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 H₀: θ = θ₀ and H₁: θ = θ₁, then the likelihood-ratio test which rejects H₀ in favour of H₁ when

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

where

$P(\Lambda(X)\leq \eta|H_0)=\alpha$

is the most powerful test of size α for a threshold η. If the test is most powerful for all $\theta_1 \in \Theta_1$, it is said to be uniformly most powerful (UMP) for alternatives in the set $\Theta_1 \,$.

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 is 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

$R_{NP}=\left\{ X: \frac{L(\theta_{0},X)}{L(\theta_{1},X)} \leq \eta\right\} .$

Any other test will have a different rejection region that we define as R_A. Furthermore define the function of region, and parameter

$P(R,\theta)=\int_R L(\theta|x)\, dx,$

where this is the probability of the data falling in region R, given parameter θ.

For both tests to have significance level α, it must be true that

$\alpha= P(R_{NP}, \theta_0)=P(R_A, \theta_0) \,.$

However it is useful to break these down into integrals over distinct regions, given by

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

and

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

Setting θ = θ₀ and equating the above two expression, yields that

$P(R_{NP} \cap R_A^c, \theta_0) = P(R_{NP}^c \cap R_A, \theta_0).$

Comparing the powers of the two tests, which are P(R_NP,θ₁) and P(R_A,θ₁), one can see that

$P(R_{NP},\theta_1) \geq P(R_A,\theta_1) \text{ if, and only if, } P(R_{NP} \cap R_A^c, \theta_1) \geq P(R_{NP}^c \cap R_A, \theta_1).$

Now by the definition of R_NP ,

$P(R_{NP} \cap R_A^c, \theta_1)= \int_{R_{NP}\cap R_A^c} L(\theta_{1}|x)\,dx \geq \frac{1}{\eta} \int_{R_{NP}\cap R_A^c} L(\theta_0|x)\,dx = \frac{1}{\eta}P(R_{NP} \cap R_A^c, \theta_0)$

$= \frac{1}{\eta}P(R_{NP}^c \cap R_A, \theta_0) = \frac{1}{\eta}\int_{R_{NP}^c \cap R_A} L(\theta_{0}|x)\,dx \geq \int_{R_{NP}^c\cap R_A} L(\theta_{1}|x)dx = P(R_{NP}^c \cap R_A, \theta_1).$

Hence the inequality holds.

Example

Let $X_1,\dots,X_n$ be a random sample from the $\mathcal{N}(\mu,\sigma^2)$ distribution where the mean μ is known, and suppose that we wish to test for $H_0:\sigma^2=\sigma_0^2$ against $H_1:\sigma^2=\sigma_1^2$. The likelihood for this set of normally distributed data is

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

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

$\Lambda(\mathbf{x}) = \frac{L\left(\sigma_1^2;\mathbf{x}\right)}{L\left(\sigma_0^2;\mathbf{x}\right)} = \left(\frac{\sigma_1^2}{\sigma_0^2}\right)^{-n/2}\exp\left\{-\frac{1}{2}(\sigma_1^{-2}-\sigma_0^{-2})\sum_{i=1}^n \left(x_i-\mu\right)^2\right\}.$

This ratio only depends on the data through $\sum_{i=1}^n \left(x_i-\mu\right)^2$. Therefore, by the Neyman-Pearson lemma, the most powerful test of this type of hypothesis for this data will depend only on $\sum_{i=1}^n \left(x_i-\mu\right)^2$. Also, by inspection, we can see that if $\sigma_1^2>\sigma_0^2$, then $\Lambda(\mathbf{x})$ is an increasing function of $\sum_{i=1}^n \left(x_i-\mu\right)^2$. So we should reject H₀ if $\sum_{i=1}^n \left(x_i-\mu\right)^2$ 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.

See also

• Statistical power

References

• Jerzy Neyman, Egon Pearson (1933). "On the Problem of the Most Efficient Tests of Statistical Hypotheses". Philosophical Transactions of the Royal Society of London. Series A, Containing Papers of a Mathematical or Physical Character 231: 289–337. doi:10.1098/rsta.1933.0009. JSTOR 91247. 
• cnx.org: Neyman-Pearson criterion

External links

• Cosma Shalizi, a professor of statistics at Carnegie Mellon University, gives an intuitive derivation of the Neyman-Pearson Lemma using ideas from economics