Paper I — Q3
(a) Let probability of obtaining Head on a biased coin be 4/5 and X be the number of heads obtained in a sequence of 25…
Let probability of obtaining Head on a biased coin be 4/5 and X be the number of heads obtained in a sequence of 25 independent tosses of the coin. The same coin is tossed again X number of times independently and we obtain Y heads. Compute Var.(X+25Y). 20 marks
Let {6, –8, 3, 2, 7, 5, 4, 9} be a random sample from a population with probability density function f(x, θ) = ½ exp(–|x–θ|), –∞<x, θ<∞. Obtain maximum likelihood estimate of θ. 5 marks
Let X₁, X₂, ..., Xₙ be a random sample from Bernoulli distribution b(1, θ), 0<θ<1. Find the lower bound for the variance of an unbiased estimator of θ based on this data. Find uniformly minimum variance unbiased estimator of θ and show that it attains Cramer-Rao lower bound. 10 marks
Let X₁, X₂, ..., Xₙ be a random sample from beta distribution of first kind β₍₁, θ₎, θ>0. Find consistent estimator of θ, and its variance also. 15 marks
हिंदी में प्रश्न पढ़ें
मान लीजिए एक अभिनत सिक्के में चित आने की प्रायिकता 4/5 है, और X सिक्के की 25 स्वतंत्र उछालों में प्राप्त चितों की संख्या को दर्शाता है। उसी सिक्के को पुनः X बार स्वतंत्र रूप से उछालने पर हमें Y चित प्राप्त होते हैं। Var.(X+25Y) की गणना कीजिए। (20 अंक)
मान लीजिए {6, −8, 3, 2, 7, 5, 4, 9}, प्रायिकता घनत्व फलन f(x, θ) = 1/2 exp(-|x-θ|), -∞ < x, θ < ∞ वाली एक समष्टि से लिया गया एक यादृच्छिक प्रतिदर्श है। θ का अधिकतम संभाविता आकलक प्राप्त कीजिए। (5 अंक)
मान लीजिए X₁, X₂, ..., Xₙ बर्नौली-बंटन b(1, θ), 0<θ<1 से लिया गया एक यादृच्छिक प्रतिदर्श है । इन आँकड़ों पर आधारित, θ के एक अनभिनत आकलक के प्रसरण के लिए निम्न परिबंध ज्ञात कीजिए । θ का एक समान न्यूनतम प्रसरण अनभिनत आकलक ज्ञात कीजिए तथा दर्शाइए कि यह क्रामर-राव निम्न परिबंध प्राप्त करता है । (10 अंक)
मान लीजिए X₁, X₂, ..., Xₙ प्रथम प्रकार के बीटा बंटन β₍₁, θ₎, θ>0 से लिया गया एक यादृच्छिक प्रतिदर्श है । θ का संगत आकलक ज्ञात कीजिए और इसका प्रसरण भी निकालिए । (15 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)
Let p = P(Head) = 4/5, q = 1/5. Then X ~ Bin(25, p), so E[X] = 25p = 20, Var(X) = 25pq = 25(4/5)(1/5) = 4.
Given X, Y | X ~ Bin(X, p). Hence E[Y | X] = pX, Var(Y | X) = Xpq.
By the law of total variance, Var(Y) = E[Var(Y | X)] + Var(E[Y | X]) = E[Xpq] + Var(pX) = pqE[X] + p²Var(X) = (4/25)(20) + (16/25)(4) = 16/5 + 64/25 = 144/25.
Also, Cov(X, Y) = E[XY] – E[X]E[Y] = E[X E(Y | X)] – E[X]·pE[X] = pE[X²] – p(E[X])² = p Var(X) = (4/5)(4) = 16/5.
Therefore, Var(X + 25Y) = Var(X) + 25²Var(Y) + 2·25·Cov(X, Y) = 4 + 625(144/25) + 50(16/5) = 4 + 3600 + 160 = 3764.
(b)
(i) The likelihood is L(θ) = (1/2)ⁿ exp(–Σ |xᵢ – θ|).
Maximizing L is equivalent to minimizing S(θ) = Σ |xᵢ – θ|.
It is a standard result that S(θ) is minimized at a sample median. The ordered sample is –8, 2, 3, 4, 5, 6, 7, 9.
Since n = 8 is even, any value between the 4th and 5th order statistics minimizes S(θ). Thus the complete set of maximum likelihood estimates is θ̂ ∈ [4, 5].
If a single value is conventionally reported, take the sample median θ̂ = (4 + 5)/2 = 4.5.
(ii) Let X₁, X₂, ..., Xₙ be iid Bernoulli(θ). The log-likelihood is log L = Σ [xᵢ log θ + (1 – xᵢ) log(1 – θ)].
The score function is ∂ log L/∂θ = Σ (xᵢ – θ)/[θ(1 – θ)].
For one observation, I(θ) = E[(X – θ)²]/[θ²(1 – θ)²] = θ(1 – θ)/[θ²(1 – θ)²] = 1/[θ(1 – θ)].
For n independent observations, Iₙ(θ) = n/[θ(1 – θ)].
