The joint moment generating function (joint mgf) is a multivariate generalization of the moment generating function.
Similarly to the univariate case, a joint mgf uniquely determines the joint distribution of its associated random vector, and it can be used to derive the cross-moments of the distribution by partial differentiation.
If you are not familiar with the univariate concept, you are advised to first read the lecture on moment generating functions.
Table of contents
Let us start with a formal definition.
Definition
Let
be a
random vector. If the
expected
value
exists
and is finite for all
real vectors
belonging to a closed rectangle
:
with
for all
,
then we say that
possesses a joint moment generating function and the function
defined
by
is
called the joint moment generating function of
.
Not all random vectors possess a joint mgf. However, all random vectors possess a joint characteristic function, a transform that enjoys properties similar to those enjoyed by the joint mgf.
As an example, we derive the joint mgf of a standard multivariate normal random vector.
Example
Let
be a
standard multivariate normal random vector. Its
support
is
and
its joint
probability density function
is
As
explained in the lecture entitled Multivariate normal
distribution, the
components of
are
mutually independent standard normal
random variables, because the joint probability density function of
can be written
as
where
is the
-th
entry of
and
is the probability density function of a standard normal random
variable:
Therefore,
the joint mgf of
can be derived as
follows:
Since
the mgf of a standard normal random variable
is
then
is defined for any
.
As a consequence,
is defined for any
.
The next proposition shows how the joint mgf can be used to derive the cross-moments of a random vector.
Proposition
If a
random vector
possesses a joint mgf
,
then
possesses finite cross-moments of order
,
for any
.
Furthermore, if you define a cross-moment of order
as
where
and
,
then
where
the derivative on the right-hand side is the
-th
order partial derivative of
evaluated at the point
.
We do not provide a rigorous proof of this
proposition, but see, e.g., Pfeiffer (1978) and
DasGupta (2010). The main intuition, however, is
quite simple. Differentiation is a linear operation and the expected value is
a linear operator. This allows us to differentiate through the expected value,
provided appropriate technical conditions (omitted here) are
satisfied:Evaluating
this derivative at the point
,
we
obtain
The following example shows how this proposition can be applied.
Example
Let's continue with the previous example. The joint mgf of a
standard normal random vector
is
The
second cross-moment of
can be computed by taking the second cross partial derivative of
:
One of the most important properties of the joint mgf is that it completely characterizes the joint distribution of a random vector.
Proposition
Let
and
be two
random vectors, possessing joint mgfs
and
.
Denote by
and
their joint distribution
functions.
and
have the same joint distribution if and only if they have the same joint
mgfs:
The reader may refer to
Feller (2008) for a rigorous proof. The informal
proof given here is almost identical to that given for the univariate case. We
confine our attention to the case in which
and
are discrete random vectors taking only finitely many values. As far as the
left-to-right direction of the implication is concerned, it suffices to note
that if
and
have the same distribution,
then
The
right-to-left direction of the implication is proved as follows. Denote by
and
the supports of
and
and by
and
their joint
probability mass functions. Define the union of the two
supports:
and
denote its members by
.
The joint mgf of
can be written
as
By
the same line of reasoning, the joint mgf of
can be written
as
If
and
have the same joint mgf,
then
for
any
belonging to a closed rectangle where the two mgfs are well-defined,
and
Rearranging
terms, we
obtain
This
equality can be verified for every
only
if
for
every
.
As a consequence, the joint probability mass functions of
and
are equal, which implies that also their joint distribution functions are
equal.
This proposition is used very often in applications where one needs to demonstrate that two joint distributions are equal. In such applications, proving equality of the joint moment generating functions is often much easier than proving equality of the joint distribution functions.
The following sections contain more details about the joint mgf.
Let
be a
random vector possessing joint mgf
.
Definewhere
is
a
constant vector and and
is
an
constant matrix.
Then, the
random vector
possesses a joint mgf
and
Using the definition of mgf, we
getIf
is defined on a closed rectangle
,
then
is defined on another closed rectangle whose shape and location depend on
and
.
Let
be a
random vector.
Let its entries
,
...,
be
mutually independent random variables possessing a mgf.
Denote the mgf of the
-th
entry of
by
.
Then, the joint mgf of
is
This fact is demonstrated as
follows:
Let
,
...,
be
mutually independent random vectors, all of dimension
.
Let
be their
sum:
Then, the joint mgf of
is the product of the joint mgfs of
,
...,
:
This
fact descends from the properties of mutually independent random vectors and
from the definition of joint
mgf:
Some solved exercises on joint moment generating functions can be found below.
Let
be a
discrete random vector and
denote its components by
and
.
Let the support of
be
and
its joint probability
mass function
be
Derive the joint moment generating function of
,
if it exists.
By the definition of moment generating
function, we
haveObviously,
the joint moment generating function exists and it is well-defined because the
above expected value exists for any
.
Let
be
a
random vector with joint moment generating
function
Derive the expected value of
.
The moment
generating function of
is
The
expected value of
is obtained by taking the first derivative of its moment generating
function:
and
evaluating it at
:
Let
be
a
random vector with joint moment generating
function
Derive the covariance between
and
.
We can use the following covariance
formula:The
moment generating function of
is
The
expected value of
is obtained by taking the first derivative of its moment generating
function:
and
evaluating it at
:
The
moment generating function of
is
To
compute the expected value of
we take the first derivative of its moment generating
function:
and
evaluating it at
:
The
second cross-moment of
is computed by taking the second cross-partial derivative of the joint moment
generating
function:
and
evaluating it at
:
Therefore,
DasGupta, A. (2010) Fundamentals of probability: a first course, Springer.
Feller, W. (2008) An introduction to probability theory and its applications, Volume 2, Wiley.
Pfeiffer, P. E. (1978) Concepts of probability theory, Dover Publications.
Please cite as:
Taboga, Marco (2021). "Joint moment generating function", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/fundamentals-of-probability/joint-moment-generating-function.
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