Paper I — Q1
(a) Let X and Y be independent random variables with exponential distribution having respective means 1/(λ₁) and 1/(λ₂), λ₁ > 0…
Let X and Y be independent random variables with exponential distribution having respective means 1/(λ₁) and 1/(λ₂), λ₁ > 0, λ₂ > 0. Find E [max (X, Y)]. 10 marks
Using Central Limit Theorem, show that limₙ → ∞ e⁻ⁿ Σₖ₌₀ⁿ (n^k)/(k!) = 1/2 10 marks
An unbiased six-sided die is thrown twice. Let X denote the smaller of the scores obtained. Then show that the probability mass function (p.m.f.) of X is given by : p_X(x) = (13-2x)/36, x = 1, 2, ..., 6 = 0, otherwise. 10 marks
Let T₁ and T₂ be two unbiased estimators of θ with Var(T₁) = Var(T₂), then show that Corr(T₁, T₂) ≥ 2e – 1, where e is the efficiency of each estimator. 10 marks
An urn contains 5 marbles of which θ are white and the others black. In order to test null hypothesis H₀ : θ = 3 versus alternative hypothesis H₁ : θ = 4, two marbles are drawn at random. H₀ is rejected if both the marbles are white, otherwise H₀ is accepted.
Show that probability of type I error in case of without replacement and with replacement schemes, both are less than 0·40, but power of the test under with replacement is higher than that of under without replacement scheme. 10 marks
हिंदी में प्रश्न पढ़ें
मान लीजिए कि X और Y स्वतंत्र एवं चरघातांकी बंटित यादृच्छिक चर हैं जिनके माध्य क्रमशः 1/(λ₁) और 1/(λ₂) हैं, जहाँ λ₁ > 0, λ₂ > 0 है। E [max (X, Y)] ज्ञात कीजिए। (10 अंक)
केन्द्रीय सीमा प्रमेय का प्रयोग करते हुए दर्शाइए कि limₙ → ∞ e⁻ⁿ Σₖ₌₀ⁿ (n^k)/(k!) = 1/2 (10 अंक)
छः फलकों वाले एक निष्पक्ष पाँसे को दो बार फेंका जाता है। मान लीजिए कि प्राप्त समंकों में छोटे समंक को X से निर्दिष्ट किया जाता है। तब दर्शाइए कि X का प्रायिकता द्रव्यमान फलन (पी.एम.एफ.) इस प्रकार दिया जाता है : p_X(x) = (13-2x)/36, x = 1, 2, ..., 6 = 0, अन्यथा। (10 अंक)
मान लीजिए θ के लिए T₁ और T₂ दो अनभिनत आकलक हैं जिनके प्रसरण Var(T₁) = Var(T₂) हैं, तब दर्शाइए कि Corr(T₁, T₂) ≥ 2e – 1, जहाँ e प्रत्येक आकलक की दक्षता है। (10 अंक)
एक कलश में 5 मार्बल हैं जिनमें से θ सफेद हैं और बाकी काले हैं। निराकरणीय परिकल्पना H₀ : θ = 3 का वैकल्पिक परिकल्पना H₁ : θ = 4 के विरुद्ध परीक्षण करने के लिए दो मार्बल यादृच्छया लिए गए हैं। यदि दोनों मार्बल सफेद आते हैं तो H₀ को अस्वीकार किया जाता है, अन्यथा H₀ को स्वीकार किया जाता है।
दर्शाइए कि प्रथम प्रकार की त्रुटि की प्रायिकता प्रतिस्थापन रहित तथा प्रतिस्थापन सहित दोनों योजनाओं में 0·40 से कम है, लेकिन परीक्षण की क्षमता प्रतिस्थापन सहित योजना में प्रतिस्थापन रहित योजना से अधिक है। (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) Expected value of the maximum
Let M = max(X, Y). For t ≥ 0, P(X > t) = e^(−λ₁t) and P(Y > t) = e^(−λ₂t). By independence, P(M ≤ t) = P(X ≤ t) P(Y ≤ t) = (1 − e^(−λ₁t))(1 − e^(−λ₂t)), so P(M > t) = e^(−λ₁t) + e^(−λ₂t) − e^(−(λ₁+λ₂)t). M is non-negative, so E(M) = ∫₀^∞ P(M > t) dt = 1/λ₁ + 1/λ₂ − 1/(λ₁ + λ₂) = (λ₁² + λ₁λ₂ + λ₂²)/[λ₁λ₂(λ₁ + λ₂)]. Check: max + min = X + Y, and min(X, Y) is exponential with rate λ₁ + λ₂, so E(M) = 1/λ₁ + 1/λ₂ − 1/(λ₁ + λ₂). For λ₁ = λ₂ = λ this equals 3/(2λ).