Hence the Cramér-Rao lower bound for any unbiased estimator of θ is 1/Iₙ(θ) = θ(1 – θ)/n.
Let T = Σ Xᵢ. Then T ~ Bin(n, θ). The estimator θ̂ = T/n is unbiased, since E[θ̂] = θ.
Also T is a complete sufficient statistic, so by the Lehmann-Scheffé theorem θ̂ is the uniformly minimum variance unbiased estimator of θ.
Its variance is Var(θ̂) = Var(T)/n² = nθ(1 – θ)/n² = θ(1 – θ)/n.
Thus θ̂ attains the Cramér-Rao lower bound.
(c)
For beta distribution of first kind β(1, θ), the density is f(x; θ) = θ(1 – x)^(θ – 1), 0 < x < 1, θ > 0.
The likelihood is L(θ) = θⁿ ∏ᵢ₌₁ⁿ (1 – xᵢ)^(θ – 1).
Hence log L = n log θ + (θ – 1)Σ log(1 – xᵢ).
The score equation is n/θ + Σ log(1 – xᵢ) = 0.
Therefore the maximum likelihood estimator, and hence a consistent estimator, is θ̂ = –n / Σᵢ₌₁ⁿ log(1 – Xᵢ).
To find its variance, define Yᵢ = –log(1 – Xᵢ).
Since the CDF of X is F_X(x) = 1 – (1 – x)^θ, we get P(Yᵢ > y) = P(Xᵢ > 1 – e^(–y)) = e^(–θy).
Thus Yᵢ ~ Exponential(rate θ). Therefore T = Σ Yᵢ ~ Gamma(shape n, rate θ).
So θ̂ = n/T. For T ~ Gamma(n, rate θ), E(1/T) = θ/(n – 1), E(1/T²) = θ²/[(n – 1)(n – 2)], n > 2.
Hence Var(1/T) = E(1/T²) – [E(1/T)]² = θ²/[(n – 1)(n – 2)] – θ²/(n – 1)² = θ²/[(n – 1)²(n – 2)].
Therefore Var(θ̂) = n²θ²/[(n – 1)²(n – 2)], n > 2.
Also, by the weak law of large numbers, T/n → 1/θ in probability, so θ̂ = n/T → θ in probability. Hence θ̂ is consistent.
How this answer will be evaluated
What earns full marks: All parts solved with correct formulas, clear steps, and final numerical answers.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Compute the variance of the sum X + 25Y using conditional variance properties. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- X ~ Bin(25,4/5): E = 20, Var = 4; Y|X ~ Bin(X,4/5)
- Total variance: Var(Y) = pqE[X]+p²Var(X) = 144/25
- Cov(X,Y) = pVar(X) = 16/5
- Var(X+25Y) = 4+3600+160 = 3764
Loses marks
- Assuming X and Y are independent
- Incorrect variance formula for binomial distribution
Earns more
- Correct calculation of E[Y|X] and Var(Y|X)
- Correct calculation of E[X+25Y] and E[(X+25Y)^2]
Extra mark
- Alternative derivation using covariance formula
- (b)(i) Find the maximum likelihood estimate of θ for the given Laplace distribution sample. 5 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Laplace likelihood; minimize Σ|xᵢ−θ|
- Median minimizes absolute deviations; ordered sample −8,…,9
- Even n: full MLE set [4,5]; conventional midpoint 4.5
Loses marks
- Using sample mean instead of median
- Incorrect differentiation of the log-likelihood
Earns more
- Correct sorting of the data points
- Explicit handling of the absolute value in the exponent
Extra mark
- Mentioning the non-differentiability at the median
- (b)(ii) Find the Cramer-Rao lower bound and the UMVUE for θ in a Bernoulli distribution. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Bernoulli score and Fisher information Iₙ(θ) = n/[θ(1−θ)]
- CRLB θ(1−θ)/n
- UMVUE θ̂ = T/n via Lehmann–Scheffé (T complete sufficient)
- Var attains the bound
Loses marks
- Incorrect Fisher Information calculation
- Failing to prove the estimator is unbiased
Earns more
- Correct derivation of the score function
- Explicit statement of the unbiasedness of the sample mean
Extra mark
- Mentioning the Cramer-Rao regularity conditions
- (c) Find a consistent estimator for θ in a Beta(1, θ) distribution and its variance. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Beta(1,θ) likelihood; score equation; MLE θ̂ = −n/Σlog(1−Xᵢ)
- Exponential transform Yᵢ = −log(1−Xᵢ); Gamma sum
- Inverse-gamma moments; Var = n²θ²/[(n−1)²(n−2)], n>2
- Consistency via WLLN
Loses marks
- Incorrect mean formula for Beta distribution
- Failing to show the estimator is consistent
Earns more
- Correct use of E(X) = 1/(1+θ) or of the exponential transform
- Clear application of the Law of Large Numbers
Extra mark
- Alternative moment estimator θ̃ = (1−X̄)/X̄ and its consistency
What "Calculate" is asking you to do
Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.
Structure that answers it
Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units
Where marks are lost
Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.
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