Paper I — Q4
(a) Consider Poisson distribution P_θ(X = j) = e^(−θ) θ^j / j! = pⱼ, j = 0, 1, 2, .... Let fⱼ be the frequency for X = j and…
Consider Poisson distribution
P_θ(X = j) = e^(−θ) θ^j / j! = pⱼ, j = 0, 1, 2, ....
Let fⱼ be the frequency for X = j and E(fⱼ) = mⱼ = npⱼ. Discuss how you obtain minimum chi-square estimate for θ. Does minimum chi-square method necessarily yield a sufficient statistic even if it exists ? 20 marks
Let the joint probability density function of X and Y be
f(x, y) = C . exp {−(4x² + 9y² − xy)},
where C is a constant. Find E(X), V(X), E(Y), V(Y) and the correlation coefficient between X and Y. 10 marks
If X₁, X₂, ..., X₆ are independent random variables such that
P(Xᵢ = -1) = P(Xᵢ = 1) = 1/2, i = 1, 2, ..., 6,
then obtain the value of
P[Σᵢ₌₁⁶ Xᵢ = 4]. 5 marks
The following data present the time (in minutes), that a commuter had to wait to catch a bus to reach his destination :
Days 1 to 10, time in minutes: 23, 25, 12, 07, 17, 16, 13, 26, 27, 12 Days 11 to 20, time in minutes: 14, 14, 16, 19, 23, 24, 12, 18, 11, 08
Use the sign-test at 0·05 level of significance to test the claim of the bus operators that commuters do not have to wait for more than 15 minutes before the bus is made available to them.
[Given Z₍₀.₀₂₅₎ = 1·96, Z₍₀.₀₅₎ = 1·645] 15 marks
हिंदी में प्रश्न पढ़ें
प्वासों बंटन
P_θ(X = j) = e^(−θ) θ^j / j! = pⱼ, j = 0, 1, 2, ....
पर विचार कीजिए । मान लीजिए कि X = j की बारम्बारता fⱼ है तथा E(fⱼ) = mⱼ = npⱼ है । आप θ का न्यूनतम काई-वर्ग आकल कैसे प्राप्त करेंगे, इसकी विवेचना कीजिए । यदि पर्याप्त प्रतिदर्शज का अस्तित्व भी है तो क्या न्यूनतम काई-वर्ग विधि पर्याप्त प्रतिदर्शज अवश्य देगा ? (20 अंक)
मान लीजिए कि X तथा Y का संयुक्त प्रायिकता घनत्व फलन
f(x, y) = C . exp {−(4x² + 9y² − xy)},
है, जहाँ C एक अचर है । E(X), V(X), E(Y), V(Y) और X और Y के बीच सहसंबंध गुणांक को ज्ञात कीजिए । (10 अंक)
यदि स्वतंत्र यादृच्छिक चर X₁, X₂, ..., X₆ इस प्रकार हैं कि
P(Xᵢ = -1) = P(Xᵢ = 1) = 1/2, i = 1, 2, ..., 6 हैं, तब
P[Σᵢ₌₁⁶ Xᵢ = 4]
का मान प्राप्त कीजिए । (5 अंक)
निम्नलिखित आँकड़े एक यात्री को उसके गंतव्य तक पहुँचने के लिए बस को पकड़ने के लिए किए गए प्रतीक्षा समय (मिनटों में) को दर्शाते हैं :
दिन 1 से 10, समय (मिनटों में): 23, 25, 12, 07, 17, 16, 13, 26, 27, 12 दिन 11 से 20, समय (मिनटों में): 14, 14, 16, 19, 23, 24, 12, 18, 11, 08
साइन-परीक्षण का 0·05 सार्थकता स्तर पर उपयोग करते हुए बस संचालकों के द्वारा दावा कि यात्रियों को बस को पकड़ने के लिए 15 मिनट से अधिक प्रतीक्षा नहीं करनी पड़ती, का परीक्षण कीजिए। [दिया गया है Z₍₀.₀₂₅₎ = 1·96, Z₍₀.₀₅₎ = 1·645] (15 अंक)
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(a) Table with header 'Treatment' and 'Block' (sub-columns 1, 2, 3, 4). Rows: Treatment 1: 73, 74, -, 71. Treatment 2: -, 75, 67, 72. Treatment 3: 73, 75, 68, -. Treatment 4: 75, -, 72, 75.
(b) Vector X_bar_1 = (-1, -1). Vector X_bar_2 = (2, 1). Matrix S_pooled = [7 -1; -1 5]. Vector X_0 = (0, 1).
