PROBLEMS
1. Let Y be any r.v.
(a) Show ϕy (t) = E[cos tY ] + iE[sin tY ]
(b) Show ϕ−y (t) = E[cos tY ] − iE[sin tY ]
(c) Show ϕ−y (t) = ϕy (−t)
2. Let Y have a symmetric distribution around 0, i.e. f (y) = f (−y)
(a) Show E(sin tY ) = 0 and ϕy (t) is real valued.
(b) Show ϕy (−t) = ϕy (t)
3. Let X and Y be ind. ident. dist. r.v.’s. Show ϕX−Y (t) = |ϕx (t)|
77
2
4. Let Yj (j = 1, 2) be independent exponential r.v.’s with
E(Yj ) = 1/λj .
Consider S = Y1 + Y2
(a) Find the c.f. of S
(b) Using (i) and the inversion theorem find the pdf of S. Hint: If
f (y) is exponential with parameter λ (E(Y ) = 1/λ), then
R ∞ −ity λ
λ
1
−λy
dt
=
λe
by the inversion
ϕ(t) = λ−it and 2π −∞ e
λ−it
formula.
78
2.7 Notes on Partial Fractions
Suppose N ∗ (s) and D ∗ (s) are polynomials in s. Suppose D ∗ (s) is a
polynomial of degree k and N ∗ (s) has degree < k.
D∗ (s) =
k
Y
i=1
(s − si ) (Roots are distinct)
and roots of N ∗ (s) do not coincide with D ∗(s) . Then
k
N ∗ (s) X Ai
=
∗
D (s)
s − si
i=1
where constants Ai are to be determined.
Since
⇒
L {e−λt } = (λ + s)−1 , e−λt = L−1 {(λ + s)−1 }
∗ X
k
(s)
N
−si t
=
L−1
A
e
i
D∗ (s)
i=1
79
N ∗ (s)
D∗ (s)
=
∗
N (s)
D∗ (s)
=
∗
N (s) =
k
X
Ai
s − si
i=1
Pk
i=1
k
X
Ai
Qk
Q
1 (s
Ai
i=1
Y
j6=i
j6=i (s
− si )
− sj )
(s − sj )
Substitute s = sr
∗
N (sr ) = Ar
Y
j6=r
N ∗ (sr )
Ar = Q
j6=r (sr − sj )
(sr − sj )
r = 1, 2, . . . , k
80
Since
D∗ (s)
=
k
Y
i=1
dD∗ (s)
ds
0∗
D (sr )
=
(s − si )
D0∗ (s) =
k Y
X
i=1 j6=i
=
Y
j6=r
(s − sj )
(sr − sj )
N ∗ (sr )
Ar = 0∗
D (sr )
r = 1, 2, . . . , k
81
Example:
Let Ti be independent r.v. having pdf qi (t) = λi e−λi t (λ1 6= λ2 ) and
consider T = T1 + T2 .
qi∗ (s)
=
λi /(λi + s) i = 1, 2
qT∗ (s)
=
q1∗ (s)q2∗ (s) = λ1 λ2 /(λ1 + s)(λ2 + s)
=
A2
A1
+
(λ1 + s) (λ2 + s)
N ∗ (s) = λ1 λ2
,
D∗ (s) = (λ1 + s)(λ2 + s), si = −λi
N ∗ (sr )
Ar = 0∗
D (sr )
,
D0∗ (s) = (λ1 + s) + (λ2 + s)
λ1 λ2
λ2 − λ 1
,
A2 =
A1 =
λ1 λ2
λ1 − λ 2
82
λ
1
λ
1
1 2
qT∗ (s) =
−
λ2 − λ 1 s + λ 1
s + λ2
Taking inverse transforms gives
λ1 λ2 −λ1 t
−λ2 t
q(t) =
−e
e
λ2 − λ 1
83
Multiple Roots
If D∗ (s) has multiple roots, the methods for finding the partial fraction
decomposition is more complicated. It is easier to use direct methods.
Example: N ∗ (s) = 1, D ∗ (s) = (a + s)2 (b + s)
1
A1
A2
B
N∗
=
=
+
+
D∗ (s)
(a + s)2 (b + s)
a + s (a + s)2
b+s
A1 (a + s)(b + s) + A2 (b + s) + B(a + s)2
1
=
2
(a + s) (b + s)
(a + s)2 (b + s)
1 = A1 (a + s)(b + s) + A2 (b + s) + B(a + s)2
(1)
1 = s2 [A1 + B] + s[A1 (a + b) + A2 + B(2a)] + [abA1 + A2 b + Ba2 ]
A1 + B = 0, A1 (a + b) + A2 + 2aB = 0
A1 (ab) + A2 b + Ba2 = 1
84
The above three equations result in solutions for (A1 , A2 , B). Another
way to solve for these constants is by substituting s = −a and s = −b in
(1); i.e.
substituting s = −b in (1)
⇒ B = (a − b)−2 ⇒ A1 = −B = −(a − b)−2
substituting s = −a in (1)
⇒ A2 = (b − a)−1
A1 = −(a − b)−2 , A2 = −(a − b)−1 , B = (a − b)−2
1
A1
A2
B
N ∗ (s)
=
=
+
+
D∗ (s)
(s + a)2 (s + b)
s + a (s + a)2
s+b
∗ N (s)
−2 −at
−1 −at
−2 −bt
=
−(a
−
b)
e
−
(a
−
b)
te
+
(a
−
b)
e
L−1
∗
D (s)
e−bt
−e−at
=
[1 + t(a − b)] +
2
(a − b)
(a − b)2
85
In general if D ∗ (s) = (s − s1 )d1 (s − s2 )d2 ...(s − sk )dk the
decomposition is
A1
A2
Ad 1
N ∗ (s)
=
+
+
.
.
.
+
D∗ (s)
s − s1
(s − s1 )2
(s − s1 )d1
+
B1
B2
Bd 2
+
+
.
.
.
+
s − s2
(s − s2 )2
(s − s2 )d2
+ ... . +
+
Z2
Zdk
Z1
+
+
.
.
.
+
(s − sk ) (s − sk )2
(s − sk )dk
86