18 CHAPTER 1. ELEMENTS OF PROBABILITY DISTRIBUTION THEORY
1.7.1 Moments and Moment Generating Functions
Definition 1.12. The nth moment (n ∈ N) of a random variable X is defined as
µ′n = E X n
The nth central moment of X is defined as
µn = E(X − µ)n ,
where µ = µ′1 = E X.
Note, that the second central moment is the variance of a random variable X, usually denoted by σ 2 .
Moments give an indication of the shape of the distribution of a random variable.
Skewness and kurtosis are measured by the following functions of the third and
fourth central moment respectively:
the coefficient of skewness is given by
γ1 =
µ3
E(X − µ)3
= 3;
3
σ
µ22
the coefficient of kurtosis is given by
γ2 =
E(X − µ)4
µ4
− 3 = 2 − 3.
4
σ
µ2
Moments can be calculated from the definition or by using so called moment generating function.
Definition 1.13. The moment generating function (mgf) of a random variable X
is a function MX : R → [0, ∞) given by
MX (t) = E etX ,
provided that the expectation exists for t in some neighborhood of zero.
More explicitly, the mgf of X can be written as
Z ∞
MX (t) =
etx fX (x)dx, if X is continuous,
−∞
X
MX (t) =
etx P (X = x)dx, if X is discrete.
x∈X
The method to generate moments is given in the following theorem.
1.7. EXPECTED VALUES
19
Theorem 1.7. If X has mgf MX (t), then
(n)
E(X n ) = MX (0),
where
dn
MX (t)|0 .
dtn
That is, the n-th moment is equal to the n-th derivative of the mgf evaluated at
t = 0.
(n)
MX (0) =
Proof. Assuming that we can differentiate under the integral sign we may write
Z
d ∞ tx
d
MX (t) =
e fX (x)dx
dt
dt −∞
Z ∞
d tx
=
e
fX (x)dx
dt
−∞
Z ∞
=
xetx fX (x)dx
−∞
= E(XetX ).
Hence, evaluating the last expression at zero we obtain
d
MX (t)|0 = E(XetX )|0 = E(X).
dt
For n = 2 we will get
d2
MX (t)|0 = E(X 2 etX )|0 = E(X 2 ).
2
dt
Analogously, it can be shown that for any n ∈ N we can write
dn
MX (t)|0 = E(X n etX )|0 = E(X n ).
dtn
Example 1.14. Find the mgf of X ∼ Exp(λ) and use results of Theorem 1.7 to
obtain the mean and variance of X.
By definition the mgf can be written as
t
MX (t) = E(e X) =
Z
∞
−∞
etx fX (x)dx.
20 CHAPTER 1. ELEMENTS OF PROBABILITY DISTRIBUTION THEORY
For the exponential distribution we have
fX (x) = λe−λx I(0,∞) (x),
where λ ∈ R+ . Here we used the notation of the indicator function IX (x) whose
meaning is as follows:
1, if x ∈ X ;
IX (x) =
0, otherwise.
That is,
fX (x) =
λe−λx , if x ∈ (0, ∞);
0,
otherwise.
Hence, integrating by the method of substitution, we get
Z ∞
Z ∞
λ
tx
−λx
provided that |t| < λ.
MX (t) =
e λe dx = λ
e(t−λ)x dx =
t−λ
0
0
Now, using Theorem 1.7 we obtain the first and the second moments, respectively:
E(X) = MX′ (0) =
(2)
E(X 2 ) = MX (0) =
Hence, the variance of X is
1
λ = ,
2
t=0
(λ − t)
λ
2λ 2
= 2.
3
t=0
(λ − t)
λ
var(X) = E(X 2 ) − [E(X)]2 =
2
1
1
− 2 = 2.
2
λ
λ
λ
Exercise 1.10. Calculate mgf for Binomial and Poisson distributions.
Moment generating functions provide methods for comparing distributions or
finding their limiting forms. The following two theorems give us the tools.
