ISS 2016 Statistics Paper-2 Solution: Question 56 (CRLB Attainability in Gamma Estimation)
Continuing the question-by-question walk through the ISS Statistics Paper-II archive, today's post takes up Question 56 from the 2016 paper. It sits right after the Cramér–Rao bound question we solved last time, and it is really testing the same idea from the opposite direction — what happens when that bound cannot be reached.
Quick Summary
Topic:Cramér–Rao Lower Bound (CRLB) attainability and estimating a reciprocal (rate) parameter in the Gamma/Exponential family
Question Reference:ISS 2016, Statistics Paper-II, Question 56
Correct Answer:Option (c) — X follows Gamma(1, 1/β)
The Question, Exactly As Asked
56. An estimator with large variance is preferred in which one of the following cases?
(a) X follows gamma (1, β)
(b) X follows U(0, θ)
(c) X follows gamma (1, 1/β)
(d) X follows gamma (0, 1/θ)
Step 1: Agree on What "Gamma(α, β)" Means
Before touching any option, fix the notation, because the whole question hinges on it. Write the Gamma distribution with shape α and scale β as
f(x; α, β) = xα−1 e−x/β / (Γ(α) βα), x > 0.
Its mean is αβ and its variance is αβ². When α = 1 this collapses to the familiar Exponential distribution, so:
Gamma(1, β) is Exponential with mean β— its pdf is (1/β) e−x/β.
Gamma(1, 1/β) is Exponential with mean 1/β, i.e.rate β— its pdf is β e−βx.
So options (a) and (c) describe the exact same family of curves (exponential), but in (a) the unknown βis the mean, while in (c) the unknown β is the reciprocal of the mean. That single difference turns out to decide the whole question.
Step 2: Option (a) — Estimating β When It Is the Mean
Take a random sample X1, X2, …, Xn from Gamma(1, β), i.e. each Xi has pdf (1/β) e−x/β. The log-likelihood for one observation is
log f = −log β − x/β.
Differentiate twice with respect to β:
d/dβ (log f) = −1/β + x/β², d²/dβ² (log f) = 1/β² − 2x/β³.
Taking expectation (using E[X] = β) gives the Fisher information for one observation as I(β) = 2β/β³ − 1/β² = 1/β². For n independent observations, the CRLB for any unbiased estimator of β is
CRLB = 1 / (n · I(β)) = β²/n.
Now check the sample mean X̄. Since E[X̄] = β, it is unbiased, and Var(X̄) = Var(X)/n = β²/n —exactly equal to the CRLB.So in case (a), the "best possible" small variance is not just a theoretical target; a perfectly ordinary estimator (the sample mean) actually achieves it. There is no reason here to settle for anything large.
Step 3: Option (c) — Estimating β When It Is the Rate
Now take X1, …, Xn from Gamma(1, 1/β), i.e. pdf β e−βx, and suppose we want to estimate the same symbol β — except now β is the rate, not the mean. The Fisher-information calculation (same steps as above, just differentiating log f = log β − βx) gives I(β) = 1/β² per observation too, so on paper the CRLB again reads β²/n. The question is whether any unbiased estimator can actually touch that number.
The complete sufficient statistic here is T = ΣXi, and because each Xi ~ Gamma(1, 1/β), the sum T ~ Gamma(n, 1/β), with pdf
fT(t) = (βn / Γ(n)) tn−1 e−βt, t > 0.
Because the mean of T is n/β (not something proportional to β), there is no linear function of T that is unbiased for β — you cannot write β = a + b·E[T] for constants a, b free of β. The only route to an unbiased estimator of β is through a reciprocal of T, and reciprocals of Gamma random variables need to be worked out very carefully using their moments.
Compute E[1/T] directly from the density:
E[1/T] = ∫0∞ (1/t) · (βn/Γ(n)) tn−1 e−βt dt = (βn/Γ(n)) ∫0∞ tn−2 e−βt dt.
The remaining integral is a Gamma(n−1, 1/β) normalising integral, equal to Γ(n−1)/βn−1 (valid only when n − 1 > 0, i.e. n ≥ 2). So
E[1/T] = βn/Γ(n) · Γ(n−1)/βn−1 = β · Γ(n−1)/Γ(n) = β/(n−1),
using Γ(n) = (n−1)Γ(n−1). This immediately tells us that β̂ = (n−1)/T satisfies E[β̂] = (n−1) · β/(n−1) = β, so β̂ is unbiased — but only when n ≥ 2 (for n = 1, E[1/T] does not even exist, because the integral ∫ t−1 e−βt dt diverges at t = 0).
To get the variance we also need E[1/T²], by the same method with one more power of t removed:
E[1/T²] = (βn/Γ(n)) ∫0∞ tn−3 e−βt dt = βn/Γ(n) · Γ(n−2)/βn−2 = β²/[(n−1)(n−2)],
which exists only when n − 2 > 0, i.e. n ≥ 3. Now,
Var(1/T) = E[1/T²] − (E[1/T])² = β²/[(n−1)(n−2)] − β²/(n−1)² = β² / [(n−1)²(n−2)],
so that
Var(β̂) = (n−1)² · Var(1/T) = β² / (n−2).
