Statistics 2023 Paper I 50 marks Prove

Q3

(a) (i) If X is a random variable with finite variance, show that lim n² P{|X| > n} = 0. n → ∞ (10 marks) (ii) In a certain recruitment test, there are multiple choice questions. There are four possible options to each question, out of which one is correct. The probability of knowing correct option for an intelligent student is 90%, while it is 20% for a weaker student. An intelligent student ticks the correct option. What is the probability that he was guessing ? (10 marks) (b) Determine whether the sequence of mutually independent random variables {Xₙ, n ≥ 1}, in which P(Xₙ = ± n^λ) = 1/(2n^(2λ)) P(Xₙ = 0) = 1 - 1/n^(2λ) (λ < 1/2) obeys Central Limit Theorem (CLT) or not. (15 marks) (c) Define Sequential Probability Ratio Test (SPRT) along with its operating characteristic function and average sample number. Determine SPRT for testing H₀ : θ = 4 against H₁ : θ = 5 in N(θ, 1) with α = 0·5 and β = 0·2. (15 marks)

हिंदी में प्रश्न पढ़ें

(a) (i) यदि X परिमित प्रसरण के साथ एक यादृच्छिक चर है, तो दिखाइए कि lim n² P{|X| > n} = 0. n → ∞ (10 अंक) (ii) किसी एक भर्ती परीक्षा में, बहुविकल्पीय प्रश्न हैं । प्रत्येक प्रश्न में चार संभव विकल्प हैं, जिनमें से एक सही है । एक बुद्धिमान छात्र के सही विकल्प जानने की प्रायिकता 90% है, जबकि एक कमजोर छात्र की केवल 20% है । एक बुद्धिमान छात्र सही विकल्प पर निशान लगाता है । इसके अनुमान से सही विकल्प पर निशान लगाने की प्रायिकता क्या है ? (10 अंक) (b) परीक्षण कीजिए कि परस्पर स्वतंत्र यादृच्छिक चरों की श्रृंखला {Xₙ, n ≥ 1}, जिसमें P(Xₙ = ± n^λ) = 1/(2n^(2λ)) P(Xₙ = 0) = 1 - 1/n^(2λ) (λ < 1/2) केंद्रीय सीमा प्रमेय (CLT) का पालन करती है या नहीं । (15 अंक) (c) अनुक्रमिक प्रायिकता अनुपात परीक्षण (SPRT) को इसके संकारक अभिलक्षण फलन एवं औसत प्रतिदर्श संख्या के साथ परिभाषित कीजिए । N(θ, 1) में H₀ : θ = 4 विरुद्ध H₁ : θ = 5 के परीक्षण के लिए अनुक्रमिक प्रायिकता अनुपात परीक्षण (SPRT), α = 0·5 तथा β = 0·2 के साथ ज्ञात कीजिए । (15 अंक)

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Directive word: Prove

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How this answer will be evaluated

Approach

Prove the limit result in (a)(i) using Markov/Chebyshev inequalities; solve (a)(ii) using Bayes' theorem with clear event definitions; for (b), verify Lindeberg condition or Lyapunov's theorem to establish CLT validity; for (c), define SPRT components then derive boundaries A, B and continuation region. Allocate ~20% time to (a)(i), ~15% to (a)(ii), ~30% to (b), and ~35% to (c) given mark distribution. Structure: state definitions → apply methods → derive results → interpret findings.

Key points expected

  • (a)(i): Application of Markov's inequality or direct variance bound to show n²P{|X|>n} ≤ E[X²I(|X|>n)] → 0
  • (a)(ii): Bayes' theorem setup with events K (knows), G (guesses), C (correct); calculation of P(G|C) = P(C|G)P(G)/P(C)
  • (b): Verification of Lindeberg condition or checking variance of sum → ∞; showing standardized sum converges to N(0,1)
  • (b): Explicit computation that Var(Sₙ) = Σn^(2λ) → ∞ and Lyapunov condition holds for λ < 1/2
  • (c): Definition of SPRT, OC function L(θ), and ASN Eθ(N); derivation of Wald's boundaries A ≈ (1-β)/α, B ≈ β/(1-α)
  • (c): Specific SPRT construction for N(θ,1): continuation region as sum(Xi - 4.5) between adjusted log-boundaries

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%10Correctly defines all random variables, events (K, G, C in a-ii), hypotheses (H₀, H₁ in c), and states conditions (finite variance in a-i, λ<1/2 in b, normal distribution in c); proper notation throughoutDefines most variables correctly but misses some conditions (e.g., forgets independence assumption in b or confuses α, β in c)Incorrect or missing definitions; confuses P{|X|>n} with P{|X|<n}; wrong hypothesis specification in SPRT
Method choice20%10Selects optimal methods: Markov inequality/variance bound for (a-i), Bayes' theorem for (a-ii), Lindeberg/Lyapunov CLT for (b), Wald's SPRT theory for (c); justifies why alternatives are inferiorUses correct general approaches but suboptimal variants (e.g., Chebyshev instead of Markov for a-i, direct verification instead of Lindeberg for b)Wrong methods entirely (e.g., tries Chebyshev with wrong scaling for a-i, uses standard CLT without checking conditions in b, treats SPRT as fixed-sample test)
Computation accuracy20%10Flawless calculations: correct limit derivation n²P{|X|>n} ≤ ∫_{|x|>n} x²dF → 0; P(G|C) = (0.25×0.1)/(0.9+0.025) = 10/185; Var(Sₙ) = n^(2λ+1)/(2λ+1); A=4, B=0.25; correct continuation inequalitiesMinor computational slips (e.g., arithmetic error in Bayes' denominator, off-by-one in variance sum, approximate boundary values without derivation)Major errors: incorrect probability values (e.g., P(C|G)=1/4 forgotten), wrong variance calculation, boundary formulas swapped or missing log terms
Interpretation20%10Explains why finite variance ensures tail decay faster than n⁻²; interprets P(G|C) ≈ 5.4% as low guessing probability despite 10% guess rate; explains why λ<1/2 ensures CLT via variance growth; interprets SPRT's efficiency vs fixed-sample testStates results without full interpretation; mentions CLT holds but doesn't explain role of λ; describes SPRT procedure without comparing to Neyman-PearsonNo interpretation; fails to explain what limit result means for tail behavior; no insight on why CLT fails/works; treats SPRT as black box
Final answer & units20%10All boxed/concluded answers: limit equals 0; P(guessing | correct) = 10/185 or 2/37 ≈ 0.054; explicit 'CLT holds' with justification; SPRT with explicit A=4, B=1/4, continuation region in terms of Sₙ and n; proper probabilistic notation throughoutCorrect final values but poorly presented; missing explicit conclusion on CLT; SPRT boundaries stated without clear decision rulesMissing or wrong final answers; incomplete SPRT specification; no clear statement of whether CLT applies

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