Paper I — Q2
(a) Let Y₁, Y₂, Y₃, ... be independent and identical Poisson random variables with parameter 1. Use central limit theorem to…
Let Y₁, Y₂, Y₃, ... be independent and identical Poisson random variables with parameter 1. Use central limit theorem to establish n! ≃ √(2π n)(n/e)ⁿ for large value of positive integer n. 20 marks
Let X₁, X₂, ..., Xₙ be a random sample such that log Xᵢ ~ N(θ, θ) distribution with θ > 0 unknown. Show that one of the solutions of the likelihood equation is the unique MLE of θ. Obtain asymptotic distribution of MLE of θ. 15 marks
State the sufficient conditions for a function φ(t) to be a characteristic function.
Investigate if the following functions are characteristic functions : 1. e⁻ᵗ⁴ 2. [1 + |t|]⁻¹ Justify your answer. (5+10 marks)
हिंदी में प्रश्न पढ़ें
माना Y₁, Y₂, Y₃, ... स्वतंत्र और सर्वसम व्यासों यादृच्छिक चर हैं जिनका प्राचल 1 है। केन्द्रीय सीमा प्रमेय का उपयोग करते हुए स्थापित कीजिए n! ≃ √(2π n)(n/e)ⁿ, जबकि धनात्मक पूर्णांक n बहुत है। (20 अंक)
माना X₁, X₂, ..., Xₙ ऐसा यादृच्छिक प्रतिदर्श है जिसका बंटन log Xᵢ ~ N(θ, θ), θ > 0 अज्ञात है। दर्शाइए कि संभाविता समीकरण का एक हल θ का एकमात्र MLE है। θ के MLE के लिए उपगामी बंटन प्राप्त कीजिए। (15 अंक)
फलन φ(t) के अभिलक्षण फलन होने के लिए पर्याप्त प्रतिबंधों को लिखिए।
जाँच कीजिए कि क्या निम्न फलन अभिलक्षण फलन हैं : 1. e⁻ᵗ⁴ 2. [1 + |t|]⁻¹ अपने उत्तर को तर्कसंगत सिद्ध कीजिए। (5+10 अंक)
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) Stirling's approximation from the central limit theorem
Let Sₙ = Y₁ + ⋯ + Yₙ. Since the Yᵢ are independent Poisson(1), Sₙ ~ Poisson(n), so P(Sₙ = n) = e⁻ⁿ nⁿ / n!.
Sₙ has mean n and variance n, so Zₙ = (Sₙ − n)/√n → N(0, 1) by the central limit theorem. Sₙ takes integer values, so use the continuity correction: P(Sₙ = n) = P(n − ½ < Sₙ ≤ n + ½) = P(−1/(2√n) < Zₙ ≤ 1/(2√n)). For large n the limits shrink to zero, so this probability is approximately φ(0) × (1/√n) = 1/√(2πn), where φ is the standard normal density.
Equating, e⁻ⁿ nⁿ / n! ≈ 1/√(2πn), so n! ≈ √(2πn) (n/e)ⁿ for large n, the ratio of the two sides tending to 1.
(b) MLE for log-normal data with log X ~ N(θ, θ)
Let zᵢ = log xᵢ, so zᵢ ~ N(θ, θ). Up to terms free of θ, the log-likelihood is l(θ) = −(n/2) log θ − S/(2θ) − nθ/2 + C, where S = Σzᵢ² (the term Σzᵢ is constant in θ). Then l′(θ) = −n/(2θ) + S/(2θ²) − n/2 = [n/(2θ²)] (S/n − θ − θ²).
The likelihood equation gives θ² + θ − S/n = 0, so θ = [−1 ± √(1 + 4S/n)]/2. Only the positive root is admissible: θ̂ = [−1 + √(1 + 4S/n)]/2. Uniqueness: the factor S/n − θ − θ² decreases strictly from S/n > 0 at θ = 0 to −∞, so l′ changes sign once, from positive to negative. Hence θ̂ is the unique maximiser, i.e. the MLE.
Asymptotic distribution: for one observation, the score is U = z²/(2θ²) − 1/(2θ) − 1/2 and U′ = 1/(2θ²) − z²/θ³. With E[z²] = θ + θ², I(θ) = −E[U′] = (θ + θ²)/θ³ − 1/(2θ²) = (1 + 2θ)/(2θ²). Hence √n (θ̂ − θ) → N(0, 2θ²/(1 + 2θ)).