Pro Tip: The step most students rush past is assuming that because the Fisher information formula I(β) = 1/β² looks identical in (a) and (c), the CRLB must be equally achievable in both. It isn't. The Cramér–Rao bound is attained only when the score function is a
Compare the two variances directly, for the same n:
CRLB (quoted, from Fisher information alone) = β²/n
Actual variance of the UMVUE of β = β²/(n−2)
Since n − 2 < n for every finite n, we get β²/(n−2) > β²/n always — the true minimum-variance unbiased estimator in case (c) is strictly larger than the CRLB, and it can never be brought down to match it. Worse still, for n = 1 or n = 2 there is no finite-variance unbiased estimator of β at all (you need n ≥ 3 just for the variance to exist). This is precisely the scenario where a large variance is not a choice but an unavoidable consequence of the parametrisation, which is exactly what the question is asking for.
Step 4: Why (b) and (d) Are Not the Answer
Option (b), X ~ U(0, θ), fails the CRLB's regularity conditions for a different reason: the support [0, θ] depends on θ, so the Cramér–Rao inequality does not even apply here. When that happens, estimators can actually do better than the regular-case bound, not worse. The UMVUE of θ, built from the maximum order statistic as T = ((n+1)/n)X(n), has variance θ²/[n(n+2)], which shrinks much faster than a regular β²/n rate as n grows — the opposite of "large variance."
Option (d), X ~ gamma(0, 1/θ), is not a valid probability distribution at all: the Gamma family requires shape parameter α > 0, and α = 0 does not define a proper density. It can be eliminated immediately on that technical ground alone.
Final Answer
The correct choice is (c) X follows gamma(1, 1/β), which matches the official ISS 2016 Statistics Paper-II answer key (option C for Series A). The reasoning: when the parameter being estimated is the rate rather than the mean of an exponential variable, the Cramér–Rao bound quoted from the Fisher information cannot be attained by any unbiased estimator — the actual UMVUE variance β²/(n−2) is always strictly larger than the CRLB β²/n, and for small samples no finite-variance unbiased estimator exists at all.
Why This Question Matters
It is tempting to treat "Fisher information" and "achievable variance" as the same idea, but Question 56 is a clean counterexample that shows they are not. Reciprocal or rate-type parameters show up constantly in reliability theory, queuing models, and survival analysis (hazard rates are rate parameters, not means), so knowing in advance that estimating them tends to inflate variance — sometimes to the point of no finite-variance estimator existing for small samples — is a genuinely useful piece of applied intuition, not just an exam trick. Students who drill this kind of reciprocal-moment derivation until it becomes automatic stop getting tripped up by CRLB questions that look identical to a standard "attains the bound" question but quietly aren't.
Frequently Asked Questions
Why does the Fisher information come out the same for Gamma(1, β) and Gamma(1, 1/β)?
Fisher information is computed from the log-likelihood's curvature with respect to whichever parameter you differentiate against, and in both cases that parameter is called β and enters the exponent linearly (as x/β or as βx). The algebra of the second derivative happens to land on 1/β² either way. What differs is not the information itself but whether an unbiased estimator exists that can convert that information into an achieved variance equal to its reciprocal.
What exactly does it mean for the Cramér–Rao bound to be "attained"?
The CRLB is attained when there exists an unbiased estimator whose variance exactly equals 1/(nI(θ)). This happens if and only if the score function ∂/∂θ (log f) can be written as a constant multiple of (T − θ) for some statistic T, i.e. the score is linear in a statistic that is itself unbiased for θ. If no such linear relationship exists, the bound is a valid lower limit but no estimator can actually reach it.
Is the UMVUE in option (c) still the best possible estimator, even though it does not reach the CRLB?
Yes. "Best possible unbiased estimator" (UMVUE) and "estimator that attains the CRLB" are not the same guarantee. The UMVUE, built here from the complete sufficient statistic T = ΣXi via the Lehmann–Scheffé approach, is guaranteed to have the smallest variance among all unbiased estimators, even though that smallest-possible variance still exceeds the CRLB.
Why does n have to be at least 3 for the variance to even exist in option (c)?
The variance calculation needs E[1/T²], and that integral only converges when the exponent on t in the integrand stays above −1 near t = 0, which requires the sample size n to satisfy n − 2 > 0. With fewer observations, the reciprocal moment diverges, so the "variance" of any unbiased estimator of the rate is effectively unbounded.
Does this issue only come up for the Exponential distribution?
No, it's a general phenomenon anytime you try to unbiasedly estimate a nonlinear (especially reciprocal) function of a natural parameter. The same inflation shows up when estimating 1/λ for a Poisson rate, 1/p for a Binomial proportion, or the precision (reciprocal variance) of a Normal population — in each case, the UMVUE of the reciprocal quantity has a variance that sits above the naive CRLB and needs a minimum sample size just to stay finite.
How should I revise this topic for the ISS exam?
Practice deriving the UMVUE and its variance from first principles for both a parameter and its reciprocal in the same exponential family, exactly as done above, rather than memorising "CRLB = 1/(nI(θ))" as a universal recipe. Once you've worked through one reciprocal-moment case by hand, spotting these options under exam time pressure becomes far easier.
If any step above felt rushed, drop a comment and I'll expand on it in a follow-up post — and if you know a fellow ISS aspirant still building up their estimation theory foundations, feel free to pass this one along to them.
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