(c) Table with 4 columns and 11 rows (including header). The columns are labeled 'Days', 'Time', 'Days', 'Time'. The data rows are as follows: Row 1: 1, 23, 11, 14. Row 2: 2, 25, 12, 14. Row 3: 3, 12, 13, 16. Row 4: 4, 07, 14, 19. Row 5: 5, 17, 15, 23. Row 6: 6, 16, 16, 24. Row 7: 7, 13, 17, 12. Row 8: 8, 26, 18, 18. Row 9: 9, 27, 19, 11. Row 10: 10, 12, 20, 08.
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) Minimum chi-square estimate of θ and sufficiency
Let mⱼ = npⱼ(θ) = n e^(−θ) θ^j/j!, with Σfⱼ = Σmⱼ = n. The minimum chi-square estimate θ̃ minimises Q(θ) = Σⱼ (fⱼ − mⱼ)²/mⱼ = Σⱼ fⱼ²/mⱼ − n (cells with small mⱼ are pooled first, e.g. a last cell j ≥ r with mⱼ ≥ 5). Since n is constant, dQ/dθ = 0 gives Σⱼ (fⱼ²/mⱼ²)(∂mⱼ/∂θ) = 0. For the Poisson, ∂mⱼ/∂θ = mⱼ(j − θ)/θ, so the equation becomes Σⱼ fⱼ²(j − θ)/mⱼ(θ) = 0, i.e. θ = [Σ j fⱼ²/mⱼ]/[Σ fⱼ²/mⱼ]. As mⱼ itself depends on θ, this is implicit and is solved numerically (iteration or Newton–Raphson from the start x̄ = Σ j fⱼ/n), checking that Q is minimum there. θ̃ is consistent, asymptotically normal and asymptotically as efficient as the MLE (BAN).
Sufficiency: no. For a Poisson sample the sufficient statistic is T = Σxᵢ = Σ j fⱼ, and the MLE x̄ = T/n is a function of T. The equation for θ̃ involves fⱼ², so θ̃ is not a function of T alone and is not sufficient in general. Counter-example, n = 19: the table f₁ = 19 (T = 19) and the table f₀ = 12, f₂ = 6, f₄ = 1 (T = 16) both give θ̃ = 1. T is minimal sufficient, so any sufficient statistic must determine T; θ̃ does not, hence θ̃ is not sufficient. (With grouped cells, θ̃ depends only on the grouped counts, which already lose information.) So the method does not necessarily yield a sufficient statistic even when one exists, whereas the MLE, where it exists, is always a function of the sufficient statistic.
(b)(i) Moments and correlation of the bivariate normal
The density is of bivariate normal form with no linear terms in the exponent, so E(X) = E(Y) = 0. Write 4x² + 9y² − xy = ½(8x² − 2xy + 18y²) = ½ zᵀΛz, z = (x, y)ᵀ, Λ = [[8, −1], [−1, 18]]. det Λ = 8 × 18 − 1 = 143, so the covariance matrix is Λ⁻¹ = (1/143)[[18, 1], [1, 8]]. V(X) = 18/143, V(Y) = 8/143, Cov(X, Y) = 1/143. ρ = (1/143)/√((18/143)(8/143)) = 1/√144 = 1/12. (C = √143/(2π) ≈ 1.903.) Matching with the standard form, 1/(2(1 − ρ²)σ₁²) = 4, 1/(2(1 − ρ²)σ₂²) = 9 and ρ/((1 − ρ²)σ₁σ₂) = 1, gives the same ρ = 1/12.
(b)(ii) P(ΣXᵢ = 4)
Let N be the number of Xᵢ equal to +1. Then N ~ Binomial(6, 1/2) and ΣXᵢ = N − (6 − N) = 2N − 6. ΣXᵢ = 4 ⇔ N = 5. Hence P = C(6,5)(1/2)⁶ = 6/64 = 3/32 = 0.09375.
(c) Sign test on the waiting times
H₀: median waiting time ≤ 15 minutes (the operators' claim) against H₁: median > 15 (one-sided). Compare each time with 15. Waits above 15 (plus signs): days 1, 2, 5, 6, 8, 9, 13, 14, 15, 16, 18, so 11. Waits below 15 (minus signs): days 3, 4, 7, 10, 11, 12, 17, 19, 20, so 9. No observation equals 15, so n = 20. Under H₀ (boundary median 15), S⁺ ~ Binomial(20, 1/2), with mean n/2 = 10 and SD √(n/4) = √5 = 2.236. Z = (S⁺ − n/2)/√(n/4) = (11 − 10)/2.236 = 0.447 (0.224 with continuity correction). For a one-sided 5% test the critical value is Z₀.₀₅ = 1.645 (1.96 = Z₀.₀₂₅ is for a two-sided test). Since 0.447 < 1.645, H₀ is not rejected. Exact check: P(S⁺ ≥ 11) = 0.412 > 0.05 (H₀ would be rejected only for S⁺ ≥ 15, as P(S⁺ ≥ 15) = 0.021). At the 5% level the sample gives no significant evidence that the wait exceeds 15 minutes, so the operators' claim is not rejected. The two-sided cut-off 1.96 leads to the same decision.
