ISS 2016 Statistics Paper-2 Solution: Question 64 (Randomised Tests for Exact Size Alpha)
Continuing our question-by-question walk through the ISS Statistics Paper-2 previous year papers, today we take up a theory question from the 2016 paper that trips up a lot of otherwise well-prepared candidates: randomised tests. It looks like a one-line definitional question, but to answer it correctly you need to actually understand why randomisation is introduced in hypothesis testing in the first place, not just memorise a line from a textbook.
Quick Summary
Topic:Randomised tests and exact size α tests in hypothesis testing
Question Reference:ISS 2016, Statistics Paper-2, Question 64
Correct Answer:Option (c) — When X is a discrete random variable and exact size α test is required
The Question, As It Appeared in the Paper
64. Randomised test is used for testing parametric hypothesis regarding distribution of a random variable X :
(a) Whenever X has a discrete distribution
(b) Whenever exact size α test is required
(c) When X is a discrete random variable and exact size α test is required
(d) When X is a continuous random variable and exact size α test is required
Step 1: What Does "Size of a Test" Actually Mean?
Before touching randomisation, let's be completely clear on one definition. In hypothesis testing, we have a null hypothesis H0 and a test (a rule that says "reject H0 if the data falls in region C, otherwise accept"). The size of the test is defined as:
α = P(reject H0 | H0 is true) = P(X ∈ C | H0)
So the size is just the probability, computed under the null hypothesis, that our rule tells us to reject. When we say we want a "level α test" or an "exact size α test," we mean we want this probability to equal a pre-specified number — commonly 0.05 or 0.01 — exactly, not just approximately or at most.
Step 2: Why a Deterministic Rule Can Fail for a Discrete X
Here is the crux of the whole question. Suppose X is a discrete random variable, and under H0 suppose we are testing using a rejection region of the form "reject if X ≥ k" for some integer k. As k decreases by 1 at a time, the probability P(X ≥ k | H0) increases, but it increases in jumps — because X only takes a discrete set of values, there's no "in-between" value of k.
Let's make this completely concrete with a small example. Suppose X ~ Binomial(5, p), and we are testing H0: p = 0.5 against H1: p > 0.5. Under H0, the probabilities are:
P(X = 0) = 1/32
P(X = 1) = 5/32
P(X = 2) = 10/32
P(X = 3) = 10/32
P(X = 4) = 5/32
P(X = 5) = 1/32
Now look at the possible rejection regions "reject if X ≥ k" and the sizes they give us:
Reject if X ≥ 5: size = P(X = 5) = 1/32 = 0.03125
Reject if X ≥ 4: size = P(X = 4) + P(X = 5) = 6/32 = 0.1875
Suppose our target significance level is α = 0.05. Notice that 0.05 lies strictly between 0.03125 and 0.1875. There is no deterministic choice of k that gives us size exactly 0.05 — the achievable sizes jump straight from 0.03125 to 0.1875 and skip right over 0.05. This is purely a consequence of X being discrete: the cumulative distribution function of a discrete random variable is a step function, so it cannot take every value between 0 and 1.
Step 3: Fixing This With a Randomised Test
The randomised test is the clever fix. Instead of a hard yes/no rule at the boundary value, we introduce a chance element exactly at the boundary. Concretely, for our example, define the test as:
If X = 5: reject H0 with probability 1 (always reject)
If X = 4: reject H0 with probability γ (toss a biased coin)
If X ≤ 3: never reject (accept H0)
The size of this randomised test is:
α = P(X = 5) + γ · P(X = 4) = 1/32 + γ · 5/32
Setting this equal to our target 0.05:
0.03125 + γ(0.15625) = 0.05
γ(0.15625) = 0.01875
γ = 0.01875 / 0.15625 = 0.12
So: whenever we observe X = 4, we reject H0 with probability exactly 0.12 (for instance, by generating a uniform random number U and rejecting if U < 0.12). This extra randomisation at the single boundary point X = 4 is precisely what lets us "fill the gap" between 0.03125 and 0.1875 and land on exactly 0.05.
Pro Tip: The entire logic of this question collapses to one idea — randomisation exists to patch the gaps left by a step-function CDF, nothing more. Don't try to memorise "randomised test = discrete," because that alone is not the complete condition (that's exactly why option (a) is a trap). The way this is usually taught is to force yourself to ask both halves of the condition every time: is the variable discrete, AND do I actually need an exact (not just "at most α") size? If the examiner only wanted a conservative test, a deterministic rule would do fine even for discrete data. Students who drill this two-part check until it's automatic rarely fall for the single-condition distractor options in this kind of question.
