Paper I — Q2
(a) Let X, Y, Z be three mutually independent standard exponential variates and W₁ = X + Y + Z, W₂ = (X + Y)/(X + Y + Z), W₃ =…
Let X, Y, Z be three mutually independent standard exponential variates and W₁ = X + Y + Z, W₂ = (X + Y)/(X + Y + Z), W₃ = X/(X + Y). Then determine the joint distribution of W₁, W₂ and W₃.
find out the marginal probability density functions of W₁, W₂ and W₃.
examine the mutual independence of W₁, W₂ and W₃, and give your comment. (10+6+4=20 marks)
Give an example to prove or disprove the following : P(lim sup Aₙ) = 0 ⇒ Σₖ₌₁^∞ P(Aₖ) < ∞, for any sequence {Aₙ, n ≥ 1} of events defined on a probability space (Ω, 𝓐, P). 15 marks
Let {Yₙ, n ≥ 1} be a sequence of random variables and Y be a degenerate random variable. Examine whether 'Yₙ converges in distribution to Y' implies 'Yₙ converges in probability to Y'. 15 marks
हिंदी में प्रश्न पढ़ें
माना X, Y, Z तीन परस्पर स्वतंत्र मानक घातीय चर हैं तथा W₁ = X + Y + Z, W₂ = (X + Y)/(X + Y + Z), W₃ = X/(X + Y). तब W₁, W₂ एवं W₃ का संयुक्त बंटन निकालिए ।
W₁, W₂ एवं W₃ के सीमांत प्रायिकता घनत्व फलन ज्ञात कीजिए ।
W₁, W₂ एवं W₃ के परस्पर स्वतंत्र होने का परीक्षण कीजिए तथा इस पर अपनी टिप्पणी दीजिए । (10+6+4=20 अंक)
निम्नलिखित को सिद्ध या अस्वीकृत करने के लिए एक उदाहरण दीजिए : P(lim sup Aₙ) = 0 ⇒ Σₖ₌₁^∞ P(Aₖ) < ∞, जहाँ {Aₙ, n ≥ 1} घटनाओं की कोई श्रृंखला है जो कि संभाव्यता अंतराल (Ω, 𝓐, P) पर परिभाषित है । (15 अंक)
माना {Yₙ, n ≥ 1} यादृच्छिक चरों की एक श्रृंखला है तथा Y एक अपभ्रष्ट यादृच्छिक चर है । परीक्षण कीजिए कि क्या 'Yₙ बंटन में Y को अभिसरित होता है', से यह निष्कर्ष निकलता है कि 'Yₙ प्रायिकता में Y को अभिसरित होता 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)(i) Joint distribution of W₁, W₂, W₃
Let S = W₁, U = W₂, V = W₃. Step 1: Invert the transformation: X = SUV, Y = SU(1 − V), Z = S(1 − U), with s > 0, 0 < u < 1, 0 < v < 1. Step 2: The Jacobian ∂(x, y, z)/∂(s, u, v) has absolute value s²u. Step 3: The joint density of X, Y, Z is e^(−(x+y+z)) = e^(−s). Step 4: The joint density of (W₁, W₂, W₃) is f(s, u, v) = s² u e^(−s), s > 0, 0 < u < 1, 0 < v < 1.
(a)(ii) Marginal densities
Step 1: f₁(s) = ∫₀¹∫₀¹ s²u e^(−s) du dv = s² e^(−s)/2, s > 0, so W₁ ~ Gamma(3, 1). Step 2: f₂(u) = u ∫₀^∞ s² e^(−s) ds = 2u, 0 < u < 1, so W₂ ~ Beta(2, 1). Step 3: f₃(v) = (∫₀^∞ s² e^(−s) ds)(∫₀¹ u du) = 2 × 1/2 = 1, 0 < v < 1, so W₃ ~ Uniform(0, 1).
(a)(iii) Mutual independence
The joint density factorises as s²u e^(−s) = [s² e^(−s)/2] × [2u] × [1] on a rectangular support, so f(s, u, v) = f₁(s) f₂(u) f₃(v). Hence W₁, W₂ and W₃ are mutually independent. Comment: the total X + Y + Z is independent of the proportions W₂ and W₃, which is a standard property of independent gamma variables.