How this answer will be evaluated
What earns full marks: Rigorous derivations, correct calculations, and clear interpretations for all parts.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Discuss how the minimum chi-square estimate of the Poisson parameter θ is obtained, and whether the method necessarily yields a sufficient statistic. 20 marks
discuss— intro → 3-4 dimensions → example → balanced close
Must cover
- Minimises Q(θ) = Σ(fⱼ − mⱼ)²/mⱼ = Σ fⱼ²/mⱼ − n and obtains Σ fⱼ²(j − θ)/mⱼ(θ) = 0 using ∂mⱼ/∂θ = mⱼ(j − θ)/θ
- States that the equation is implicit in θ and solved numerically (start from x̄), with small cells pooled
- Answers that the method does not necessarily give a sufficient statistic: T = Σ j fⱼ is sufficient, θ̃ is not a function of T, supported by an example
Loses marks
- Claiming that θ̃ equals x̄ or is always sufficient
- Writing the estimating equation without deriving it from Q(θ)
- Answering the sufficiency question without any argument or example
Earns more
- Uses the given notation mⱼ = npⱼ
- States that θ̃ is BAN, asymptotically equivalent to the MLE x̄
- Notes that the MLE is a function of the sufficient statistic whenever it exists
Extra mark
- Mentions Neyman's modified chi-square (fⱼ in the denominator) as a variant
- (b)(i) Find E(X), V(X), E(Y), V(Y) and the correlation coefficient for the joint density f(x, y) = C exp{−(4x² + 9y² − xy)}. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Recognises the bivariate normal form with zero means (no linear terms)
- Writes the exponent as ½ zᵀΛz with Λ = [[8, −1], [−1, 18]] (or matches the standard form) and inverts to the covariance matrix (1/143)[[18, 1], [1, 8]]
- V(X) = 18/143, V(Y) = 8/143 and ρ = 1/12
Loses marks
- Treating X and Y as independent, giving V(X) = 1/8 and V(Y) = 1/18
- Wrong sign of ρ (the xy term in the exponent is +xy, so ρ > 0)
Earns more
- Gives C = √143/(2π)
- Cross-checks ρ = 1/12 by matching with the standard bivariate normal exponent
- (b)(ii) Find P(X₁ + … + X₆ = 4) for independent variables taking −1 and 1 with probability 1/2 each. 5 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Defines N as the number of +1 values, N ~ Binomial(6, 1/2), and ΣXᵢ = 2N − 6
- ΣXᵢ = 4 gives N = 5
- P = C(6,5)/2⁶ = 3/32
Loses marks
- Solving 2N − 6 = 4 incorrectly (for example N = 4)
- Using a total other than 2⁶ = 64 equally likely sequences
Earns more
- Equivalent counting: 6 favourable sequences out of 64
- (c) Use the sign test at the 5% level to test the operators' claim that commuters wait no more than 15 minutes. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Hypotheses H₀: median ≤ 15 against H₁: median > 15 (one-sided)
- Signs: 11 plus and 9 minus, no ties, so n = 20
- Z = (11 − 10)/√5 = 0.447 (0.224 with continuity correction) against Z₀.₀₅ = 1.645; H₀ not rejected, claim stands
Loses marks
- Miscounting the signs (correct: 11 plus, 9 minus, no ties)
- Wrong direction of H₁ (the claim is median ≤ 15, so H₁ is median > 15)
- No decision stated in terms of the operators' claim
Earns more
- Notes that 1.96 is the two-sided cut-off and is not needed for the one-sided test
- States the Binomial(20, 1/2) null distribution of the number of plus signs
- Exact check P(S⁺ ≥ 11) = 0.412
Extra mark
- Exact critical region S⁺ ≥ 15 (P = 0.021)
What "Discuss" is asking you to do
Lay the issue out from more than one side — how it arose, what is claimed for it, what is held against it, and where it now stands. UPSC attaches discuss to broad topics with several live dimensions, so coverage of the dimensions earns more than the strength of your opinion.
Structure that answers it
Set the issue up → the case as it is made → the case against → the dimension both sides leave out → where the balance now lies
Where marks are lost
Listing facts with no thread between them, or arguing one side throughout and calling it a discussion.
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