(a) Let X and Y be independent random variables with exponential distribution having respective means $\frac{1}{\lambda_1}$ and $\frac{1}{\lambda_2}$, $\lambda_1 0, \lambda_2 0$. Find E [max (X, Y)]. (10 marks)

(b) Using Central Limit Theorem, s

Question

(a) Let X and Y be independent random variables with exponential distribution having respective means $\frac{1}{\lambda_1}$ and $\frac{1}{\lambda_2}$, $\lambda_1 > 0, \lambda_2 > 0$. Find E [max (X, Y)]. (10 marks)

(b) Using Central Limit Theorem, show that
$$\lim_{n 	o \infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!} = \frac{1}{2}$$ (10 marks)

(c) 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 :
$$p_X(x) = \frac{13-2x}{36}, \quad x = 1, 2, ..., 6$$
$$= 0, \quad 	ext{otherwise.}$$ (10 marks)

(d) 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)

(e) 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)

UPSC Answer Check · Accepted Answer

Prove each of the five results systematically, allocating approximately 2 minutes per mark (20 minutes total). For (a), use the identity E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)] or direct integration; for (b), recognize the Poisson sum and apply CLT with continuity correction; for (c), enumerate favorable outcomes for minimum value; for (d), apply Cauchy-Schwarz and efficiency definition; for (e), compute hypergeometric vs binomial probabilities. Present each proof with clear statement of assumptions, step-by-step derivation, and boxed final result.
- (a) Correct setup using E[max(X,Y)] = ∫∫ max(x,y)f_X(x)f_Y(y)dxdy or equivalent identity with min(X,Y) ~ Exp(λ₁+λ₂)
- (b) Identification of Poisson(n) probability mass function and application of CLT with continuity correction to show P(S_n ≤ n) → Φ(0) = 1/2
- (c) Enumeration of outcomes where min equals k: (k,k), (k,j) for j>k, (i,k) for i>k, yielding count (13-2k) for each k = 1,...,6
- (d) Use of Var(T₁+T₂) ≥ 0 and efficiency definition e = [CRLB]/Var(T₁) to establish the inequality Corr(T₁,T₂) ≥ 2e-1
- (e) Type I error: P(reject|H₀) = C(3,2)/C(5,2) = 0.3 (without replacement) vs (3/5)² = 0.36 (with replacement); Power: P(reject|H₁) = C(4,2)/C(5,2) = 0.6 vs (4/5)² = 0.64, showing higher power with replacement

Dimension	Weight	Max marks	Excellent	Average	Poor
Setup correctness	20%	10	Correctly identifies distributions and parameters for all parts: joint PDF for (a), Poisson(n) for (b), uniform discrete for (c), unbiasedness condition for (d), hypergeometric vs binomial for (e); states all assumptions explicitly	Identifies most distributions correctly but misses some parameter specifications or assumptions; minor errors in setup for one sub-part	Major errors in identifying distributions; confuses with/without replacement in (e); incorrect parameterization of exponential or Poisson distributions
Method choice	20%	10	Optimal methods: E[max] = E[X]+E[Y]-E[min] for (a), CLT with continuity correction for (b), combinatorial enumeration for (c), Cauchy-Schwarz with efficiency definition for (d), correct hypothesis testing framework for (e)	Correct but suboptimal methods (e.g., direct double integration for (a) without simplification); omits continuity correction in (b); correct approach but inefficient execution	Incorrect methods: attempts direct integration without exploiting independence for (a), misses CLT application for (b), uses probability multiplication incorrectly for (c), ignores efficiency definition for (d)
Computation accuracy	20%	10	Flawless calculations: E[max(X,Y)] = 1/λ₁ + 1/λ₂ - 1/(λ₁+λ₂); correct limit evaluation; exact counts 11,9,7,5,3,1 summing to 36; algebraic manipulation yielding Corr ≥ 2e-1; precise values 0.3, 0.36, 0.6, 0.64	Minor arithmetic errors in one sub-part (e.g., miscounts outcomes in (c) by 1-2); correct method but final numerical error; partial credit for correct intermediate steps	Major computational errors: incorrect integration limits, wrong factorial calculations, probability values not summing to 1, incorrect comparison of power values in (e)
Interpretation	20%	10	Clear interpretation: explains why min of exponentials is exponential with summed rate; interprets Poisson sum as P(S_n ≤ n) for S_n ~ Poisson(n); explains minimum die score intuition; interprets efficiency bound meaning; explains why with-replacement has higher power due to constant variance	Provides correct results with minimal interpretation; some explanation of statistical meaning but incomplete across all parts	No interpretation provided; fails to explain why results hold; confuses type I error with power in verbal explanation
Final answer & units	20%	10	All five answers clearly stated and boxed: (a) closed-form expression in λ₁,λ₂; (b) limit equals 1/2; (c) verified PMF with support; (d) established inequality; (e) numerical verification of claims with explicit comparison	Most answers stated clearly but one missing or incomplete; correct answers but poor presentation without final boxed statements	Missing final answers for multiple parts; incorrect final statements; fails to verify claims in (e) with actual numbers

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