(b) Example to disprove P(lim sup Aₙ) = 0 ⇒ Σ P(Aₖ) < ∞
The statement is false. The first Borel-Cantelli lemma gives only the converse: Σ P(Aₖ) < ∞ ⇒ P(lim sup Aₙ) = 0. Counterexample: take Ω = (0, 1) with Lebesgue measure and Aₙ = (0, 1/n). Step 1: P(Aₙ) = 1/n, so Σ P(Aₙ) = Σ 1/n = ∞. Step 2: The sets decrease, so lim sup Aₙ = ∩ Aₙ = ∩ (0, 1/n) = ∅. Step 3: P(lim sup Aₙ) = 0 even though the series diverges. Hence the implication does not hold in general. It does hold when the Aₙ are independent, by the second Borel-Cantelli lemma.
(c) Convergence in distribution to a degenerate limit
Yes, it implies convergence in probability. Let Y = c with probability one, so F_Y(x) = 0 for x < c and 1 for x ≥ c. Let Fₙ be the distribution function of Yₙ. Step 1: Fix ε > 0. The points c − ε and c + ε are continuity points of F_Y, so Fₙ(c − ε) → 0 and Fₙ(c + ε) → 1. Step 2: P(|Yₙ − c| ≥ ε) ≤ P(Yₙ ≤ c − ε) + P(Yₙ > c + ε/2) = Fₙ(c − ε) + 1 − Fₙ(c + ε/2). Step 3: Both Fₙ(c − ε) → 0 and Fₙ(c + ε/2) → 1, so the bound tends to 0. Hence Yₙ → c in probability. For a non-degenerate limit this implication fails in general.
How this answer will be evaluated
What earns full marks: Rigorous derivations, correct theorems, clear counterexamples, precise notation.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a)(i) Derive the joint density of W1, W2, W3. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Inverse map X=SUV, Y=SU(1-V), Z=S(1-U)
- Jacobian s^2 u
- Joint density s^2 u e^(-s) on s>0, 0<u<1, 0<v<1
Loses marks
- Wrong Jacobian; omitting the support.
Earns more
- Inverse transformation, Jacobian and density with support.
Extra mark
- Check that the density integrates to 1.
- (a)(ii) Calculate the marginal densities of W1, W2, W3. 6 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- W1 ~ Gamma(3,1): s^2 e^(-s)/2
- W2 ~ Beta(2,1): 2u
- W3 ~ Uniform(0,1)
Loses marks
- Marginals without integration.
Earns more
- All three marginals with integration shown.
Extra mark
- Names of the distributions.
- (a)(iii) Examine mutual independence and comment. 4 marks
examine— intro → how/why with reasoning → evidence → conclusion
Must cover
- Joint density equals product of marginals
- Rectangular support
- Comment: sum independent of ratios
Loses marks
- Asserting independence without factorisation.
Earns more
- Factorisation shown with comment.
Extra mark
- Link to independent gamma property.
- (b) Justify with a counterexample whether P(limsup An)=0 implies the series converges. 15 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- Statement is false
- Counterexample An=(0,1/n) on (0,1)
- Series diverges while limsup is empty
- First Borel-Cantelli gives only the converse
Loses marks
- Proving the statement; example with a non-diverging series.
Earns more
- A valid counterexample with both computations and reference to Borel-Cantelli.
Extra mark
- Second Borel-Cantelli for independent events.
- (c) Examine whether convergence in distribution to a degenerate limit implies convergence in probability. 15 marks
examine— intro → how/why with reasoning → evidence → conclusion
Must cover
- Yes, for a constant limit
- Continuity points c±eps of F_Y
- Bound P(|Yn-c|>=eps) by Fn terms tending to 0
- Fails in general for non-degenerate limit
Loses marks
- Claiming it always fails; no proof.
Earns more
- Correct claim with a proof using distribution functions.
Extra mark
- Counterexample for non-degenerate limit.
What "Derive" is asking you to do
Reach the stated expression from a starting relation, justifying every step. The destination is printed in the question, so only the route earns marks, and the assumptions you work under are part of that route.
Structure that answers it
Assumptions and notation defined → starting relation or governing equation → each step with its justification → the required expression → limiting case or boundary check
Where marks are lost
Writing the standard result first and fitting three lines to it, which an examiner reads at a glance. Marks also go on assumptions left unstated — lossless medium, small amplitude, errors independent with zero mean — and on symbols used before they are defined, even when the question says usual notations.
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.
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