ISS 2016 Statistics Paper-2 Solution: Question 39 (Neyman-Pearson Lemma Explained)
Continuing our question-by-question walk through the ISS Statistics Paper-2 previous year papers, today's stop is Question 39 from the 2016 exam. This one is purely conceptual — no computation at all — and it tests whether you can state the Neyman-Pearson Lemma precisely rather than just recognise its name. A huge number of aspirants lose marks here simply because they mix up the direction of an inequality, so we will go through every line of the statement slowly.
Quick Summary
Topic:Neyman-Pearson Lemma — most powerful critical region for a simple hypothesis
Question Reference:ISS 2016, Statistics Paper-2, Question 39
Correct Answer:Option (a) — 1, 2 and 3
The Question as It Appeared in the Paper
Consider the following statements with respect to Neyman-Pearson Lemma if C is a critical region of size α and k is a constant.
L0/L1 ≤ k inside C
L0/L1 ≥ k outside C
C is a most powerful critical region of size α for testing θ = θ0 against θ = θ1
Which of the above are correct?
(a) 1, 2 and 3
(b) 2 and 3 only
(c) 1 and 3 only
(d) 3 only
Step 1: What Problem Is the Neyman-Pearson Lemma Actually Solving?
Before touching the three statements, it helps to remember exactly what situation this lemma addresses. Suppose you have a random sample from a population whose density is f(x, θ), and you want to test a simple null hypothesis against a simple alternative hypothesis:
H0 : θ = θ0 versus H1 : θ = θ1
"Simple" here means each hypothesis specifies the distribution completely — there is no range of values, just one fixed value of θ on each side. Among all the possible tests (equivalently, all the possible critical regions C in the sample space) that keep the Type I error fixed at level α, we want the one test that has the highest power, i.e. the smallest Type II error, at θ = θ1. That best test is called the most powerful (MP) test, and the region on which we reject H0 is the most powerful critical region of size α. The Neyman-Pearson Lemma tells us exactly how to construct it.
Step 2: Stating the Lemma Precisely
Let L0 = L(θ0) be the likelihood of the sample under H0, and L1 = L(θ1) be the likelihood of the sample under H1. The Neyman-Pearson Lemma says: a region C in the sample space is a most powerful critical region of size α for testing H0 against H1 if there exists a non-negative constant k such that:
L0/L1 ≤ k for every sample point inside C, and
L0/L1 ≥ k for every sample point outside C, and
P(sample point falls in C | H0) = α
In words: we reject H0 (i.e. the sample point lands "inside C") exactly when the likelihood ratio L0/L1 is small — that is, when the data is relatively more likely under H1 than under H0. We accept H0 when that ratio is large.
Pro Tip: The single most common mistake on this exact question is flipping the direction of the two inequalities, because students instinctively write "≤ k" for both lines instead of checking which region (inside or outside C) each one belongs to. The way this is usually taught is to anchor it with a one-line story: "inside C we reject H0, so the evidence favours H1, so L0 (numerator) must be small relative to L1 — hence ≤ k." Once you can reconstruct the inequality direction from that single sentence instead of memorising it, this type of statement-based question becomes almost mechanical.
Step 3: Checking Statements 1 and 2
Statement 1 says L0/L1 ≤ k inside C. Comparing this with the lemma above, this is exactly the first condition of the Neyman-Pearson Lemma, written with no change at all. So Statement 1 is correct.
Statement 2 says L0/L1 ≥ k outside C. This is exactly the second condition of the lemma. So Statement 2 is also correct.
Both statements reproduce the lemma's defining inequalities word for word, just split across two separate lines instead of being combined into one. There is no trick here beyond reading carefully — many candidates second-guess themselves and think one of the two must be "the wrong one" simply because MCQs often work that way, but that intuition fails on this particular question.
Step 4: Checking Statement 3
Statement 3 claims that C, built this way, is "a most powerful critical region of size α for testing θ = θ0 against θ = θ1." This is not a separate assumption — it is the conclusion of the Neyman-Pearson Lemma itself. If a region C satisfies conditions 1 and 2 together with the size condition P(C | H0) = α, then the lemma guarantees C is the most powerful critical region of that size. So Statement 3 correctly describes what the lemma proves, and it too is correct.
Step 5: Final Answer
Since Statements 1, 2 and 3 are all faithful restatements of the Neyman-Pearson Lemma — two giving the defining inequalities and the third giving the conclusion they lead to — all three are correct. That matches option (a) 1, 2 and 3.
This agrees exactly with the official answer key for Series A of the ISS 2016 Statistics Paper-2 (STATISTICS-II), which lists option (a) as the key for Question 39.
Why This Question Matters
The Neyman-Pearson Lemma is the foundation stone of the entire theory of hypothesis testing — virtually every later result about uniformly most powerful (UMP) tests, likelihood ratio tests, and SPRT builds directly on it. ISS Paper-2 repeatedly tests this lemma in slightly different disguises: sometimes as a direct statement question like this one, sometimes buried inside a numerical problem asking you to actually construct the critical region for a given f(x, θ). Students who drill this kind of statement-matching exercise until the inequality direction becomes automatic, rather than something to be re-derived under exam pressure, tend to pick up these marks very reliably — this is a habit worth building early rather than fixing in the last month before the exam.
Frequently Asked Questions
Is the Neyman-Pearson Lemma only applicable to simple hypotheses?
Yes, in its original form the lemma applies strictly to testing a simple null against a simple alternative, where both θ0 and θ1 are single fixed values. For composite hypotheses, you instead look for a uniformly most powerful (UMP) test, which may or may not exist and typically requires the monotone likelihood ratio property.
Why is the ratio written as L0/L1 and not L1/L0?
Either convention can be used as long as you are consistent, but the standard ISS/UPSC convention uses L0/L1, which is why the inequality is "small inside C" rather than "large inside C." If you memorise the ratio the other way round, simply reverse both inequality signs — the underlying logic (reject H0 when data favours H1) stays the same.
What exactly does "size α" mean for the critical region C?
The size of a critical region is the probability of rejecting H0 when H0 is actually true, i.e. P(sample point falls in C | H0) = α. This is exactly the Type I error probability, and it is fixed in advance (commonly 0.05 or 0.01) before the most powerful region is constructed.
How is the constant k actually determined in a numerical problem?
k is found by first expressing the likelihood ratio L0/L1 as a function of a test statistic, then using the size condition P(reject H0 | H0) = α to solve for the cut-off value of that statistic (and hence k) from the sampling distribution under H0.
Does the most powerful test depend on which value θ1 is chosen for the alternative?
In general, yes — the most powerful critical region constructed via the Neyman-Pearson Lemma can depend on the specific θ1 chosen. When the resulting critical region turns out to be the same for every θ1 in a one-sided alternative, that test becomes uniformly most powerful (UMP) for the composite alternative, which is a stronger and more useful property.
Is this topic likely to repeat in future ISS Paper-2 attempts?
Very likely. Statistical inference, and the Neyman-Pearson Lemma in particular, is one of the highest-weightage topics across ISS Statistics Paper-2 papers year after year, appearing both as direct theory questions like this one and as computational problems on constructing MP or UMP tests.
If anything here felt unclear, or you spotted a step worth discussing further, drop a comment below — and if a fellow aspirant is also working through the ISS previous year papers, consider sharing this post with them.
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