Let 
be a
parametric family of densities for a random observable
X, where X may be real valued, vector valued,
continuous, or discrete, or neither. (Of course,
if X is discrete you may prefer to think of
as a probability mass function,
but we can just think of it as a density of
the discrete type or whatever.)
We assume the dominating measure 
is
-finite. (Don't worry about what
-finite means if you don't know.)
Basically, if X is continuous then
is Lebesgue measure and 
=
dx below; if X is discrete then
is counting measure and 
= 
where the summation is over all
all possible values of x.) Thus, for instance
Now we have by the definition of a probability density that
Thus, differentiating both sides
Assuming we can interchange differentiation and integration,
Note that if f is strictly positive then
and (
) may be rewritten as
This identity will turn out to be useful for many purposes.
How can we justify the interchange of differentiation and
integration in ()? Well, assume that the
parameter space is open and for each 
there is a
neighborhood 
and ``dominating'' function
such that
Then, assuming the usual measurability,
the conditions for Theorem will hold
with (a,b) of the theorem being the neighborhood
. As will be seen in the example
below, it is often easier to verify the identity
() using ad hoc methods rather than
establishing () and ().
The following result is very useful for many of the standard densities - those of so called ``exponential type.''
Proof. We will make use of the elementary inequality
Perhaps the easiest way to verify this is to
use the Taylor expansion 
=
, valid for all
. Then 
=
and all
the summands are nonnegative when 
.
Suppose 
is such that
is contained in the neigborhood where 
is
finite. Then () gives us
Similarly, by simple monotonicity properties of the exponential function,
The results in (), (),
(), and () can be combined into
Since the sum of two nonnegative numbers is larger
than either of them, we can
combine () and ()
into the single inequality
Let G(x) the the r.h.s. of ().
Since 
is in the neighborhood
where is finite,
And
Therefore, 
works as a dominating function
in Theorem and we have
which completes the proof.
This last result can be extended to the following.
Proof. Define the measure
Now it follows from the proof of Proposition
that
is finite in a neighborhood of when n = 1.
Assuming 
in a neighborhood
of , it follows by the proof of Proposition
that the same holds for 
.
Thus, by induction, for each n,
in a neighborhood of .
Hence, applying Proposition , it follows
that 
exists for each n
and may be computed by differentiating under the integral
sign. This completes the proof.
