Statistics 2025 Paper I 50 marks Compulsory Prove

Q1

(a) Let E, F and G be three pairwise independent events such that P(E∩F) = 0·1 and P(F∩G) = 0·3. Prove that P(Eᶜ∪G) ≥ 11/12. 10 marks (b) If X and Y are non-negative independent random variables and their joint moment generating function is given by M_{X,Y}(t₁,t₂) = e^{t₁+2e^{t₂}-3}; t₁ > 0, t₂ > 0, then show that 2P(X+Y=2) = 9P(X+Y=0). 10 marks (c) If X₁, X₂, ... be a sequence of i.i.d. U(0, 1) random variables, then find the value of Lim_{n→∞} P(∑_{i=1}^{n} X_i ≤ n/2 + √(n/144)) [use √3 = 1·74, Φ(0·29) = 0·6141, Φ(1) = 0·8413]. 10 marks (d) Let X₁, X₂, ..., Xₙ be a random sample from normal N(μ, σ²) distribution. Obtain sufficient statistic for parameters (μ, σ²) when both the parameters are unknown. If σ² is known, what will be sufficient statistic for parameter μ ? 10 marks (e) A random variable X has the following distribution under H₀ and H₁ : x : 1 2 3 4 5 6 f₀(x) : 0·01 0·01 0·01 0·01 0·01 0·95 f₁(x) : 0·05 0·04 0·03 0·02 0·01 0·85 Find the best test of size 0·03 and its probability of type-II error for testing H₀ : f = f₀ versus H₁ : f = f₁. Is it unbiased test ? Why ? 10 marks

हिंदी में प्रश्न पढ़ें

(a) मान लीजिए E, F और G तीन युग्मतः स्वतंत्र घटनाएँ इस प्रकार हैं कि P(E∩F) = 0·1 और P(F∩G) = 0·3 है। सिद्ध कीजिए कि P(Eᶜ∪G) ≥ 11/12 है। 10 (b) यदि X और Y ऋणेतर स्वतंत्र यादृच्छिक चर हैं तथा जिनका संयुक्त आघूर्ण जनक फलन M_{X,Y}(t₁,t₂) = e^{t₁+2e^{t₂}-3}; t₁ > 0, t₂ > 0 है, तब दर्शाइए कि, 2P(X+Y=2) = 9P(X+Y=0) है। 10 (c) यदि X₁, X₂, ... स्वतंत्र और सर्वथा बंटित U(0, 1) यादृच्छिक चरों का एक अनुक्रम है, तब Lim_{n→∞} P(∑_{i=1}^{n} X_i ≤ n/2 + √(n/144)) का मान ज्ञात कीजिए। [प्रयोग कीजिए, √3 = 1·74, Φ(0·29) = 0·6141, Φ(1) = 0·8413] 10 (d) मान लीजिए कि X₁, X₂, ..., Xₙ प्रसामान्य N(μ, σ²) बंटन से लिया गया एक यादृच्छिक प्रतिदर्श है। जब दोनों प्राचल अज्ञात हैं, तो प्राचलों (μ, σ²) के लिए पर्याप्त प्रतिदर्शज निकालिए। यदि σ² ज्ञात है, तो प्राचल μ के लिए पर्याप्त प्रतिदर्शज क्या होगा ? 10 (e) H₀ तथा H₁ के अंतर्गत एक यादृच्छिक चर X का बंटन निम्नलिखित है : x : 1 2 3 4 5 6 f₀(x) : 0·01 0·01 0·01 0·01 0·01 0·95 f₁(x) : 0·05 0·04 0·03 0·02 0·01 0·85 H₀ : f = f₀ विरुद्ध H₁ : f = f₁ का परीक्षण करने के लिए 0·03 आमाप वाला श्रेष्ठतम परीक्षण तथा उसकी द्वितीय प्रकार की त्रुटि की प्रायिकता प्राप्त कीजिए। क्या यह परीक्षण अनभिनत है ? क्यों ? 10

