ISS 2016 Statistics Paper-2 Solution: Question 48 (SPRT Always Terminates)
Continuing our question-by-question walk through the ISS Statistics Paper-2 previous year papers, today we take up Question 48 from the 2016 paper. This one sits in the sequential analysis portion of the syllabus and asks something that trips up a lot of students: does a Sequential Probability Ratio Test (SPRT) always end, or could it, in theory, go on sampling forever?
Quick Summary
Topic:Sequential Analysis — Sequential Probability Ratio Test (SPRT) and its termination property
Question Reference:ISS 2016, Statistics Paper-2, Question 48
Correct Answer:Option (a) — Both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement-I
The Question, As Asked
48. Consider the following statements:
Statement-I: An SPRT always provides a decision.
Statement-II: Probability of terminating an SPRT is one.
Which one of the following is correct in respect of the above two statements?
(a) Both Statement-I and Statement-II are true and Statement-II is the correct explanation for Statement-I
(b) Both Statement-I and Statement-II are true but Statement-II is not the correct explanation for Statement-I
(c) Statement-I is true but Statement-II is false
(d) Statement-I is false but Statement-II is true
Step 1: What an SPRT Actually Is
Before touching the statements, let's build the SPRT from scratch, since the whole question hinges on understanding what "terminating" means here.
Suppose we want to test a simple null hypothesis H0: θ = θ0 against a simple alternative H1: θ = θ1, based on a sequence of independent and identically distributed (i.i.d.) observations X1, X2, X3, … drawn one at a time from a density f(x, θ).
Unlike a fixed-sample-size test, where you decide the sample size n in advance and then make a decision, an SPRT does not fix n beforehand. Instead, after every single new observation, it checks whether there is already enough evidence to stop and decide, or whether it should keep sampling.
At stage n, after observing X1, X2, …, Xn, we compute the likelihood ratio:
λn = [f(X1,θ1) f(X2,θ1) … f(Xn,θ1)] ÷ [f(X1,θ0) f(X2,θ0) … f(Xn,θ0)]
This is simply the joint likelihood under H1 divided by the joint likelihood under H0, using only the first n observations. Two constants A and B (with B < 1 < A) are fixed in advance using the desired Type I error α and Type II error β, roughly as A ≈ (1−β)/α and B ≈ β/(1−α).
The SPRT decision rule at each stage n is:
If λn ≥ A, stop sampling and reject H0 (accept H1).
If λn ≤ B, stop sampling and accept H0.
If B < λn < A, do not decide yet — draw one more observation Xn+1 and repeat.
Step 2: Turning the Likelihood Ratio into a Random Walk
Working with a product of n terms directly is clumsy, so the standard trick is to take logarithms. Define:
Yi = log[f(Xi,θ1) ÷ f(Xi,θ0)], for i = 1, 2, 3, …
and let Zn = log λn = Y1 + Y2 + … + Yn.
Since X1, X2, X3, … are i.i.d., the Yi's are also i.i.d. random variables. Notice what Zn is: it is a running sum of i.i.d. increments, exactly like the position of a particle doing a random walk on the real line, taking one random step Yi at a time.
In this language, the SPRT decision rule becomes: keep walking as long as Zn stays strictly between log B and log A; stop the moment Zn steps outside that interval, and the side you exit on tells you which hypothesis to accept.
Step 3: Why the Walk Is Guaranteed to Exit the Strip
This is the heart of the question, so let's go slowly. The interval (log B, log A) is a fixed, finite strip on the real line. The question is: is it certain, with probability 1, that the random walk Zn eventually leaves this strip, or could it wander inside the strip forever without ever crossing either boundary?
Here is where the Strong Law of Large Numbers (SLLN) does the real work. Since Y1, Y2, … are i.i.d. with some finite mean E(Y), the SLLN tells us that:
Zn / n → E(Y), almost surely, as n → ∞.
Now, as long as f(x, θ0) and f(x, θ1) are genuinely different densities, which is exactly what "testing H0 against H1" presumes, otherwise there would be nothing to test, a standard argument using Jensen's inequality shows E(Y) is strictly negative when θ0 is the true value, and strictly positive when θ1 is the true value. In other words, E(Y) ≠ 0 under either hypothesis.
Since Zn/n converges almost surely to a nonzero constant, Zn itself must drift off to +∞ (if the true value is θ1) or to −∞ (if the true value is θ0), almost surely. A sequence that drifts to ±∞ cannot stay trapped inside the fixed finite strip (log B, log A) forever; it must cross one of the two boundaries at some finite stage.
