Paper I — Q2
(a) Let a random variable X have exponential distribution with mean 1/θ, θ > 0. To test H₀ : θ = 3 against H₁ : θ = 2, construct…
Let a random variable X have exponential distribution with mean 1/θ, θ > 0. To test H₀ : θ = 3 against H₁ : θ = 2, construct sequential probability ratio test. Show that probability of terminating the test at the first stage when null hypothesis is true is 1 – 8/27 ((A–B)/AB), where B and A, B < A, are stopping bounds. 20 marks
Each Sunday a fisherman visits one of three possible locations near his home : he goes to the sea with probability 1/2, to a river with probability 1/4, or to a lake with probability 1/4. If he goes to the sea there is an 80% chance that he will catch fish; corresponding figures for the river and the lake are 40% and 60% respectively.
Find the probability that, on a given Sunday, he catches fish.
If, on a particular Sunday, he comes home without catching anything, determine the most likely place that he has been to. (5+10=15 marks)
Let X₁ < X₂ < X₃ be the order statistics from uniform population having probability density function f(x; θ) = 1/θ, 0 < x < θ. Show that 4X₁ is an unbiased estimator of θ. 15 marks
हिंदी में प्रश्न पढ़ें
मान लीजिए कि एक यादृच्छिक चर X का बंटन चरघातांकी है जिसका माध्य 1/θ, θ > 0 है। H₀ : θ = 3 का H₁ : θ = 2 के विरुद्ध परीक्षण करने के लिए अनुक्रमिक प्रायिकता अनुपात परीक्षण की रचना कीजिए। यदि निराकरणीय परिकल्पना सत्य है, तो दर्शाइए कि प्रथम चरण में परीक्षण निरस्त होने की प्रायिकता 1 – 8/27 ((A–B)/AB) है, जहाँ B और A, B < A, समाप्ति सीमाएँ हैं। (20 अंक)
प्रत्येक रविवार को एक मछुआरा अपने घर के पास तीन संभावित स्थानों में से किसी एक स्थान पर जाता है : वह प्रायिकता 1/2 के साथ समुद्र को, प्रायिकता 1/4 के साथ एक नदी को, या प्रायिकता 1/4 के साथ एक सरोवर को जाता है। यदि वह समुद्र को जाता है, तो उसके मछली पकड़ने का संयोग 80% है; अनुरूपी संख्याएँ नदी और सरोवर के लिए क्रमशः 40% और 60% हैं।
दिए गए एक रविवार के दिन वह मछली पकड़े, इस बात की प्रायिकता ज्ञात कीजिए।
यदि किसी दिए गए रविवार के दिन वह बिना मछली पकड़े घर वापस आता है, तो वह जहाँ से वापस आया, उस अधिकतम संभावित स्थान का निर्धारण कीजिए। (5+10=15 अंक)
मान लीजिए कि एकसमान समष्टि जिसका प्रायिकता घनत्व फलन f(x; θ) = 1/θ, 0 < x < θ है, से X₁ < X₂ < X₃ क्रम प्रतिदर्शज लिए गए हैं। दर्शाइए कि 4X₁, θ का एक अनभिनत आकलक है। (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) SPRT for H₀: θ = 3 against H₁: θ = 2 and first-stage termination
Here f(x; θ) = θe^(−θx), x > 0. For observations x₁, …, xₙ the likelihood ratio is λₙ = Π f(xᵢ; 2)/f(xᵢ; 3) = (2/3)ⁿ e^(Σxᵢ). Wald's SPRT with bounds B < A (approximately A = (1 − β)/α, B = β/(1 − α)): continue sampling while B < λₙ < A; stop and reject H₀ if λₙ ≥ A; stop and accept H₀ if λₙ ≤ B. Equivalently, continue while n ln(3/2) + ln B < Σxᵢ < n ln(3/2) + ln A.
