What is the point of a likelihood ratio test? Suppose you have:

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*$n$ data points, i.i.d. $X_i \forall i \in 1,2,3,...n$

*$H_0: X_i \sim \mathcal{N}(0,1)$

*$H_1: X_i \sim \mathcal{N}(0,4)$
You know the distribution of $X_i$ under both $H_o$ and $H_1$
The problem is to choose $H_0$ or $H_1$ that results in a given False Positive Rate (Type 1 error) under $H_0$
The likelihood ratio test as described here goes as follows:

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*Define the likelihood ratio $R = \frac{\mathcal{L}(H_1 \mid X)}{\mathcal{L}(H_0 \mid X)}$ = $\frac{P(X;H_1)}{P(X; H_0)}$

*Set $R$ equal to some threshold $\xi$

*Decide a level of Type 1 error $\alpha$ under $H_0$ that you want

*Find the threshold $\xi$ such that $P(R > \xi ; H_0) = \alpha$

*Reject $H_0$ if $R > \xi$
This seems like a lot of that that didn't need to be. We could directly have set $P(X;H_0) = \alpha$ to be the decision boundary. That is to say if $P(X;H_0)$ is lesser than $\alpha$ then reject the null hypothesis. Why the extra steps of coming via $\xi$? What purpose does the likelihood ratio test serve?

I don't understand your first suggestion: What does $P(X;H_0) = \alpha$ mean?
(A probability measure can only be applied to 'events'; like the event $\{R>\zeta\}$ for the likelihood ratio test.)
To answer the question on purpose of the test: the point of the likelihood ratio test is that for simple hypothesis it gives the most power for a given size (Neyman-Pearson Lemma, see Likelihood Ratio Test wikipedia page for what seems like a good explanation to me).
My intuition for this is that it gives 'the most bang for your buck'. I think of size as money I can spend and power as something I want to buy (e.g. rice); but rice is sold at a variety of shops each with finite capacity; to get the most rice for a set amount of money I should buy from the cheapest places I can. (Here cheap price corresponds to high likelihood ratio.)
This algorithm you presented gives a pretty clean way to express that idea mathematically (in my opinion - 5 simple lines is not too bad!).