Pro Tip: Students often try to justify SPRT termination by vaguely saying the likelihood ratio keeps growing, without pinning down why. The precise argument always needs two ingredients together: i.i.d. increments, so the SLLN applies, and E(Y) not equal to zero, so the walk actually drifts instead of just hovering near zero forever. Drilling this two-step argument until it becomes automatic, which is how it is typically taught in a structured sequential-analysis module, is what separates a guessed answer from one you can defend under questioning.
This is precisely Wald's termination theorem for the SPRT: under the mild condition that the two hypothesized distributions are distinct, P(N < ∞) = 1, where N is the random stage at which the test stops. In words: the probability that an SPRT terminates in a finite number of steps is exactly one.
Step 4: Connecting This Back to the Two Statements
Now re-read the two statements with the random-walk picture in mind.
Statement-II says: "Probability of terminating an SPRT is one." We just derived exactly this result, P(N < ∞) = 1. So Statement-II is true, and it is a genuine, provable theorem (Wald's theorem), not just a plausible-sounding claim.
Statement-I says: "An SPRT always provides a decision." Think about what "providing a decision" means here: the test eventually stops and tells you to accept H0 or accept H1, rather than sampling forever without ever reaching a verdict. But that is exactly the event {N < ∞}, and Step 3 showed this event has probability one. So Statement-I is also true, and it is true precisely because Statement-II holds; "the test always gives a decision" is just another way of saying "the test terminates with probability one."
So we have: both statements are true, and Statement-II is not merely true alongside Statement-I, it is the actual logical reason Statement-I is true. That makes option (a)the correct choice.
Final Answer and Key Check
Working through the random-walk argument independently gives option (a): both statements are true, and Statement-II correctly explains Statement-I. This matches the official ISS 2016 Statistics Paper-2 answer key, which also records option (a) as correct for Question 48.
Why This Question Matters
Sequential analysis questions in the ISS exam rarely ask you to compute a number; they ask you to reason carefully about a theorem's logical structure, which is exactly what this question does. A candidate who has only memorised "SPRT terminates with probability 1" as an isolated fact can get confused when asked whether that fact explains something else, versus merely being true alongside it. Understanding the random-walk proof, rather than the one-line statement of the result, is what lets you confidently separate options (a) and (b) in a question like this instead of guessing between them.
Frequently Asked Questions
What exactly does it mean for an SPRT to "terminate"?
Termination means the test eventually reaches stage N (a finite, though random, number of observations) at which the likelihood ratio λN crosses one of the two boundaries A or B, so a final decision (accept H0 or accept H1) is made. It does not mean the test stops after a fixed, pre-announced sample size.
Is it possible for an SPRT to never stop, even in theory?
In principle, yes, the random walk could wander inside the strip forever without crossing a boundary, but Wald's theorem tells us this outcome has probability exactly zero whenever the two densities being compared are genuinely different. So it is not impossible in a strict logical sense, but it happens with probability zero, which in these exams is treated as "the SPRT always terminates."
Why do we need the condition that the two densities are different?
If f(x, θ0) and f(x, θ1) were identical, then Yi = log 1 = 0 for every observation, so Zn would stay at zero forever and never leave the strip. The "distinct distributions" condition is exactly what guarantees E(Y) is nonzero, which is what forces the random walk to drift and eventually exit.
How is this different from the Neyman-Pearson Lemma questions in the same paper?
The Neyman-Pearson Lemma deals with a single fixed-sample-size test and asks which critical region is most powerful at a given sample size. SPRT questions, like this one, deal with a sequential procedure where the sample size itself is random, and the focus shifts to properties of the stopping rule, such as whether and when the test concludes.
Do I need to reproduce the full random-walk proof in the actual exam?
No, since this is an objective-type paper you only need to select the correct option. But working through the underlying proof, as done above, is what makes you fast and confident at spotting the right answer among closely worded statement-based options, rather than relying on rote recall.
Where does this topic fit in the ISS Statistics Paper-2 syllabus?
This falls under Sequential Analysis, a recurring theme in Inference-heavy sections of Statistics Paper-2 alongside estimation theory and testing of hypotheses, and SPRT-based statement questions have appeared in multiple past years, so this derivation is worth having fully internalised rather than half-remembered.
If anything in this derivation feels unclear, especially the step connecting the Strong Law of Large Numbers to the random walk drifting to infinity, drop a comment below and we will work through it together. And if you know a fellow ISS aspirant who is also grinding through these previous year papers, consider sharing this with them; working through one question a day, properly, adds up faster than it seems.
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