Step 4: Why Continuous X Never Needs This
Now consider the flip side. If X is a continuous random variable, its CDF F(x) = P(X ≤ x) is a continuous function that increases smoothly from 0 to 1, with no jumps. By the intermediate value theorem, for any target α between 0 and 1, we can always find a critical value c such that P(X ≥ c | H0) = α exactly. There is never a "gap" to fill, so a deterministic critical region always suffices to hit any exact size α we want. This is why option (d) is incorrect — continuous random variables are precisely the case where randomisation is never needed.
Step 5: Eliminating the Other Options
With Steps 2 through 4 in hand, we can now evaluate every option precisely:
(a) Whenever X has a discrete distribution — Incorrect. Discreteness alone is not sufficient. If we are happy with a conservative test (size at most α, not exactly α), a plain deterministic rule works perfectly well even for discrete X. Randomisation is only forced on us when we additionally insist on an exact size.
(b) Whenever exact size α test is required — Incorrect. As shown in Step 4, if X is continuous, we can always achieve an exact size α test deterministically. Randomisation isn't needed here at all.
(c) When X is a discrete random variable and exact size α test is required —Correct.This is the only option that captures both necessary conditions together, as demonstrated by the worked example above.
(d) When X is a continuous random variable and exact size α test is required — Incorrect, and in fact the opposite of the truth: continuous random variables are exactly the case where randomisation is unnecessary.
So the correct answer is option (c), which matches the official ISS 2016 answer key for Test Booklet Series A, where Question 64 is listed as (C).
Why This Question Matters
Randomised tests sit at the foundation of the Neyman-Pearson theory of hypothesis testing, which is one of the most heavily tested areas across ISS, IIT JAM (MS) and GATE (ST) statistics syllabi. The concept resurfaces repeatedly — in deriving the most powerful test via the Neyman-Pearson Lemma for discrete distributions, in UMP (uniformly most powerful) test constructions, and in power function calculations where the power itself depends on the randomisation constant γ. A candidate who genuinely understands why randomisation patches the gap in a discrete CDF rather than just recalling a rule will find a whole cluster of related MCQs across different years far easier, because the underlying logic is identical every time — only the specific distribution and numbers change.
Frequently Asked Questions
What exactly is a randomised test in statistics?
A randomised test is a hypothesis testing rule where, for at least one outcome of the data (typically the boundary value of the test statistic), the decision to reject or accept the null hypothesis is not fixed but is instead made using an auxiliary random experiment, such as tossing a biased coin with a specified probability γ. It is used alongside the usual "always reject" and "always accept" regions.
Why can't we just round the significance level instead of randomising?
Rounding or choosing the nearest achievable size would give you a conservative test (size less than or equal to α) or an oversized test (size greater than α), but not a test whose size is exactly equal to the pre-specified α. Many theoretical results, including the Neyman-Pearson Lemma for discrete distributions, are stated specifically for tests of exact size α, so exact equality genuinely matters in those derivations.
Does randomisation ever apply to continuous random variables?
No. Since the CDF of a continuous random variable is continuous and takes every value between 0 and 1, a deterministic critical region can always be chosen to achieve any target size α exactly. Randomisation is a tool specifically for handling the "jumps" in a discrete distribution's CDF, so it is never required for purely continuous distributions.
How do I compute the randomisation probability γ in a problem?
First find the two closest deterministic sizes that bracket your target α (by trying consecutive values of the critical constant, as we did with X ≥ 5 and X ≥ 4 in the worked example). Then set up the equation "size from the always-reject part" plus "γ times the probability of the boundary value" equals α, and solve for γ. The resulting γ will always lie between 0 and 1 if your target α genuinely lies between the two bracketing sizes.
Is this topic important for the ISS Paper-2 exam specifically?
Yes, testing of hypotheses (including the Neyman-Pearson approach, UMP tests, and randomised tests for discrete distributions) is a recurring, high-weightage area in ISS Statistics Paper-2, and similar questions appear in multiple years of the previous year papers, sometimes phrased as direct definitional MCQs like this one and sometimes embedded inside numerical problems that require you to actually compute γ.
What's the most common mistake candidates make on this type of question?
The most common mistake is picking option (a), because "randomised test = discrete distribution" is a half-remembered rule that sounds right. The fix is to always check both conditions explicitly: is the variable discrete, and is an exact (not merely conservative) size genuinely demanded by the problem? Missing the second condition is what makes distractor options like (a) and (b) tempting.
If anything in this derivation feels unclear, especially the computation of γ in the worked example, drop a comment below and we'll walk through it further. And if you found this useful, do share it with a fellow ISS aspirant who's also working through the Paper-2 previous year papers — question-by-question practice like this adds up fast.
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