Suppose you have:

  1. $n$ data points, i.i.d. $X_i \forall i \in 1,2,3,...n$
  2. $H_0: X_i \sim \mathcal{N}(0,1)$
  3. $H_1: X_i \sim \mathcal{N}(0,4)$

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:

  1. Define the likelihood ratio $R = \frac{\mathcal{L}(H_1 \mid X)}{\mathcal{L}(H_0 \mid X)}$ = $\frac{P(X;H_1)}{P(X; H_0)}$
  2. Set $R$ equal to some threshold $\xi$
  3. Decide a level of Type 1 error $\alpha$ under $H_0$ that you want
  4. Find the threshold $\xi$ such that $P(R > \xi ; H_0) = \alpha$
  5. 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?


1 Answer 1


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!).

  • $\begingroup$ $P(X;H_0) = \alpha$ is indeed illegible. I have made an update to the post. Basically I wanted to say that if the probability of observing the observed data under the assumption of the null hypothesis is lesser than $\alpha$ then reject the null. Why do you go through the entire $\xi$ business? $\endgroup$ Commented Jul 22, 2022 at 9:55
  • $\begingroup$ Because you also want to have power - i.e. a large probability of catching the alternative if it turns out to be true. If you reject some region where probability of observing data under null is small it may turn out to be even smaller under the alternative. The point is to come up with a region that is unlikely under null while being maximally likely under alternative. Does that intuition help? $\endgroup$ Commented Jul 23, 2022 at 16:49
  • $\begingroup$ I guess to flesh out that analogue with the rice...let's say that at different shops they sell different quantities of rice at different prices - maybe 100g for £1 in place A, 5kg for £3 in place B. You want to buy from place which gives you most rice mass per unit money ie place B - as opposed to place with cheapest price ie place A (which I think is the analogue of what you are suggesting) Not sure how enlightening that is though! $\endgroup$ Commented Jul 23, 2022 at 16:53
  • $\begingroup$ I am afraid I do not see how the example is relevant. If I go in with an $\alpha$ in mind then I have set $\beta$ as well since $\alpha$ uniquely identifies a cutoff point (denoted as $q_{\alpha/2}$ in case of a standard normal). So, what I am saying is that the moment I decide what the type 1 error rate has to be the cutoff and $\beta$ are decided as well. So what additional purpose is served by the likelihood ratio test $\endgroup$ Commented Jul 27, 2022 at 11:23

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