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 take the form of

for the constant given by

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.

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