# 2.3 The neyman-pearson criterion  (Page 2/3)

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## Neyman-pearson criterion

$\max\{{P}_{D}\},\text{such that}{P}_{F}\le$

The maximization is over all decision rules (equivalently, over all decision regions ${R}_{0}$ , ${R}_{1}$ ). Using different terminology, the Neyman-Pearson criterionselects the most powerful test of size (not exceeding)  .

Fortunately, the above optimization problem has an explicit solution. This is given by the celebrated Neyman-Pearson lemma , which we now state. To ease the exposition, our initial statement of this result onlyapplies to continuous random variables, and places a technical condition on the densities. A more general statement is givenlater in the module.

## Neyman-pearson lemma: initial statement

Consider the test ${}_{0}:(x, {f}_{0}(x))$ ${}_{1}:(x, {f}_{1}(x))$ where ${f}_{i}(x)$ is a density. Define $(x)=\frac{{f}_{1}(x)}{{f}_{0}(x)}$ , and assume that $(x)$ satisfies the condition that for each $\in \mathbb{R}$ , $(x)$ takes on the value  with probability zero under hypothesis ${}_{0}$ . The solution to the optimization problem in is given by $(x)=\frac{{f}_{1}(x)}{{f}_{0}(x)}\underset{{}_{0}}{\overset{{}_{1}}{}}$ where  is such that ${P}_{F}=\int {f}_{0}(x)\,d x=$ If $=0$ , then  . The optimal test is unique up to a set of probability zero under ${}_{0}$ and ${}_{1}$ .

The optimal decision rule is called the likelihood ratio test . $(x)$ is the likelihood ratio , and  is a threshold . Observe that neither the likelihood ratio nor the threshold depends on the a priori probabilities $({}_{i})$ . they depend only on the conditional densities ${f}_{i}$ and the size constraint  . The threshold can often be solved for as a function of  , as the next example shows.

Continuing with , suppose we wish to design a Neyman-Pearson decision rule withsize constraint  . We have

$(x)=\frac{\frac{1}{\sqrt{2\pi }}e^{-\left(\frac{(x-1)^{2}}{2}\right)}}{\frac{1}{\sqrt{2\pi }}e^{-\left(\frac{x^{2}}{2}\right)}}=e^{x-\frac{1}{2}}$
By taking the natural logarithm of both sides of the LRT and rarranging terms, the decision rule is not changed, and weobtain $x\underset{{}_{0}}{\overset{{}_{1}}{}}\ln +\frac{1}{2}\equiv$ Thus, the optimal rule is in fact a thresholding rule like we considered in . The false-alarm probability was seen to be ${P}_{F}=Q()$ Thus, we may express the value of  required by the Neyman-Pearson lemma in terms of  : $=Q()^{(-1)}$

## Sufficient statistics and monotonic transformations

For hypothesis testing involving multiple or vector-valued data, direct evaluation of the size( ${P}_{F}$ ) and power ( ${P}_{D}$ ) of a Neyman-Pearson decision rule would require integrationover multi-dimensional, and potentially complicated decision regions. In many cases, however, this can be avoided bysimplifying the LRT to a test of the form $t\underset{{}_{0}}{\overset{{}_{1}}{}}$ where the test statistic $t=T(x)$ is a sufficient statistic for the data. Such a simplified form is arrived at by modifying both sides of the LRT withmontonically increasing transformations, and by algebraic simplifications. Since the modifications do not change thedecision rule, we may calculate ${P}_{F}$ and ${P}_{D}$ in terms of the sufficient statistic. For example, the false-alarm probability may be written

${P}_{F}=(\text{declare}{}_{1})=\int {f}_{0}(t)\,d t$
where ${f}_{0}(t)$ denotes the density of $t$ under ${}_{0}$ . Since $t$ is typically of lower dimension than $x$ , evaluation of ${P}_{F}$ and ${P}_{D}$ can be greatly simplified. The key is being able to reduce theLRT to a threshold test involving a sufficient statistic for which we know the distribution .

## Common variances, uncommon means

Let's design a Neyman-Pearson decision rule of size  for the problem ${}_{0}:(x, (0, ^{2}I))$ ${}_{1}:(x, (1, ^{2}I))$ where $> 0$ , $^{2}> 0$ are known, $0=\left(\begin{array}{c}0\\ \\ 0\end{array}\right)$ , $1=\left(\begin{array}{c}1\\ \\ 1\end{array}\right)$ are $N$ -dimensional vectors, and $I$ is the $N$ $N$ identity matrix. The likelihood ratio is

$(x)=\frac{\prod_{n=1}^{N} \frac{1}{\sqrt{2\pi ^{2}}}e^{-\left(\frac{({x}_{n}-)^{2}}{2^{2}}\right)}}{\prod_{n=1}^{N} \frac{1}{\sqrt{2\pi ^{2}}}e^{-\left(\frac{{x}_{n}^{2}}{2^{2}}\right)}}=\frac{e^{-\sum_{n=1}^{N} \frac{({x}_{n}-)^{2}}{2^{2}}}}{e^{-\sum_{n=1}^{N} \frac{{x}_{n}^{2}}{2^{2}}}}=e^{\frac{1}{2^{2}}\sum_{n=1}^{N} 2{x}_{n}-^{2}}=e^{\frac{1}{^{2}}(-\left(\frac{N^{2}}{2}\right)+\sum_{n=1}^{N} {x}_{n})}$
To simplify the test further we may apply the natural logarithm and rearrange terms to obtain $t\equiv \sum_{n=1}^{N} {x}_{n}\underset{{}_{0}}{\overset{{}_{1}}{}}\frac{^{2}}{}\ln +\frac{N}{2}\equiv$
We have used the assumption $> 0$ . If $< 0$ , then division by  is not a monotonically increasing operation, and the inequalitieswould be reversed.
The test statistic $t$ is sufficient for the unknown mean. To set the threshold  , we write the false-alarm probability (size) as ${P}_{F}=(t> )=\int {f}_{0}(t)\,d t$ To evaluate ${P}_{F}$ , we need to know the density of $t$ under ${}_{0}$ . Fortunately, $t$ is the sum of normal variates, so it is again normally distributed. In particular, we have $t=Ax$ , where $A=1^T$ , so $(t, (A0, A^{2}IA^T)=(0, N^{2}))$ under ${}_{0}$ . Therefore, we may write ${P}_{F}$ in terms of the Q-function as ${P}_{F}=Q(\frac{}{\sqrt{N}})$ The threshold is thus determined by $=\sqrt{N}Q()^{(-1)}$ Under ${}_{1}$ , we have $(t, (A1, A^{2}IA^T)=(N, N^{2}))$ and so the detection probability (power) is ${P}_{D}=(t> )=Q(\frac{-N}{\sqrt{N}})$ Writing ${P}_{D}$ as a function of ${P}_{F}$ , the ROC curve is given by ${P}_{D}=Q(Q({P}_{F})^{(-1)}-\frac{\sqrt{N}}{})$ The quantity $\frac{\sqrt{N}}{}$ is called the signal-to-noise ratio . As its name suggests, a larger SNR corresponds to improved performance of the Neyman-Pearsondecision rule.
In the context of signal processing, the foregoing problem may be viewed as the problem of detecting aconstant (DC) signal in additive white Gaussian noise : ${}_{0}:{x}_{n}={w}_{n},n=1,,N$ ${}_{1}:{x}_{n}=A+{w}_{n},n=1,,N$ where $A$ is a known, fixed amplitude, and $({w}_{n}, (0, ^{2}))$ . Here $A$ corresponds to the mean  in the example.

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