Theorem 1.8. Let FX (x) and FY (y) be two cdfs whose all moments exist. Then
1. If FX and FY have bounded support, then FX (u) = FY (u) for all u iff
E(X n ) = E(Y n ) for all n = 0, 1, 2, . . ..
2. If the mgfs of X and Y exist and are equal, i.e., MX (t) = MY (t) for all t in
some neighborhood of zero, then FX (u) = FY (u) for all u.
1.7. EXPECTED VALUES
21
Theorem 1.9. Suppose that {X1 , X2 , . . .} is a sequence of random variables, each
with mgf MXi (t). Furthermore, suppose that
lim MXi (t) = MX (t), for all t in a neighborhood of zero,
i→∞
and MX (t) is an mgf. Then, there is a unique cdf FX whose moments are determined by MX (t) and, for all x where FX (x) is continuous, we have
lim FXi (x) = FX (x).
i→∞
This theorem means that the convergence of mgfs implies convergence of cdfs.
Example 1.15. We know that the Binomial distribution can be approximated by a
Poisson distribution when p is small and n is large. Using the above theorem we
can confirm this fact.
The mgf of Xn ∼ Bin(n, p) and of Y ∼ Poisson(λ) are, respectively:
MXn (t) = [pet + (1 − p)]n ,
t
MY (t) = eλ(e −1) .
We will show that the mgf of X tends to the mgf of Y , where λ = np.
We will need the following useful result given in the lemma:
Lemma 1.1. Let a1 , a2 , . . . be a sequence of numbers converging to a, that is,
limn→∞ an = a. Then
an n
lim 1 +
= ea .
n→∞
n
Now, we can write
n
MXn (t) = pet + (1 − p)
n
1
t
= 1 + np(e − 1)
n
n
λ(et − 1)
= 1+
n
t
−→ eλ(e −1) = MY (t).
n→∞
Hence, by Theorem 1.9 the Binomial distribution converges to a Poisson distribution.
22 CHAPTER 1. ELEMENTS OF PROBABILITY DISTRIBUTION THEORY
1.8 Functions of Random Variables
If X is a random variable with cdf FX (x), then any function of X, say g(X) = Y
is also a random variable. The question then is “what is the distribution of Y ?”
The function y = g(x) is a mapping from the induced sample space of the random
variable X, X , to a new sample space, Y, of the random variable Y , that is
g(x) : X → Y.
The inverse mapping g −1 acts from Y to X and we can write
g −1 (A) = {x ∈ X : g(x) ∈ A} where A ⊂ Y.
Then, we have
P (Y ∈ A) = P (g(X) ∈ A)
= P {x ∈ X : g(x) ∈ A}
= P X ∈ g −1 (A) .
The following theorem relates the cumulative distribution functions of X and Y =
g(X).
Theorem 1.10. Let X have cdf FX (x), Y = g(X) and let domain and codomain
of g(X), respectively, be
X = {x : fX (x) > 0}, and Y = {y : y = g(x) f or some x ∈ X }.
(a) If g is an increasing function on X then FY (y) = FX g −1(y) for y ∈ Y.
(b) If g is a decreasing function on X , then FY (y) = 1 − FX g −1(y) for y ∈ Y.
Proof. The cdf of Y = g(X) can be written as
FY (y) = P (Y ≤ y)
= P (g(X) ≤ y)
= P {x ∈ X : g(x) ≤ y}
Z
=
fX (x)dx.
{x∈X :g(x)≤y}
(a) If g is increasing, then
{x ∈ X : g(x) ≤ y} = {x ∈ X : g −1 g(x) ≤ g −1(y)} = {x ∈ X : x ≤ g −1 (y)}.
1.8. FUNCTIONS OF RANDOM VARIABLES
So, we can write
FY (y) =
Z
23
fX (x)dx
{x∈X :g(x)≤y}
=
Z
fX (x)dx
{x∈X :x≤g −1 (y)}
=
Z
g −1 (y)
fX (x)dx
g −1 (y) .
