# Generalized Neyman-Pearson Lemma

We will review the generalized Neyman-Pearson lemma, see [CK01].

Let ${(\Omega, \mathcal{F}, \mu)}$ be a probability space. ${\mathcal{G}, \mathcal{H} \subset L^1(\mathcal{\mu})}$ are two subspaces of ${\mu}$-integrable random variables. Let ${\chi}$ be defined by

$\displaystyle \chi = \{X: \Omega/\mathcal{F} \rightarrow [0,1]/\mathcal{B}([0,1])\}$

and

$\displaystyle \chi_x = \{X\in \chi: \mathbb{E}[H X] \le x, \ \forall H \in \mathcal{H}\}.$

The problem is to find

$\displaystyle V(x) = \sup_{X\in \chi_x} \inf_{G\in \mathcal{G}} \mathbb{E} [GX].$

In other words, it is to maximize

$\displaystyle \gamma(x) := \inf_{G\in \mathcal{G}} \mathbb{E}[ GX]$

over the constraints

$\displaystyle s(x) = \sup_{H\in \mathcal{H}} \mathbb{E} [HX]\le x, \ \forall H\in \mathcal{H}.$

Assumption 1 ${\mathcal{G}\subset L^1(\mu)}$ is convex and closed under ${\mu}$-a.e.

The following theorem is a straightforward generalization of the result of [CK01].

Theorem 1 Suppose Assumption 1holds. Then, there exists ${(\hat G, \hat H, \hat z, \hat X) \in \mathcal{G} \times \overline{Co(\mathcal{H})} \times (0,\infty) \times \chi_x}$, such that

1. For all ${(x, G) \in \chi_x \times \mathcal{G}}$, following holds:

$\displaystyle \mathbb{E} [\hat G X] \le \mathbb{E} [\hat G \hat X] \le \mathbb{E} [ G \hat X].$

This implies the exitence of the saddle point, i.e.

$\displaystyle V(x) = \mathbb{E} \hat G \hat X] = \sup_{X\in \chi_x} \inf_{G\in \mathcal{G}} \mathbb{E} [GX] = \inf_{G\in \mathcal{G}} \sup_{X\in \chi_x} \mathbb{E} [GX].$

2. Let ${\tilde V(z) = \inf_{G\times \overline{Co(\mathcal{H})}} \mathbb{E} [(G- zH)^+]}$. Then,

$\displaystyle V(z) = \inf_{z>0} \{xz + \tilde V(z)\}.$

Moreover, ${\hat z}$ is given by

$\displaystyle \hat z = \arg\min_{z>0} \{xz + \tilde V(z)\}.$

3. For all ${z\ge 0}$, there exists ${(\hat G_z, \hat H_z) \in G\times \overline{Co(\mathcal{H})}}$ such that

$\displaystyle \tilde V(z) = \mathbb{E} [(\hat G_z- z \hat H_z)^+].$

${(\hat G, \hat H)}$ can be taken by ${\hat G = \hat G_{\hat z}}$ and ${\hat H = \hat H_{\hat z}}$.

4. ${\hat X}$ can take the form of

$\displaystyle \hat X = I_{\{\hat G > \hat z \hat H\}} + B I_{\{\hat z \hat H = \hat G\}}, \ \ \ \ \ (1)$

where ${B\in L^0(\mathcal{F})}$ is chosen to satisfy

$\displaystyle \mathbb{E}[\hat H \hat X] = x. \ \ \ \ \ (2)$

5. In such a way, we have

$\displaystyle \mathbb{E}[\hat G \hat X] = \mathbb{E}[(\hat G - \hat z \hat H)^+ ] + \hat z x, \ \mu-a.e. \ \ \ \ \ (3)$

and ${\sup_{\mathcal{H}} \mathbb{E} [ H \hat X] = x}$.