ISS 2016 Statistics Paper-2 Solution: Question 27 (Consistent Estimators for Cauchy Location)
Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, today's post tackles Question 27 from the 2016 paper. This one looks like a simple "pick the true statement" question, but it hides one of the most elegant (and most misunderstood) results in estimation theory — and it trips up a huge number of aspirants who assume "more data always means a better average."
Quick Summary
Topic:Consistency of estimators — sample mean vs. sample median for the Cauchy distribution
Question Reference:ISS 2016, Statistics Paper-2, Question 27
Correct Answer:Option (b) — the sample median is a consistent estimator of μ, but the sample mean is not
The Question, As Asked
If X1, X2, X3, ..., Xn is a random sample of size n from a Cauchy distribution with pdf
f(x) = 1 / (π[1 + (x − μ)2]); −∞ < x < ∞,
then which one of the following is true for estimating μ?
(a) Sample mean is a consistent estimator but not sample median
(b) Sample median is a consistent estimator but not sample mean
(c) Both sample mean and sample median are consistent estimators
(d) Neither sample mean nor sample median is a consistent estimator
Step 1: What Makes This Distribution Different From a Normal or Exponential
Before touching any formula, notice the shape of f(x). The denominator grows only like x2, so the tails of this curve die out much more slowly than, say, a Normal distribution's tails (which die out like e−x²). This is what statisticians call a heavy-tailed distribution, and it has a genuinely strange consequence: if you try to compute E(X) = ∫ x·f(x) dx over the whole real line, the integral does not converge to a finite number — it diverges. In plain words, the Cauchy distribution has no finite mean and no finite variance at all.
What it does have is a clean, well-defined centre of symmetry. Since f(x) depends on x only through (x − μ)2, the curve is a mirror image of itself around x = μ. That makes μ simultaneously the population median and the mode, even though it can never be called the "population mean" in the usual integral sense. This single fact — a well-defined median but an undefined mean — is the whole key to the question.
Step 2: Why the Sample Mean Fails
A natural first guess is "just average the observations, X̄ = (1/n)ΣXi, and it will settle down near μ as n grows." For almost every distribution you meet in a first statistics course, that guess is correct — it's the Weak Law of Large Numbers. But the standard proof of the Weak Law needs the population to have a finite mean and variance so that Var(X̄) = σ²/n can be shown to shrink to 0. We already saw that the Cauchy distribution has neither. So the usual proof doesn't apply, and we have to check directly what happens to X̄.
The clean way to do this is through the characteristic function (CF), φX(t) = E(eitX), which exists for every distribution even when the mean does not. It is a standard, well-known result in distribution theory that a Cauchy(μ, 1) random variable has characteristic function
φX(t) = eiμt − |t|
Now build the sample mean's characteristic function using the independence of X1, ..., Xn. For independent variables, the CF of a sum is the product of individual CFs, and scaling a variable by a constant c scales the argument of its CF by c:
φX̄(t) = E(eit·(1/n)ΣXi) = ∏i=1n E(ei(t/n)Xi) = [φX(t/n)]n
Substitute the Cauchy CF with t replaced by t/n:
φX̄(t) = [eiμ(t/n) − |t/n|]n = en·(iμt/n − |t|/n) = eiμt − |t|
Compare this last expression with φX(t) = eiμt − |t| written above — they are identical. This means X̄, the average of n independent Cauchy(μ,1) observations, has exactly the same distribution as a single Cauchy(μ,1) observation, no matter how large n is. Averaging a thousand observations spreads out just as much as looking at one single observation. Since the distribution of X̄ never tightens around μ as n → ∞, X̄ does not converge in probability to μ, so by definition the sample mean is not a consistent estimator of μ.
Pro Tip: The trap in this question is assuming "large sample ⇒ sample mean gets better," which is true for well-behaved distributions but false here. Whenever a question mentions the Cauchy distribution, your first move should be to check whether the population mean/variance even exist before reasoning about X̄ at all — this reflex is exactly what gets drilled into students until it becomes automatic in a structured problem-solving routine, the kind of habit that a place like Sunrise Classes spends real classroom time building rather than just mentioning once and moving on.
