ISS 2016 Statistics Paper-2 Solution: Question 6 (Random Effects Model Covariance)
Continuing our question-by-question walk through the ISS 2016 Statistics Paper-2 previous year paper, today's post takes on Question 6 — a deceptively short-looking item on the one-way random effects model that turns into a nice lesson on reading covariance statements carefully. If you've been following along since Question 1, you'll notice this question sits right after the ANOVA and Gauss-Markov questions, and it uses exactly the same "components of variance" thinking.
Quick Summary
Topic: One-way random effects model — variance, covariance and unbiased estimation
Question Reference: ISS 2016, Statistics Paper-2, Question 6 (Test Booklet Series A)
Official Key Status: Marked X — item dropped from scoring by the examining body
Our Derivation: Statement 1 is true; Statements 2 and 3, as printed, are both false — so none of the four listed options can be fully correct, which explains the drop
The Question, Exactly As It Appears In The Paper
Suppose b1, b2, b3, …, bk are independent N(0, σ²) random variables and εij are independent N(0, τ²) random variables for i = 1, 2, 3, …, k and j = 1, 2, 3, …, n. Suppose we observe only Xij = bi + εij for i = 1, 2, 3, …, k and j = 1, 2, 3, …, n. Then which of the following assertions are true?
Var(Xij) = σ² + τ² for all i, j
Cov(Xij, Xi'j') = 0 for all i, j except when i = i' and j = j'
(Xij − Xij')²/2 is an unbiased estimator of σ² for j ≠ j'
Select the correct answer using the code given below:
(a) 1 and 2 only
(b) 1 and 3 only
(c) 2 and 3 only
(d) 1, 2 and 3
Step 1: Understand What Model We're Actually Given
Before touching any statement, write out clearly what the setup is saying, because the whole question hinges on keeping two different "sources of randomness" separate.
We have k groups, indexed by i = 1, …, k. Each group has its own random group-effect bi, and all the bi's are independent of each other, each drawn from N(0, σ²). Think of σ² as the between-group variance — how much groups differ from each other on average.
Within each group i, we take n observations, indexed by j = 1, …, n. Each observation has its own error term εij, and all these error terms — across every i and every j — are independent of each other and of every bi, each drawn from N(0, τ²). Think of τ² as the within-group (error) variance.
What we actually observe is Xij = bi + εij. Every observation in group i shares the same random group-effect bi, but has its own private error εij. This is the standard one-way random effects model, and it's the natural place where the "F-ratio" ideas from the ANOVA questions earlier in this paper actually come from.
Step 2: Checking Statement 1 — the Variance
Since Xij = bi + εij, and bi is independent of εij, variance of a sum of independent random variables is just the sum of variances:
Var(Xij) = Var(bi) + Var(εij) = σ² + τ²
This holds for every i and every j, because every bi has variance σ² and every εij has variance τ² — none of them are special. So Statement 1 is true, and there is nothing hidden here.
Step 3: Checking Statement 2 — the Covariance, Case by Case
This is where most students rush and get it wrong, because there are actually three distinct cases to check, not two. Let's write out the general covariance:
Cov(Xij, Xi'j') = Cov(bi + εij, bi' + εi'j') = Cov(bi, bi') + Cov(bi, εi'j') + Cov(εij, bi') + Cov(εij, εi'j')
Because every b is independent of every ε, the two cross terms in the middle are always zero, no matter what i, j, i', j' are. So we're left with:
Cov(Xij, Xi'j') = Cov(bi, bi') + Cov(εij, εi'j')
Now split into the three cases that actually occur:
Case A — i ≠ i' (different groups):Since the b's are independent across groups, Cov(bi, bi') = 0. Since the ε's are independent across everything, Cov(εij, εi'j') = 0 too. So the total covariance is 0.
Case B — i = i', but j ≠ j' (same group, different observations):Now Cov(bi, bi) = Var(bi) = σ². But Cov(εij, εij') = 0 because the errors within the same group are still independent of each other when j ≠ j'. So the total covariance is σ², not zero.
Case C — i = i' and j = j' (the same observation):This is just Var(Xij) = σ² + τ², from Step 2.
So the true picture is: covariance is 0 only when i ≠ i'; it is σ² when i = i' but j ≠ j'; and it is σ² + τ² only when the observation is identical to itself. Statement 2, as printed, says the covariance is zero everywhere except the single case "i = i' and j = j'" — but that leaves Case B completely unaccounted for. Two observations from the same group but different individuals still share a nonzero covariance of σ², coming purely from the common random effect bi they both contain. So Statement 2, exactly as worded, is false — it misses a whole case.
Pro Tip: Whenever a question defines an observation as "shared random effect + private error," always split covariance checking into three buckets — different groups, same group but different unit, and the exact same observation — rather than just two. This three-way split is exactly the habit that is drilled into students working through random/mixed-effects problems until it becomes automatic, because it's precisely the kind of "in-between" case (same group, different unit) that a rushed reading of an MCQ statement will skip.