(b) Limit by the Central Limit Theorem
Let Y₁, Y₂, … be independent Poisson(1) variables and Sₙ = Y₁ + … + Yₙ. Then Sₙ ~ Poisson(n), so P(Sₙ = k) = e⁻ⁿ nᵏ/k! and P(Sₙ ≤ n) = e⁻ⁿ Σₖ₌₀ⁿ nᵏ/k!. Each Yᵢ has mean 1 and variance 1, so E(Sₙ) = Var(Sₙ) = n. By the Central Limit Theorem (Lindeberg–Lévy), Zₙ = (Sₙ − n)/√n converges in distribution to N(0, 1). The limit distribution function Φ is continuous, so convergence holds at every point, in particular at 0: P(Sₙ ≤ n) = P(Zₙ ≤ 0) → Φ(0) = 1/2. Hence lim (n→∞) e⁻ⁿ Σₖ₌₀ⁿ nᵏ/k! = 1/2.
(c) p.m.f. of the smaller score
Let U and V be the two scores; the 36 outcomes (u, v) are equally likely. X = min(U, V). P(X ≥ x) = P(U ≥ x) P(V ≥ x) = ((7 − x)/6)² = (7 − x)²/36, x = 1, …, 6, and P(X ≥ 7) = 0. P(X = x) = P(X ≥ x) − P(X ≥ x + 1) = [(7 − x)² − (6 − x)²]/36 = (13 − 2x)/36. Counting check: X = x when both scores equal x (1 outcome) or one equals x and the other is larger (2(6 − x) outcomes); total 2(6 − x) + 1 = 13 − 2x. So p_X(x) = 11/36, 9/36, 7/36, 5/36, 3/36, 1/36 for x = 1, …, 6, and 0 otherwise; the sum is 36/36 = 1.
(d) Lower bound on the correlation of two unbiased estimators
Let Var(T₁) = Var(T₂) = σ² > 0 and ρ = Corr(T₁, T₂). Let V₀ be the minimum possible variance of an unbiased estimator of θ (the Cramér–Rao bound, or the variance of the MVUE). The efficiency of each estimator is e = V₀/σ², so V₀ = eσ². T = (T₁ + T₂)/2 is also unbiased for θ, since E(T) = θ. Its variance is Var(T) = ¼[Var(T₁) + Var(T₂) + 2Cov(T₁, T₂)] = ¼(2σ² + 2ρσ²) = σ²(1 + ρ)/2. No unbiased estimator has variance below V₀, so Var(T) ≥ V₀: σ²(1 + ρ)/2 ≥ eσ² ⇒ (1 + ρ)/2 ≥ e ⇒ ρ ≥ 2e − 1. Hence Corr(T₁, T₂) ≥ 2e − 1. (Since ρ ≤ 1, this also shows e ≤ 1.)
(e) Type I error and power with and without replacement
Under H₀ (θ = 3) the urn has 3 white and 2 black marbles; under H₁ (θ = 4), 4 white and 1 black. H₀ is rejected when both marbles drawn are white. Without replacement (hypergeometric): α = C(3,2)/C(5,2) = 3/10 = 0.30 Power = 1 − β = C(4,2)/C(5,2) = 6/10 = 0.60 With replacement (independent draws): α = (3/5)² = 9/25 = 0.36 Power = (4/5)² = 16/25 = 0.64 Both α = 0.30 and α = 0.36 are less than 0.40. The power with replacement, 0.64, exceeds the power without replacement, 0.60.