Step 3: Why the Sample Median Succeeds
Since the mean is off the table, we turn to the median. For any continuous distribution with density f and population median m (defined by F(m) = 1/2), there is a general large-sample result: the sample median Mn, based on n independent observations, satisfies
√n (Mn − m) → N(0, 1 / [4 f(m)2]) in distribution, as n → ∞
Here is the intuition behind that formula, built step by step. Let Fn be the empirical (sample-based) CDF, i.e. Fn(x) = (fraction of observations ≤ x). By definition, Fn(Mn) ≈ 1/2. At the true median m, the ordinary CLT applied to the indicator variables 1(Xi ≤ m) — each with mean F(m) = 1/2 and variance F(m)(1 − F(m)) = 1/4 — gives
√n (Fn(m) − 1/2) → N(0, 1/4)
Now relate the gap in Fn to the gap in Mn. Because Fn is (locally) increasing with slope roughly f(m) near x = m, a small shift Mn − m in the x-direction corresponds to a shift of about f(m)·(Mn − m) in the Fn-direction. Equating this to the Fn(m) − 1/2 gap above and rearranging gives
√n (Mn − m) ≈ [√n (Fn(m) − 1/2)] / f(m) → N(0, 1/4) / f(m), i.e. N(0, 1/[4f(m)2])
This confirms the formula. The crucial feature is that the asymptotic variance 1/[4n f(m)2] shrinks to 0 as n → ∞ for any distribution with f(m) > 0 — regardless of whether the distribution has a finite mean or not, because this whole argument never once used E(X) or Var(X). It only used the density value at the median.
Now plug in the Cauchy density at x = μ. From f(x) = 1/(π[1 + (x−μ)2]), setting x = μ gives (x−μ)2 = 0, so
f(μ) = 1/(π[1 + 0]) = 1/π
Therefore the asymptotic variance of the sample median is
1 / [4n·(1/π)2] = π2 / (4n)
As n → ∞, π²/(4n) → 0, and the mean of the limiting distribution is μ itself. So Mn converges in probability to μ — the sample median is a consistent estimator of μ, and in fact it is the natural tool of choice for this distribution.
Step 4: Matching the Answer Choices
Putting Steps 2 and 3 together: the sample mean is not consistent for μ, while the sample median is consistent for μ. That is exactly option (b) — "Sample median is a consistent estimator but not sample mean." Checking against the official ISS 2016 Statistics Paper-2 answer key, Question 27 is indeed marked as (b), so our derivation and the official key agree completely.
Why This Question Matters
This question is a favourite in ISS-level papers precisely because it breaks the instinct that "sample mean is always the best unbiased, consistent choice." Real distributions with heavy tails — financial return series, certain error models in physics, and the Cauchy distribution itself as a teaching example — behave nothing like the Normal distribution, and the mean can be a genuinely unusable estimator for them. Recognising when a "safe default" statistic quietly stops working is a core skill the exam is testing, not just the ability to memorise a formula.
Frequently Asked Questions
Why doesn't the Cauchy distribution have a finite mean?
The tails of the Cauchy density decay only like 1/x² for large |x|, so the integral ∫x·f(x)dx over the whole real line does not converge — it behaves like a divergent logarithmic integral at infinity. Because the integral has no finite value, E(X) is undefined, and by extension the variance is undefined too.
What does it actually mean for an estimator to be "consistent"?
An estimator Tn, built from a sample of size n, is consistent for a parameter θ if Tn converges in probability to θ as n → ∞ — informally, the estimator's distribution becomes more and more concentrated around the true value as you collect more data. It is a large-sample property; a consistent estimator can still be biased or noisy for small n.
Is the sample mean ever useful when the population has heavy tails?
Not for the Cauchy case specifically — the reproducing property shown in Step 2 means averaging never helps, no matter how large the sample gets. For other heavy-tailed but finite-variance distributions, the sample mean can still be consistent, just with a slower rate of convergence than in the Normal case.
Why is the sample median unaffected by the extreme values a Cauchy sample can produce?
The median only cares about the ordering of observations, not their magnitudes, so a handful of extremely large or small values (which the Cauchy distribution produces often, due to its heavy tails) cannot drag it away from the centre. The sample mean, in contrast, is a straight arithmetic average and gets pulled directly by those same extreme values.
Does this "median beats mean" result hold for every symmetric distribution, or is Cauchy special?
The sample median is consistent for the centre of any continuous, symmetric distribution with positive density there — that part is general. What is special about Cauchy is the other half of the comparison: for most symmetric distributions (Normal, Logistic, Uniform) the sample mean is consistent too, just usually more efficient than the median; Cauchy is one of the rare cases where the mean fails outright.
How is this kind of question usually framed in ISS Paper-2?
It typically appears as a single "which statement is true" MCQ testing whether you know the special properties of a named distribution — Cauchy for mean/median consistency, exponential for memorylessness, and so on — rather than asking for a full derivation. Being able to reconstruct the underlying argument quickly, as done above, is what lets you answer confidently instead of guessing from memory.
If any step above felt unclear — especially the characteristic function argument in Step 2 or the median variance derivation in Step 3 — drop your doubt in the comments below, and feel free to share this post with a fellow ISS aspirant who might be stuck on the same question.
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