At the first stage λ₁ = (2/3)e^X, which exceeds 2/3 for every X > 0. Then λ₁ ≥ A ⇔ X ≥ ln(3A/2), and λ₁ ≤ B ⇔ X ≤ ln(3B/2) (possible only if B > 2/3). Under H₀, X has density 3e^(−3x), so P(X ≥ c) = e^(−3c). The test continues when ln(3B/2) < X < ln(3A/2): P(continue) = e^(−3 ln(3B/2)) − e^(−3 ln(3A/2)) = (2/(3B))³ − (2/(3A))³ = (8/27)(1/B³ − 1/A³). Hence P(terminate at stage 1 | H₀) = 1 − (8/27)(1/B³ − 1/A³) = 1 − (8/27)(A³ − B³)/(A³B³). The printed form 1 − (8/27)(A − B)/(AB) is this result with the bounds written as their cubes (A³ → A, B³ → B); the cube arises from the rate 3. For example, A = 5, B = 0.8 gives 0.4237. If B ≤ 2/3 the lower bound cannot stop at stage 1, and the probability is 8/(27A³).
(b)(i) Probability of catching fish
Let S, R, L be sea, river, lake and F = catches fish. P(S) = 1/2, P(R) = P(L) = 1/4; P(F | S) = 0.8, P(F | R) = 0.4, P(F | L) = 0.6. By the total probability theorem, P(F) = (1/2)(0.8) + (1/4)(0.4) + (1/4)(0.6) = 0.40 + 0.10 + 0.15 = 0.65 = 13/20.
(b)(ii) Most likely place after a day without fish
P(Fᶜ) = 1 − 0.65 = 0.35 = 7/20. By Bayes' theorem, P(place | Fᶜ) = P(place) P(Fᶜ | place)/P(Fᶜ): Sea: (1/2)(0.2)/0.35 = 0.10/0.35 = 2/7 ≈ 0.286 River: (1/4)(0.6)/0.35 = 0.15/0.35 = 3/7 ≈ 0.429 Lake: (1/4)(0.4)/0.35 = 0.10/0.35 = 2/7 ≈ 0.286 The three posterior probabilities sum to 1. The river has the largest, 3/7, so it is the most likely place. The sea is visited most often, but its high catch rate makes a blank day there less likely than at the river.
(c) Unbiasedness of 4X₁
X₁ is the minimum of three independent U(0, θ) observations. For 0 < x < θ, P(X₁ > x) = (1 − x/θ)³, so the density is f₁(x) = (3/θ)(1 − x/θ)². E(X₁) = ∫₀^θ x (3/θ)(1 − x/θ)² dx. Put y = x/θ, dx = θ dy: E(X₁) = 3θ ∫₀¹ y(1 − y)² dy = 3θ(1/2 − 2/3 + 1/4) = 3θ(1/12) = θ/4. Hence E(4X₁) = 4 × θ/4 = θ for every θ > 0, so 4X₁ is an unbiased estimator of θ. Check: X₁/θ ~ Beta(1, 3), whose mean is 1/(1 + 3) = 1/4.
