byMarco Taboga, PhD
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 thecross-moments of the distribution by partial differentiation.
If you are not familiar with the univariate concept, you are advised to first read the lecture onmoment generating functions.
Table of contents
Let us start with a formal definition.
Definition Let be a
random vector. If theexpected 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 thejoint moment generating function of
.
Not all random vectors possess a joint mgf. However, all random vectors possess ajoint 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. Itssupport
is
and itsjoint probability density function
is
As explained in the lecture entitledMultivariate 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 isthen
is defined for any
. As a consequence,
is defined for any
.
The next proposition shows how the joint mgf can be used to derive thecross-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) andDasGupta (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
![[eq25]](/image.pl?url=https%3a%2f%2fwww.statlect.com%2fimages%2fjoint-moment-generating-function__54.png&f=jpg&w=240)
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.
PropositionLet and
be two
random vectors, possessing joint mgfs
and
. Denote by
and
theirjoint distribution functions.
and
have the same joint distribution if and only if they have the same joint mgfs:
![[eq33]](/image.pl?url=https%3a%2f%2fwww.statlect.com%2fimages%2fjoint-moment-generating-function__70.png&f=jpg&w=240)
The reader may refer toFeller (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
theirjoint 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:![[eq54]](/image.pl?url=https%3a%2f%2fwww.statlect.com%2fimages%2fjoint-moment-generating-function__127.png&f=jpg&w=240)
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:![[eq57]](/image.pl?url=https%3a%2f%2fwww.statlect.com%2fimages%2fjoint-moment-generating-function__138.png&f=jpg&w=240)
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 itsjoint probability mass function be
Derive the joint moment generating function of, if it exists.
By the definition of moment generating function, we have
Obviously, the joint moment generating function exists and it is well-defined because the above expected value exists for any.
Letbe a
random vector with joint moment generating function
Derive the expected value of.
Themoment generating function of is
The expected value of
is obtained by taking the first derivative of its moment generating function:
and evaluating it at
:
Letbe a
random vector with joint moment generating function
Derive thecovariance 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 secondcross-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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