# Neyman-Pearson Lemma

The Neyman-Pearson Lemma is a fundamental result in the theory of hypothesis testing and can also be restated in a form that is foundational to classification problems in machine learning. Even though the Neyman-Pearson lemma is a very important result, it has a simple proof. Let’s go over the theorem and its proof.

## Performance of a classifier

Before stating the lemma, we need to know some basic metrics for judging the performance of a binary classifier. The binary classification problem is this: given an observation, $$X$$ we want to predict the class of this observation, $$Y \in \{1,2\}$$. A classifier is a function $$h$$ that predicts the class of a given observation. There are four probabilities that we care about when predicting classes

$P(h(X) = 1 | Y=1)$ $P(h(X) = 2 | Y=1)$ $P(h(X) = 1 | Y=2)$ $P(h(X) = 2 | Y=2)$

There are different names for these probabilities. The Neyman-Pearson lemma is concerned with two of them. The first, $$P(h(X)=2|Y=2)$$, is often called the discovery rate (DR) or the power. We will call it the DR. The second, $$P(h(X)=2|Y=1)$$, is often called the false alarm rate (FAR) or Type I Error. We will call it the FAR. Notice that with the FAR and the DR we can easily determine the other probabilities (since two pairs must add up to 1).

## Neyman-Pearson Lemma

We are now ready to state the lemma. Denote the conditional densities $$f(X|Y=i) = f_i(x)$$. For $$t > 0$$ define a classifier (with parameter $$t$$) by

$h_t(x) = \begin{cases} 1 \text{ if } f_2(x) \leq t f_1(x) \\ 2 \text{ if } f_2(x) > t f_2(x) \end{cases}.$

Let $$h(x) \rightarrow \{1,2\}$$ be any classification function. Then $$DR(h) \leq DR(h_t)$$ if $$FAR(h) \leq FAR(h_t)$$.

## Proof

First, we notice that we can partition the space into 4 parts:

$A = \{ x : h(x) = h_t(x) = 2 \}$ $B = \{ x : h(x) = 2, h_t(x) = 1 \}$ $C = \{ x : h(x) = 1, h_t(x) = 2 \}$ $D = \{ x : h(x) = h_t(x) = 1 \}.$

Then,

$FAR(h) = P(h(X)=2|Y=1) = P(X \in A \cup B | Y = 1)$ $FAR(h_T) = P(h_(X)=2|Y=1) = P(X \in A \cup C | Y = 1).$

By assumption, $$FAR(h) \leq FAR(h_t)$$, so

$P(X \in A \cup B | Y = 1) \leq P(X \in A \cup C | Y = 1)$

which implies

$P(X \in B | Y = 1) \leq P(X \in C | Y = 1)$

or equivalently

$\int_B f_1(x) dx \leq \int_C f_1(x) dx.$

By the definitions of $$h_t, B, C$$ we have that $$f_2 \leq t f_1$$ on $$B$$ and $$f_2 > t f_1$$ on $$C$$. From this we know

$P(X \in B | Y = 2) = \int_B f_2(x) dx$ $\leq t \int_B f_1(x) dx$ $\leq \int_C t f_1(x) dx$ $\leq \int_C f_2(x) dx$ $= P(X \in C | Y = 2).$

Applying this result, we have

$DR(h) = P(h(x) =2 | Y=2)$ $= P(X \in A \cup B| Y = 2)$ $\leq P(X \in A \cup C| Y = 2)$ $= P(h_t(X) = 2 | Y = 2)$ $= DR(h_t).$

This is what we wanted to prove.

## Hypothesis testing

When thinking about hypothesis tests, the Neyman-Pearson lemma applies to tell us that a likelihood ratio test is the the most powerful hypothesis test for a given significance level $$\alpha$$. The reason why likelihood ratio tests aren’t the only tests we use is that the relationship between $$\alpha$$ and $$t$$ is not always clear (i.e. we don’t always know how to calibrate the test).

## Generalization

Something I’m interested in looking into is the sequential probability ratio test. This is a generalization of the Neyman-Pearson lemma for sequential hypothesis testing (sequential analysis). Here are a few links if this interests you:

Published on June 5, 2019