How this answer will be evaluated
Framework: SPRT (Sequential Probability Ratio Test)
What earns full marks: Complete derivations with correct notation, clear interpretation, and proper statistical reasoning throughout.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Construct the SPRT for H₀: θ = 3 against H₁: θ = 2 for an exponential variable with mean 1/θ, and find the probability of stopping at the first stage under H₀. 20 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Likelihood ratio λₙ = (2/3)ⁿ e^(Σxᵢ) and the rule: continue while B < λₙ < A, reject H₀ at λₙ ≥ A, accept H₀ at λₙ ≤ B
- First-stage stopping events X ≥ ln(3A/2) and X ≤ ln(3B/2), using P(X ≥ c) = e^(−3c) under H₀
- Result 1 − (8/27)(1/B³ − 1/A³), reconciled with the printed form by reading A, B as the cubes of the bounds
Loses marks
- Likelihood ratio inverted, or θ taken as the mean instead of the rate
- Ignoring one of the two stopping events (λ₁ ≥ A or λ₁ ≤ B)
- Stating the printed result without deriving it from P(X ≥ c) = e^(−3c)
Earns more
- Equivalent continuation region in terms of Σxᵢ
- Notes that the lower bound can stop at stage 1 only if B > 2/3
- States Wald's approximations A = (1 − β)/α and B = β/(1 − α)
Extra mark
- Numerical illustration of the first-stage probability for chosen A and B
- (b)(i) Find the probability that the fisherman catches fish on a given Sunday. 5 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Partition of the Sunday into sea, river and lake with probabilities 1/2, 1/4, 1/4
- Total probability: (1/2)(0.8) + (1/4)(0.4) + (1/4)(0.6)
- P(F) = 0.65 = 13/20
Loses marks
- Averaging the three catch rates without the location weights
- Arithmetic slip in the weighted sum
Earns more
- Clear definition of the events S, R, L, F and the conditional probabilities
- Correct arithmetic 0.40 + 0.10 + 0.15
- (b)(ii) Find the most likely place he visited, given that he came home without catching anything. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- P(no fish) = 1 − 0.65 = 0.35 used as the denominator in Bayes' theorem
- Posterior probabilities: sea 2/7, river 3/7, lake 2/7
- Conclusion: the river is the most likely place, with posterior probability 3/7
Loses marks
- Using the catch probabilities 0.8, 0.4, 0.6 instead of the no-catch probabilities
- Choosing the sea because it has the largest prior probability or the highest catch rate
Earns more
- Uses P(no fish | place) = 0.2, 0.6, 0.4
- Checks that the posterior probabilities sum to 1
Extra mark
- Explains why the sea, with prior 1/2, is not the answer (its high catch rate)
- (c) Show that 4X₁ is an unbiased estimator of θ, where X₁ is the smallest order statistic of a sample of three from the uniform distribution on (0, θ). 15 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Distribution of the minimum: P(X₁ > x) = (1 − x/θ)³ and density (3/θ)(1 − x/θ)²
- E(X₁) = ∫₀^θ x f₁(x) dx = θ/4 with the substitution y = x/θ
- E(4X₁) = θ for all θ > 0, hence unbiased
Loses marks
- Using the density 1/θ for X₁ or treating X₁ as the sample mean
- Wrong limits or algebra in the integration
Earns more
- Beta(1, 3) shortcut: X₁/θ has mean 1/4
- States the definition of unbiasedness, E(T) = θ for all θ
Extra mark
- Notes that (4/3)X₃ is also unbiased, with smaller variance (θ²/15 against 3θ²/5)
What "Construct" is asking you to do
Build the required object — a velocity diagram, a sequential test, a control chart, a geometrical figure — step by step, so the sequence is visible on the page. The steps are marked, not only the finished thing.
Structure that answers it
Data and requirement → scale or basis chosen, stated → construction steps in order → the finished construction, labelled → quantities read off, or the result it yields
Where marks are lost
A diagram drawn without a stated scale, so nothing can be scaled off it and the quantities that follow lose their support. In statistics, writing down the procedure without fixing its defining constants — stopping bounds in terms of the two error probabilities, or the control limits — leaves it unmarkable.
Practice this exact question
Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
Evaluate my answer →More from Statistics 2022 Paper I
- Q1 (a) Let X and Y be independent random variables with exponential distribution having resp…
- Q2 (a) Let a random variable X have exponential distribution with mean 1/θ, θ > 0. To test H…
- Q3 (a) (i) How large a sample must be taken in order that the probability will be at least 0…
- Q4 (a) Consider Poisson distribution P_θ(X = j) = e^(−θ) θ^j / j! = pⱼ, j = 0, 1, 2, .... Le…
- Q5 (a) Define general linear model with usual assumptions. If y₁ = β₁ + u₁, y₂ = –β₁ + β₂ +…