−∞
= FX
(b) Now, if g is decreasing, then
{x ∈ X : g(x) ≤ y} = {x ∈ X : g −1 g(x) ≥ g −1 (y)} = {x ∈ X : x ≥ g −1 (y)}.
So, we can write
FY (y) =
Z
fX (x)dx
{x∈X :g(x)≤y}
=
=
Z
fX (x)dx
{x∈X :x≥g −1 (y)}
Z ∞
fX (x)dx
= 1 − FX g −1 (y) .
g −1 (y)
Example 1.16. Find the distribution of Y
U([0, 1]). The cdf of X is

 0, for
x, for
FX (x) =

1, for
= g(X) = − log X, where X ∼
x < 0;
0 ≤ x ≤ 1;
x > 1.
For x ∈ [0, 1] the function g(x) = − log x is defined on Y = (0, ∞) and it is
decreasing.
For y > 0, y = − log x implies that x = e−y , i.e., g −1 (y) = e−y and
FY (y) = 1 − FX g −1 (y) = 1 − FX e−y = 1 − e−y .
Hence we may write
FY (y) = 1 − e−y I(0,∞) .
This is exponential distribution function for λ = 1.
24 CHAPTER 1. ELEMENTS OF PROBABILITY DISTRIBUTION THEORY
For continuous rvs we have the following result.
Theorem 1.11. Let X have pdf fX (x) and let Y = g(X), where g is a monotone
function. Suppose that fX (x) is continuous on its support X = {x : fX (x) >
0} and that g −1(y) has a continuous derivative on support Y = {y : y =
g(x) f or some x ∈ X }. Then the pdf of Y is given by
d
fY (y) = fX g −1 (y) | g −1 (y)|IY .
dy
Proof.
d
FY (y)
dy
d
−1
F
g
(y)
,
X
dy =
d
−1
1
−
F
g
(y)
,
X
dy
d −1
−1
fX g (y) dy g (y),
d −1
=
−fX g −1 (y) dy
g (y),
fY (y) =
if g is increasing;
if g is decreasing.
if g is increasing;
if g is decreasing.
which gives the thesis of the theorem.
Example 1.17. Suppose that Z ∼ N (0, 1). What is the distribution of Y = Z 2 ?
For Y > 0, the cdf of Y = Z 2 is
FY (y) = P (Y ≤ y)
= P (Z 2 ≤ y)
√
√
= P (− y ≤ Z ≤ y)
√
√
= FZ ( y) − FZ (− y).
The pdf can now be obtained by differentiation:
d
FY (y)
dy
d
√
√ =
FZ ( y) − FZ (− y)
dy
1
1
√
√
= √ fZ ( y) + √ fZ (− y)
2 y
2 y
fY (y) =
Now, for the standard normal distribution we have
1
2
fZ (z) = √ e−z /2 , ∞ < z < ∞.
2π
1.8. FUNCTIONS OF RANDOM VARIABLES
25
This gives,
1 (−√y)2 /2
1
1 (√y)2 /2
fY (y) = √ √ e
+√ e
2 y
2π
2π
1 1 −y/2
=√ √ e
, 0 < y < ∞.
y 2π
This is a well known pdf function, which we will use in statistical inference. It is
called chi squared random variable with one degree of freedom and it is denoted
by χ21 .
Note that g(Z) = Z 2 is not a monotone function, but the range of Z, (−∞, ∞),
can be partitioned so that it is monotone on its sub-sets.
Exercise 1.11. The pdf obtained in Example 1.17 is also pdf of a Gamma rv for
some specific values of its parameters. What are these values?
Exercise 1.12. Suppose that Z ∼ N (0, 1). Find the distribution of Y = µ + σZ
for constant µ and σ.
Exercise 1.13. Let X be a random variable with moment generating function MX .
(i) Show that the moment generating function of Y = a + bX, where a and b are
constants, is given by
MY (t) = eta MX (tb).
(ii) Derive the moment generating function of Y ∼ N (µ, σ 2 ). Hint: First find
MZ (t) for a standard normal rv Z.