ISS 2016 Statistics Paper-2 Solution: Question 55 (Cramer-Rao Bound for Cauchy Distribution)
Continuing our question-by-question walkthrough of the ISS Statistics Paper-2 archive, today we take on Question 55 from the 2016 paper. This one brings together two ideas that often get tested separately -- the Cramer-Rao inequality and the Cauchy distribution -- and asks you to combine them under exam pressure. As always, every algebraic and calculus step is shown in full, so you can follow along even if you are seeing this integral for the first time.
Quick Summary
Topic:Cramer-Rao Lower Bound (CRLB) and Fisher Information for the Cauchy distribution
Question Reference:ISS 2016, Statistics Paper-2, Question 55
Correct Answer:Option (b), 2/n
The Question As Asked
Cramer Rao lower bound of variance for the parameter θ of the distribution with pdf
f(x, θ) = (1/π) · 1/[1 + (x − θ)²], where −∞ < x < ∞
is:
(a) 1/n
(b) 2/n
(c) 1/n²
(d) 2/n²
Step 1: Recognise the Distribution and Write the CRLB Formula
The density f(x, θ) = (1/π) · 1/[1 + (x − θ)²] is the standard Cauchy distribution with location parameter θ(and scale fixed at 1). We have a random sample X₁, X₂, ..., Xₙ of size n, each distributed this way.
For an unbiased estimator T of θ, the Cramer-Rao inequality says:
Var(T) ≥ 1 / [n · I(θ)]
where I(θ) is the Fisher information contained in a single observation, defined as:
I(θ) = E[ (∂/∂θ log f(X, θ))² ]
So the whole problem reduces to finding I(θ) for one observation, and then the CRLB for the sample of size n is simply 1/[n I(θ)].
Step 2: Take the Log of the Density
Start with:
f(x, θ) = (1/π) · [1 + (x − θ)²]−1
Taking the natural log on both sides:
log f(x, θ) = −log π − log[1 + (x − θ)²]
This is the only form we need -- we never have to touch the original f(x, θ) again, only its log.
Step 3: Differentiate With Respect to θ (The Score Function)
Let u = x − θ purely to keep the algebra clean (remember, x is fixed/data and θ is the variable we differentiate with respect to, so du/dθ = −1).
log f = −log π − log(1 + u²)
Differentiating term by term:
∂/∂θ log f = − [1/(1+u²)] · ∂/∂θ(1+u²)
Now ∂/∂θ(1+u²) = 2u · (du/dθ) = 2u · (−1) = −2u. Substituting this back:
∂/∂θ log f = − [1/(1+u²)] · (−2u) = 2u / (1+u²)
So, writing u back as (x − θ), the score function is:
∂/∂θ log f(x, θ) = 2(x − θ) / [1 + (x − θ)²]
Step 4: Square the Score and Set Up the Fisher Information Integral
By definition:
I(θ) = E[(score)²] = ∫−∞∞ [2u / (1+u²)]² · f(u) du
where we are integrating over u = x − θ (a simple change of variable that shifts the Cauchy density to be centred at 0, so f(u) = (1/π) · 1/(1+u²)).
Squaring the score function:
[2u/(1+u²)]² = 4u² / (1+u²)²
Multiplying by the density:
I(θ) = ∫−∞∞ [4u² / (1+u²)²] · [1 / (π(1+u²))] du = (4/π) ∫−∞∞ u² / (1+u²)³ du
Pro Tip: The trickiest part of this question is not the differentiation -- it's recognising that the leftover integral ∫ u²/(1+u²)³ du is a textbook candidate for the substitution u = tanφ, because 1 + tan²φ = sec²φ collapses the cube in the denominator instantly. The way this integral is usually drilled in classroom practice is to memorise this exact substitution pattern whenever you see (1+u²) raised to a power in the denominator with polynomial u in the numerator -- it turns a seemingly hard calculus problem into pure trigonometric simplification within two lines.
Step 5: Solve the Integral Using u = tanφ
Let u = tanφ, so that du = sec²φ dφ, and 1 + u² = 1 + tan²φ = sec²φ.
As u ranges from −∞ to ∞, φ ranges from −π/2 to π/2.
