How to work out the relationship between error $\alpha$ and threshold $\eta$ Reading up on the Neyman Pearson Lemma, I have a question about how to obtain the probability for a type I error, $\alpha$, when we establish a threshold $\eta$ for the liklihood ratio $\Lambda(x)=\frac{ L(x \mid \theta _0)}{ L (x\mid\theta _1)} \leq \eta$.
The probability of a type I error (rejecting the null-hypothesis, under the condition, that it is actually true) is given by the lemma by 
$$\Pr(\Lambda(X)\leq \eta\mid H_0)=\alpha$$
As I understand it, we can either choose an $\eta$, and then calculate the probability of a type 1 error, or (preferrably) find an expression for $\Pr(\Lambda(X)\leq \eta\mid H_0)$ and basically find a function $\eta : \alpha \mapsto \eta(\alpha)$. Either way, I need to find a way to solve $\Pr(\Lambda(X)\leq \eta\mid H_0)$.
When I tried to solve this, I found that $\Pr(\Lambda(X)\leq \eta\mid H_0)$ looks like a cumulative distribution function of $\Lambda$. This makes me wonder: How one could obtain how $\Lambda$ is distributed?
Or to rephrase: How can I use the Neyman-Pearson lemma to perform a hypothesis test between two models, with a given type ii error-probability of $\alpha=0.05$.
 A: This question is resolved, because it contains a misunderstanding of  the term $\Pr(\Lambda(X)\le\eta | H_0)$. This term is not meant to be a cdf for $\Lambda$, but a probability for $X$ fulfilling the constraint of $\Lambda(X)\le\eta$. So there is no need to think of the likelihood ratio to be distributed in any way.
A: Neyman Pearson's lemma states that if you fix $\alpha$ or the false alarm probability ($P_{FA}$) you can calculate the threshold, $\eta$, as a solution to the following equation:
\begin{equation}
\alpha = P_{FA}=\int_{\eta}^{\infty} P_r (Z|H_0)d Z,
\end{equation}
where $Z=\Lambda(X)$. 
So to answer your question it is essential in many cases to determine the PDF of the test.
Example: for detecting a DC level of value $A$ in white Gaussian noise (WGN) the test will read
\begin{equation}
\Lambda(X)=\frac{1}{N} \sum_{N=0}^{N-1}x_n >  \eta.
\end{equation}
In this case the test $\Lambda(X)$ is Gaussian distributed witn variance $\frac{\sigma^2}{N}$ under both hypotheses and with zero mean under hypothesis $H_0$ and with mean $A$ under $H_1$.
Using this you can calculate the threshold for a given probability of false alarm using Neyman Pearson's Lemma.