Step 4: Checking Statement 3 — the Unbiased Estimator
Fix a group i and pick two different observations within it, j ≠ j'. Look at the difference:
Xij − Xij' = (bi + εij) − (bi + εij') = εij − εij'
Notice that bi cancels out completely, because both observations come from the same group and therefore carry the same group effect. This is the whole point of taking a within-group difference — it deliberately removes σ² from the picture and leaves only the error variance τ².
Since εij and εij' are independent N(0, τ²) variables, their difference has variance:
Var(εij − εij') = Var(εij) + Var(εij') = τ² + τ² = 2τ²
Also, E(εij − εij') = 0, so E[(εij − εij')²] = Var(εij − εij') = 2τ². Dividing by 2:
E[(Xij − Xij')² / 2] = 2τ² / 2 = τ²
So (Xij − Xij')²/2 is indeed an unbiased estimator — but of τ², the within-group error variance, not of σ², the between-group variance, as Statement 3 claims. This is a completely different quantity: σ² can never be recovered from a within-group difference, precisely because the whole point of that difference was to cancel σ² out. So Statement 3, as worded, is false.
Step 5: Putting It Together — and Why the Item Was Dropped
Our rigorous derivation gives: Statement 1 true, Statement 2 false (it omits Case B), Statement 3 false (it names the wrong parameter). That means the only statement that survives is Statement 1 alone. But look at the four answer choices again — (a) 1 and 2, (b) 1 and 3, (c) 2 and 3, (d) 1, 2 and 3. Every single option pairs Statement 1 with at least one of the other two, and none of them offers "1 only." Mathematically, there is no option among the four that is completely correct.
This is exactly the kind of internal inconsistency that leads an examining body to withdraw a question after the fact. Checking the official ISS 2016 Statistics Paper-2 answer key confirms this directly:Question 6 in Series A is marked "X"— meaning it was dropped from scoring altogether, and every candidate was awarded marks for it regardless of what they selected. Interestingly, the same underlying question (renumbered for each shuffled booklet) is marked "X" in Series B, C and D as well, confirming this wasn't a one-off printing glitch in a single series but a genuine flaw the examiners themselves recognised.
So there is no "correct option" to report here in the conventional sense. What we can report with full confidence is the actual mathematics: Var(Xij) = σ² + τ² always; the covariance structure has three distinct cases, not two; and any within-group difference-based estimator can only ever recover τ², never σ².
Why This Question Matters
Even though this particular item earned no marks for anyone, the underlying idea is worth far more than the one mark it would have carried. Random effects models — where an observation is built as "shared group effect + private noise" — are the backbone of nested/hierarchical designs, split-plot experiments, and variance components estimation, all of which show up repeatedly across ISS Paper-2 in later years. The habit of decomposing a compound random variable before touching any variance or covariance formula is exactly what separates a fast, confident answer from a guess, and it's a habit that's usually built by working through several of these decompositions rather than memorising a single result.
Frequently Asked Questions
What exactly is a "one-way random effects model"?
It's a model where data is grouped (say, into k groups of n observations each), and each group is assumed to have its own randomly drawn effect on top of a common baseline, rather than a fixed, known effect. Here that random group effect is bi, and each individual observation also carries its own independent noise term εij on top of it.
Why does subtracting two observations from the same group remove σ²?
Because both observations in that group share the exact same random effect bi. When you subtract one from the other, bi appears with a plus sign in one term and a minus sign in the other, so it cancels out algebraically, leaving only the difference of the two independent error terms.
What does an "X" in an ISS official answer key actually mean?
It means the item was removed from scoring entirely, usually because the examiners themselves identified an ambiguity, printing error, or a genuine flaw in the options after the exam — sometimes following representations from candidates. Every candidate receives credit for that item automatically, regardless of which option they marked or whether they left it blank.
If a question is dropped, should I still study it while preparing?
Yes, absolutely. The scoring outcome tells you nothing about how useful the underlying concept is for the rest of the syllabus. In this case, distinguishing between-group and within-group variance is tested again and again in ANOVA, variance components, and mixed-model questions, so the concept itself remains high-value even though this particular item carried no marks.
How can I tell, during an actual exam, whether a covariance statement in an MCQ has a hidden gap like this one?
Write out the general covariance formula symbolically first, then enumerate every distinct case implied by the indices involved (here: different groups, same group with different units, and the identical observation) before comparing to what the statement claims. If the statement's claimed set of exceptions doesn't match your own enumerated case list exactly, that's your signal something is off.
Is this related to the ANOVA questions earlier in the same paper?
Yes, directly. The two-way and one-way ANOVA F-ratio questions earlier in this very paper compare "variance between groups" against "variance within groups" — which are precisely σ² and τ² from this question. Understanding this decomposition makes those F-ratio computations far more intuitive rather than just a formula to plug numbers into.
If you worked through this one differently, spotted a subtlety we missed, or just want to compare notes on the covariance case-split, drop a comment below — and if this helped clear things up, consider sharing it with a fellow ISS aspirant working through the same paper.
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