Substitute into the integral:
u² / (1+u²)³ du = tan²φ / sec&sup6;φ · sec²φ dφ = tan²φ / sec&sup4;φ dφ
Now write tan²φ = sin²φ/cos²φ and sec⁴φ = 1/cos⁴φ, so dividing by sec⁴φ is the same as multiplying by cos⁴φ:
tan²φ / sec⁴φ = (sin²φ/cos²φ) · cos⁴φ = sin²φ · cos²φ
So the integral becomes:
∫−π/2π/2 sin²φ cos²φ dφ
Step 6: Evaluate Using the Double Angle Formula
Use the identity sin(2φ) = 2 sinφ cosφ, so sin²(2φ) = 4 sin²φ cos²φ, which gives:
sin²φ cos²φ = sin²(2φ) / 4
Next, use sin²(2φ) = [1 − cos(4φ)] / 2 (the standard power-reduction identity), so:
sin²φ cos²φ = [1 − cos(4φ)] / 8
Now integrate term by term from −π/2 to π/2:
∫−π/2π/2 [1 − cos(4φ)] / 8 dφ = (1/8) · [φ − sin(4φ)/4] evaluated from −π/2 to π/2
At φ = π/2: value is π/2 − sin(2π)/4 = π/2 − 0 = π/2
At φ = −π/2: value is −π/2 − sin(−2π)/4 = −π/2 − 0 = −π/2
Subtracting: π/2 − (−π/2) = π
So the integral equals (1/8) · π = π/8.
Step 7: Assemble the Fisher Information
Recall from Step 4:
I(θ) = (4/π) · ∫−∞∞ u²/(1+u²)³ du = (4/π) · (π/8)
The π in the numerator and denominator cancel exactly:
I(θ) = 4/8 = 1/2
This is the Fisher information contained in a single Cauchy observation with respect to its location parameter θ.
Step 8: Apply the Cramer-Rao Formula for a Sample of Size n
For n independent observations, the Cramer-Rao lower bound is:
CRLB = 1 / [n · I(θ)] = 1 / [n · (1/2)] = 2/n
Final Answer: Option (b), 2/n.This matches the official answer key, which records option (b) for Question 55 of the 2016 Statistics Paper-2 (Series A).
Why This Question Matters
Fisher information and the Cramer-Rao bound appear almost every year in the ISS Statistics papers, usually dressed up in a different distribution each time -- Poisson, exponential, normal, uniform, and here, Cauchy. The Cauchy case is a favourite because it is deceptively simple to set up but forces you through a genuine definite-integral calculation rather than a one-line plug-and-chug. Students who drill this kind of derivation -- log-density, score function, square and integrate -- until the trigonometric substitution becomes automatic tend to solve this entire family of questions (regardless of which distribution shows up) within two or three minutes in the actual exam, which is exactly the kind of speed this paper rewards given its strict negative marking.
Frequently Asked Questions
What exactly does the Cramer-Rao lower bound tell us?
It gives the smallest possible variance that any unbiased estimator of a parameter can have, based only on the information content of the distribution itself. If an estimator's variance equals this bound, it is called an efficient estimator, because no unbiased estimator can do better.
Why does the Cauchy distribution come up so often in these questions?
The Cauchy distribution is a classic trap distribution in statistics because it has no finite mean or variance, yet its Fisher information for the location parameter is perfectly well defined and finite. This makes it a popular way to test whether students confuse the distribution has no mean with we cannot analyse its estimators -- the two are unrelated.
Is the sample mean a good estimator of θ for a Cauchy distribution?
No, and this is a well-known surprising result: the sample mean of n Cauchy observations has exactly the same distribution as a single observation, so increasing the sample size does not reduce its variance at all. The sample median performs far better and is the usual recommended estimator in practice.
Why do we work with the log of the density instead of the density directly?
Differentiating a product or a complicated rational function directly is messy. Taking the log first converts the expression into sums and simple powers, which differentiate far more cleanly, and this log-derivative (the score function) is exactly the quantity the Fisher information formula is built around.
Does the Cramer-Rao bound always decrease as 1/n?
For independent and identically distributed samples, yes -- the Fisher information from n independent observations is simply n times the information from one observation, so the bound always scales as 1/n once you have found the single-observation information I(θ).
What should I do if my answer does not match the official key on a similar question?
Re-derive the score function from scratch and check the sign of each derivative term carefully, since a single sign error in the score function gets squared away and can silently give you the wrong Fisher information. If the mismatch persists after rechecking, it is worth comparing your integral setup line by line against a fully worked solution like this one.
If you worked through this differently or spotted a quicker route to the same answer, drop it in the comments below -- and if a fellow ISS aspirant is stuck on Fisher information questions like this one, consider sharing this with them.
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