# Neyman–Pearson lemma

In statistics, the Neyman–Pearson lemma, named after Jerzy Neyman and Egon Pearson, states that when performing a hypothesis test between two point hypotheses H0θ = θ0 and H1θ = θ1, then the likelihood-ratio test which rejects H0 in favour of H1 when

${\displaystyle \Lambda (x)={\frac {L(\theta _{0}\mid x)}{L(\theta _{1}\mid x)}}\leq \eta }$

where

${\displaystyle P(\Lambda (X)\leq \eta \mid H_{0})=\alpha }$

is the most powerful test of size α for a threshold η. If the test is most powerful for all ${\displaystyle \theta _{1}\in \Theta _{1}}$, it is said to be uniformly most powerful (UMP) for alternatives in the set ${\displaystyle \Theta _{1}\,}$.

In practice, the likelihood ratio is often used directly to construct tests — see Likelihood-ratio test. However it can also be used to suggest particular test-statistics that might be of interest or to suggest simplified tests — for this, one considers algebraic manipulation of the ratio to see if there are key statistics in it related to the size of the ratio (i.e. whether a large statistic corresponds to a small ratio or to a large one).

## Proof

Define the rejection region of the null hypothesis for the NP test as

${\displaystyle R_{NP}=\left\{x:{\frac {L(\theta _{0}|x)}{L(\theta _{1}|x)}}\leq \eta \right\}.}$

Any other test will have a different rejection region that we define as ${\displaystyle R_{A}}$. Furthermore, define the probability of the data falling in region R, given parameter ${\displaystyle \theta }$ as

${\displaystyle P(R,\theta )=\int _{R}L(\theta |x)\,dx,}$

For both tests to have size ${\displaystyle \alpha }$, it must be true that

${\displaystyle \alpha =P(R_{NP},\theta _{0})=P(R_{A},\theta _{0})\,.}$

It will be useful to break these down into integrals over distinct regions:

${\displaystyle P(R_{NP},\theta )=P(R_{NP}\cap R_{A},\theta )+P(R_{NP}\cap R_{A}^{c},\theta ),}$

and

${\displaystyle P(R_{A},\theta )=P(R_{NP}\cap R_{A},\theta )+P(R_{NP}^{c}\cap R_{A},\theta ).}$

Setting ${\displaystyle \theta =\theta _{0}}$ and equating the above two expression yields that

${\displaystyle P(R_{NP}\cap R_{A}^{c},\theta _{0})=P(R_{NP}^{c}\cap R_{A},\theta _{0}).}$

Comparing the powers of the two tests, ${\displaystyle P(R_{NP},\theta _{1})}$ and ${\displaystyle P(R_{A},\theta _{1})}$, one can see that

${\displaystyle P(R_{NP},\theta _{1})\geq P(R_{A},\theta _{1})\iff P(R_{NP}\cap R_{A}^{c},\theta _{1})\geq P(R_{NP}^{c}\cap R_{A},\theta _{1}).}$

Now by the definition of ${\displaystyle R_{NP}}$,

${\displaystyle P(R_{NP}\cap R_{A}^{c},\theta _{1})=\int _{R_{NP}\cap R_{A}^{c}}L(\theta _{1}|x)\,dx\geq {\frac {1}{\eta }}\int _{R_{NP}\cap R_{A}^{c}}L(\theta _{0}|x)\,dx={\frac {1}{\eta }}P(R_{NP}\cap R_{A}^{c},\theta _{0})}$
${\displaystyle ={\frac {1}{\eta }}P(R_{NP}^{c}\cap R_{A},\theta _{0})={\frac {1}{\eta }}\int _{R_{NP}^{c}\cap R_{A}}L(\theta _{0}|x)\,dx\geq \int _{R_{NP}^{c}\cap R_{A}}L(\theta _{1}|x)dx=P(R_{NP}^{c}\cap R_{A},\theta _{1}).}$

Hence the inequality holds.

## Example

Let ${\displaystyle X_{1},\dots ,X_{n}}$ be a random sample from the ${\displaystyle {\mathcal {N}}(\mu ,\sigma ^{2})}$ distribution where the mean ${\displaystyle \mu }$ is known, and suppose that we wish to test for ${\displaystyle H_{0}:\sigma ^{2}=\sigma _{0}^{2}}$ against ${\displaystyle H_{1}:\sigma ^{2}=\sigma _{1}^{2}}$. The likelihood for this set of normally distributed data is

${\displaystyle L\left(\sigma ^{2};{\mathbf {x} }\right)\propto \left(\sigma ^{2}\right)^{-n/2}\exp \left\{-{\frac {\sum _{i=1}^{n}\left(x_{i}-\mu \right)^{2}}{2\sigma ^{2}}}\right\}.}$

We can compute the likelihood ratio to find the key statistic in this test and its effect on the test's outcome:

${\displaystyle \Lambda ({\mathbf {x} })={\frac {L\left(\sigma _{0}^{2};{\mathbf {x} }\right)}{L\left(\sigma _{1}^{2};{\mathbf {x} }\right)}}=\left({\frac {\sigma _{0}^{2}}{\sigma _{1}^{2}}}\right)^{-n/2}\exp \left\{-{\frac {1}{2}}(\sigma _{0}^{-2}-\sigma _{1}^{-2})\sum _{i=1}^{n}\left(x_{i}-\mu \right)^{2}\right\}.}$

This ratio only depends on the data through ${\displaystyle \sum _{i=1}^{n}\left(x_{i}-\mu \right)^{2}}$. Therefore, by the Neyman–Pearson lemma, the most powerful test of this type of hypothesis for this data will depend only on ${\displaystyle \sum _{i=1}^{n}\left(x_{i}-\mu \right)^{2}}$. Also, by inspection, we can see that if ${\displaystyle \sigma _{1}^{2}>\sigma _{0}^{2}}$, then ${\displaystyle \Lambda ({\mathbf {x} })}$ is a decreasing function of ${\displaystyle \sum _{i=1}^{n}\left(x_{i}-\mu \right)^{2}}$. So we should reject ${\displaystyle H_{0}}$ if ${\displaystyle \sum _{i=1}^{n}\left(x_{i}-\mu \right)^{2}}$ is sufficiently large. The rejection threshold depends on the size of the test. In this example, the test statistic can be shown to be a scaled Chi-square distributed random variable and an exact critical value can be obtained.