How this answer will be evaluated
What earns full marks: Complete derivations with all steps shown, correct notation, and clear interpretation of results
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Find E[max(X, Y)] for independent exponential variables with means 1/λ₁ and 1/λ₂. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Distribution of the maximum by independence: P(M ≤ t) = (1 − e^(−λ₁t))(1 − e^(−λ₂t))
- Correct integration of the survival function (or joint density) to get E(M)
- Final result E[max(X, Y)] = 1/λ₁ + 1/λ₂ − 1/(λ₁ + λ₂)
Loses marks
- Treating max(X, Y) as exponential, or taking E[max] = max(1/λ₁, 1/λ₂)
- Adding the two means without subtracting 1/(λ₁ + λ₂)
Earns more
- Alternative route: max + min = X + Y with min(X, Y) exponential with rate λ₁ + λ₂
- Check for λ₁ = λ₂ = λ giving 3/(2λ)
Extra mark
- Extension to the maximum of n independent exponential variables
- (b) Using the Central Limit Theorem, show that e⁻ⁿ Σₖ₌₀ⁿ nᵏ/k! → 1/2 as n → ∞. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Identifies the sum of n independent Poisson(1) variables as Poisson(n), so the given sum equals P(Sₙ ≤ n)
- Standardises with E(Sₙ) = Var(Sₙ) = n and applies the CLT: (Sₙ − n)/√n → N(0, 1)
- Concludes P(Sₙ ≤ n) → Φ(0) = 1/2
Loses marks
- Using Stirling's formula or direct summation instead of the CLT
- Not recognising the sum as a Poisson(n) distribution function at n
Earns more
- Notes that Φ is continuous, so the distribution function converges at 0
- States the additive (reproductive) property of the Poisson distribution
Extra mark
- Numerical illustration, e.g. the sum is about 0.503 for n = 10000
- (c) Show that the p.m.f. of the smaller score X in two throws of a fair die is (13 − 2x)/36 for x = 1, …, 6. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Defines X = min(U, V) over 36 equally likely outcomes
- Derives p_X(x) = (13 − 2x)/36 from P(X ≥ x) = ((7 − x)/6)² or by counting 2(6 − x) + 1 outcomes
- States p_X(x) = 0 otherwise and checks that 11, 9, 7, 5, 3, 1 (over 36) sum to 1
Loses marks
- Confusing the minimum with the maximum (which gives (2x − 1)/36)
- Omitting the factor 2 for the two orderings of unequal scores
Earns more
- Lists the probabilities for x = 1 to 6
- Gives the counting argument as a second method
- (d) Show that Corr(T₁, T₂) ≥ 2e − 1 for two unbiased estimators of θ with equal variance and efficiency e. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Defines efficiency e = V₀/σ², where V₀ is the minimum (Cramér–Rao) variance of unbiased estimators
- Forms the unbiased estimator (T₁ + T₂)/2 and finds its variance σ²(1 + ρ)/2
- Uses Var((T₁ + T₂)/2) ≥ V₀ to obtain ρ ≥ 2e − 1
Loses marks
- Assuming T₁ and T₂ are independent, so the covariance term is dropped
- Using a wrong definition of efficiency or a biased combination of T₁ and T₂
Earns more
- Correct use of Cov(T₁, T₂) = ρσ²
- Notes that ρ ≤ 1 implies e ≤ 1
Extra mark
- Equality holds when (T₁ + T₂)/2 attains the minimum variance V₀
- (e) Show that both type I error probabilities are below 0.40 and that the power with replacement exceeds the power without replacement. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Sets up H₀: 3 white, 2 black and H₁: 4 white, 1 black, with rejection when both marbles are white
- Without replacement: α = C(3,2)/C(5,2) = 0.30 and power = C(4,2)/C(5,2) = 0.60
- With replacement: α = (3/5)² = 0.36 and power = (4/5)² = 0.64, so both α < 0.40 and 0.64 > 0.60
Loses marks
- Confusing type I error with power or type II error
- Using with-replacement probabilities for the without-replacement scheme
Earns more
- Uses the hypergeometric model for the scheme without replacement
- States power = 1 − β = P(reject H₀ | H₁)
What "Prove" is asking you to do
Establish that the statement holds for every case it claims, not for one representative case. The argument must be closed: each line follows from a definition, a hypothesis, or a named theorem you are entitled to use.
Structure that answers it
Given and to prove, restated → theorem or construction to be used, named → the argument line by line → conclusion stated as proved
Where marks are lost
Testing one example, which illustrates but proves nothing. On an if and only if claim, proving one direction and stopping forfeits that half outright, and degenerate cases — zero, the empty set, the equality case — have to be disposed of rather than assumed away.
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