Directive word: Prove

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How this answer will be evaluated

Approach

This question demands rigorous mathematical proofs and derivations across five sub-parts. Begin with (a) by establishing bounds using pairwise independence and probability inequalities; for (b) extract marginal MGFs to identify distributions and compute probabilities; for (c) apply CLT with given variance 1/12; for (d) use factorization theorem for sufficient statistics; for (e) construct Neyman-Pearson test via likelihood ratios. Allocate time proportionally: ~18 min each for (a), (b), (c) and ~12 min each for (d), (e).

Key points expected

  • For (a): Apply P(Eᶜ∪G) = 1 - P(E∩Gᶜ) and use pairwise independence to bound P(E∩G) ≤ 1/12, yielding the inequality
  • For (b): Factor joint MGF to identify X~Poisson(1) and Y~Poisson(2), then compute P(X+Y=2) and P(X+Y=0) using sum of Poissons
  • For (c): Apply CLT with E[X_i]=1/2, Var(X_i)=1/12, standardize to get Φ(0.29)=0.6141 as limit
  • For (d): State factorization theorem, show (X̄, S²) is sufficient for (μ,σ²), and X̄ alone when σ² known
  • For (e): Compute likelihood ratios f₁/f₀, order critical region by ratios, find size 0.03 test using x=4,5,6, compute β=0.88, verify unbiasedness

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%10Correctly identifies all given conditions: pairwise independence in (a), MGF factorization in (b), i.i.d. uniform setup with mean 1/2 and variance 1/12 in (c), normal distribution parameters in (d), and hypothesis specification with pmf values in (e)Identifies most conditions but misses subtle points like non-negativity constraint in (b) or confuses parameters in (d)Misidentifies key conditions or uses wrong distributions, such as treating (c) as exact rather than asymptotic, or confusing H₀ and H₁ in (e)
Method choice20%10Selects optimal methods: Bonferroni-type bounds for (a), MGF uniqueness theorem for (b), CLT for (c), factorization theorem for (d), Neyman-Pearson lemma for (e); justifies each choiceUses correct but suboptimal methods, or applies correct methods without clear justification; may use direct enumeration instead of CLT in (c)Chooses inappropriate methods like Chebyshev instead of CLT in (c), or attempts integration for discrete distributions in (e)
Computation accuracy20%10All calculations precise: P(E∩G) bound ≤1/12 in (a), Poisson probabilities 9e⁻³ and 2e⁻³ in (b), standardized value 0.29 leading to 0.6141 in (c), correct sufficient statistic forms in (d), likelihood ratios 5,4,3,2,1,17/19 and critical region {4,5,6} with β=0.88 in (e)Minor computational slips like arithmetic errors in probability calculations or incorrect standardization constant in (c)Major errors: wrong variance in CLT, incorrect Poisson parameter identification, or critical region of wrong size in (e)
Interpretation20%10Interprets results meaningfully: explains why bound is tight in (a), recognizes X+Y~Poisson(3) in (b), discusses convergence in (c), explains dimension reduction in (d), and explicitly verifies unbiasedness via power comparison in (e)Provides results with minimal interpretation; states answers without explaining statistical significance or test propertiesMisinterprets results, such as claiming test is unbiased without verification, or misunderstanding what sufficient statistics represent
Final answer & units20%10All five answers clearly stated: inequality proved with ≥11/12 in (a), equality 2P(X+Y=2)=9P(X+Y=0) verified in (b), limit 0.6141 in (c), (X̄,∑(X_i-X̄)²) and X̄ in (d), critical region {4,5,6}, β=0.88, test is biased in (e)Most answers present but some incomplete or poorly formatted; may miss stating final conclusion for unbiasedness in (e)Missing or incorrect final answers; leaves proofs incomplete without numerical conclusions where required

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