We will review useful Neyman-Pearson lemma, see [FS04]. It is often used to resolve optimization problem arised from efficient hedging problems.
Let and
be two probability measures on
. We say
is absolutely continuous w.r.t.
and write
, if for all
By Radon-Nikodym theorem, if and only if there exists
-measurable function
satisfying
We often write , and refer this to density or Radon-Nikodym derivative.
Hence, if , then
. In general, if
does not hold, then we have Lesbegue decomposition instead: there exists
with
satisfying
for some -measurable function
. In this case, we can generalize Radon-Nikodym derivative by
Proposition 1 (Neyman-Pearson Lemma) Let
and
be two probability measures on
. Suppose
satisfies
for some fixed
. Then, for arbitrary
,
Proof: Let . Define
Then,
, and
on
. Also,
. Now, we have
This completes the proof.
Remark The above Proposition 1 generalized the Neyman-Pearson Lemma given by [FS04], in which .
Let be the collection of all
-measurable indicator functions. Proposition 1 provides a partial solution to the following optimization problem [NP]: Given
, find
, which maximizes
under constraint
By 1, if there exists satisfies
with
, then
solves the problem [NP]. Now, we generalize Neyman-Pearson Lemma from indicator function to the randomized function. Let
be the space of all randomized test function
.
Proposition 2 Let
,
and
be three probability measures on
, and
. Consider optimization problem [GNP]: Given fixed
, find
, which maximizes
under constraint
Then, there exisits
solves the above problem [GNP], and
must satisfy
and some random variable
. In particular,
can simply take a non-random as of
For the optimization problem [GNP], optimizer given by (2) with
of (3) and (4) may not be unique one.
Proposition 3 Assume there exists
with continuous cdf under
is continuous. Then, there exists an optimizer
both for the problem [NP] and [GNP], i.e. the value of the problem [NP] and [GNP] are equal to each other.
Proof: Let be given by (3). We can construct
in each of the following three cases.
- If
, then
may be of (2) with
.
- If
, then
may be of (2) with
.
- If
, then
. Let
. Then,
satisfies
-
as
, and
as
.
-
is continuous.
Hence, there exists
such that
. Now,
may be of (2) with
.
-
One can use Proposition 1 and 2 to verify given above is the optimizer for both problems.