Neyman-Pearson Theorem

The Likelihood-Ratio Test is given by

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Proof

Then we want to find the Detector that maximizes

$$J=P_D+\lambda (P_{FA}-\alpha )$$

Where $\alpha$ is $\alpha =max(P_{FA})$, $P_{FA}$ is the Probability of False Alarm, and $P_D$ is the Probability of Detection.

We have the threshold

$$\Lambda (x)\lessgtr \gamma$$

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We then get

$$\begin{array}{r}J={\int }_{{\Omega }_1}\underset{g(x)}{\underbrace{[p(x;H_1)+\lambda p(x;H_0)]}}dx-\lambda \alpha \end{array}$$

And we have

$${\Omega }_1^{\ast }=\{x\in {\mathbb{R}}^N:g(x)>0\}$$

Which is the optimum detector. From this we get

\begin{align} p(x;H_{1})+\lambda p(x;H_{0}) &>0 \\ p(x;H_{1})&> -\lambda p(x;H_{0}) \\ \Lambda(x)=\frac{p(x;H_{1})}{p(x;H_{0})} &>-\lambda \end{align} $$Where we find $\gamma$ by

\gamma: \int {\Omega{1}(\gamma)}p(x;\mathcal{H}_{0}) , dx = \alpha