(c) Characteristic functions
(i) By Bochner's theorem, φ is a characteristic function if and only if φ(0) = 1, φ is continuous at 0, and φ is positive definite: Σⱼ Σₖ φ(tⱼ − tₖ) aⱼ āₖ ≥ 0 for all n, real t₁,…,tₙ and complex a₁,…,aₙ. A convenient sufficient test is Pólya's criterion: φ is real, even, continuous, φ(0) = 1, and convex and non-increasing on (0, ∞).
(ii)
- φ(t) = e^(−t⁴) = 1 − t⁴ + ⋯ has φ″(0) = 0. For a characteristic function whose second derivative exists at 0, φ″(0) = −E[X²], so E[X²] = 0 and X = 0 almost surely, giving φ ≡ 1. This contradicts φ(t) ≠ 1 for t ≠ 0. So e^(−t⁴) is not a characteristic function.
- φ(t) = 1/(1 + |t|) is real, even, continuous, φ(0) = 1, and for t > 0, φ′ = −1/(1 + t)² < 0 and φ″ = 2/(1 + t)³ > 0. It is convex and decreasing to 0, so by Pólya's criterion 1/(1 + |t|) is a characteristic function.
How this answer will be evaluated
What earns full marks: Rigorous derivation with all steps shown; correct asymptotic forms; precise justification for characteristic functions.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Use the central limit theorem on iid Poisson(1) variables to establish Stirling's approximation, 20 marks 20 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Sn ~ Poisson(n) and P(Sn = n) = e^-n n^n / n!
- CLT with continuity correction: P(n-1/2 < Sn <= n+1/2) tends to 1/sqrt(2 pi n)
- Equate and rearrange to n! ≈ sqrt(2 pi n)(n/e)^n
Loses marks
- Skipping the continuity correction without justification
- Circular use of Stirling's formula
Earns more
- Correct mean and variance of Sn
- Statement that the ratio tends to 1
Extra mark
- Mention of the local limit theorem as the rigorous route
- (b) Show one solution of the likelihood equation is the unique MLE of θ and find its asymptotic distribution, 15 marks 15 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Log-likelihood and equation θ² + θ − S/n = 0
- Positive root θ̂ = [−1 + √(1+4S/n)]/2 with the uniqueness argument
- Asymptotic law N(0, 2θ²/(1+2θ)) from I(θ) = (1+2θ)/(2θ²)
Loses marks
- Keeping the negative root
- Wrong Fisher information
Earns more
- Using E[z²] = θ + θ² correctly
- Noting the term Σz is free of θ
Extra mark
- Second-derivative check of the maximum
- (c)(i) (i) State sufficient conditions for φ(t) to be a characteristic function, 5 marks 5 marks
enumerate— list the items in order → one line each → no commentary
Must cover
- φ(0) = 1 and continuity at 0
- Positive definiteness (Bochner)
- Pólya's criterion as a convenient sufficient test
Loses marks
- Omitting continuity
- Stating only boundedness
Earns more
- Correct statement of the positive-definite sum
Extra mark
- Note that Bochner's conditions are also necessary
- (c)(ii) (ii) Investigate whether exp(−t⁴) and 1/(1+|t|) are characteristic functions, 10 marks 10 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- exp(−t⁴) is not one, via φ''(0) = 0 forcing X = 0
- 1/(1+|t|) is one, via Pólya's criterion
- Justification with the derivatives or convexity
Loses marks
- Conclusion without justification
- Claiming exp(−t⁴) is valid
Earns more
- Verifying evenness, continuity and φ(0) = 1
- Contradiction φ ≡ 1 stated clearly
Extra mark
- Alternative: 1/(1+|t|) as a mixture of triangular or Laplace-type functions
What "Derive" is asking you to do
Reach the stated expression from a starting relation, justifying every step. The destination is printed in the question, so only the route earns marks, and the assumptions you work under are part of that route.
Structure that answers it
Assumptions and notation defined → starting relation or governing equation → each step with its justification → the required expression → limiting case or boundary check
Where marks are lost
Writing the standard result first and fitting three lines to it, which an examiner reads at a glance. Marks also go on assumptions left unstated — lossless medium, small amplitude, errors independent with zero mean — and on symbols used before they are defined, even when the question says usual notations.
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