Statistics 2025 Paper I 50 marks Solve

Q2

(a) Let X be a continuous random variable having probability density function f(x) = { (2/25)(x+2), -2 ≤ x ≤ 3; 0, otherwise. Find the cumulative distribution function of Y = X² and hence find probability density function of Y. 20 marks (b) The joint probability mass function of two random variables (X, Y) be P(X = x, Y = y) = { (x+1)Cᵧ · ¹⁶Cₓ · (1/6)ʸ · (5/6)ˣ⁺¹⁻ʸ · (1/2)¹⁶, y = 0, 1, 2,..., x+1; x = 0, 1, 2,..., 16; 0, otherwise. Evaluate the following: (i) E(X), Var.(X); (ii) E(Y), Var.(Y); (iii) Cov. (X, Y). 5+5+5=15 marks (c) Let the joint probability density function of (X, Y) be f(x,y) = { 2e^{-(x+y)}, 0 < x < y < ∞; 0, otherwise. Compute the following: (i) P(Y<1); (ii) P(λX<Y), λ>1; (iii) P(Y>3X | Y>2X). 5+5+5=15 marks

हिंदी में प्रश्न पढ़ें

(a) मान लीजिए X एक संतत यादृच्छिक चर है जिसका प्रायिकता घनत्व फलन f(x) = { (2/25)(x+2), -2 ≤ x ≤ 3; 0, अन्यथा है। Y = X² का संचयी बंटन फलन ज्ञात कीजिए और इस प्रकार Y का प्रायिकता घनत्व फलन प्राप्त कीजिए। 20 (b) दो यादृच्छिक चरों (X, Y) का संयुक्त प्रायिकता द्रव्यमान फलन P(X = x, Y = y) = { (x+1)Cᵧ · ¹⁶Cₓ · (1/6)ʸ · (5/6)ˣ⁺¹⁻ʸ · (1/2)¹⁶, y = 0, 1, 2,..., x+1; x = 0, 1, 2,..., 16; 0, अन्यथा है। निम्नलिखित का मान निकालिए : (i) E(X), Var.(X); (ii) E(Y), Var.(Y); (iii) Cov. (X, Y). 5+5+5=15 (c) मान लीजिए कि (X, Y) का संयुक्त प्रायिकता घनत्व फलन निम्नवत है : f(x,y) = { 2e^{-(x+y)}, 0 < x < y < ∞; 0, अन्यथा. निम्नलिखित की गणना कीजिए : (i) P(Y<1); (ii) P(λX<Y), λ>1; (iii) P(Y>3X | Y>2X). 5+5+5=15

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Approach

Solve this multi-part numerical problem by allocating approximately 40% time to part (a) given its 20 marks weightage, and 30% each to parts (b) and (c). Begin with clear identification of the distribution type for each part, show complete derivation steps for transformations in (a), recognize the compound/binomial structure in (b), and carefully handle the constrained region of integration in (c). Conclude each sub-part with boxed final answers and appropriate probability statements.

Key points expected

  • Part (a): Correctly identify support of Y=X² as [0,9] with special handling for Y∈[0,4) where two X-values map to same Y; derive piecewise CDF F_Y(y) = P(X²≤y) by splitting into X∈[-√y,√y] and account for f(x)=0 for x<-2
  • Part (a): Differentiate CDF to obtain PDF f_Y(y) with proper case distinction for y∈[0,4) and y∈[4,9], verifying total probability equals 1
  • Part (b): Recognize X~Binomial(16, 1/2) and Y|X=x~Binomial(x+1, 1/6), then apply laws of total/conditional expectation and variance including E(Y)=E[E(Y|X)] and Var(Y)=E[Var(Y|X)]+Var[E(Y|X)]
  • Part (b): Compute Cov(X,Y)=E[XY]-E[X]E[Y] using E[XY]=E[X·E(Y|X)]=E[X·(x+1)/6] with proper summation or known binomial moments
  • Part (c): Set up correct double integrals over region 0<x<y<∞ with Jacobian handling; for P(Y<1) integrate x from 0 to y then y from 0 to 1
  • Part (c): For P(λX<Y) with λ>1, identify region as 0<x<y/λ<y and evaluate; for conditional probability P(Y>3X|Y>2X) use definition P(Y>3X)/P(Y>2X) with proper region identification

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%10Correctly identifies all supports: for (a) Y∈[0,9] with critical point at 4 where mapping changes; for (b) recognizes hierarchical structure X~Bin(16,½) and Y|X~Bin(X+1,⅙); for (c) accurately sketches region 0<x<y<∞ and identifies integration limits for each probabilityIdentifies basic supports but misses critical point at y=4 in (a), or recognizes binomial in (b) but confuses parameters, or sets up integrals in (c) with minor limit errorsWrong support identification in (a) leading to incorrect CDF ranges; fails to recognize conditional structure in (b); completely wrong region description in (c) such as x<y<∞ without lower bound
Method choice20%10Uses transformation method with proper Jacobian consideration for (a); applies Adam's law/law of total expectation and variance elegantly for (b); employs appropriate change of variables or direct integration with correct order for (c) conditional probabilitiesUses correct general methods but with inefficient approaches like direct CDF differentiation without simplification in (a), or computes moments by direct summation instead of using conditional expectation properties in (b)Attempts to use CDF method incorrectly by ignoring multiplicity of inverse in (a); tries direct joint summation for all moments in (b) leading to intractable calculations; uses wrong method for conditional probability in (c) such as assuming independence
Computation accuracy20%10All integrations and differentiations executed flawlessly: piecewise CDF with correct coefficients (2/25)(√y+2)²/2 etc.; exact binomial moments E[X]=8, Var(X)=4, E[Y]=17/12, Var(Y)=185/144; exponential integrals yielding clean answers like 1-2e⁻¹+e⁻² for (c)(i)Correct method but arithmetic errors in coefficients, such as wrong normalization constant or sign errors in integration; correct binomial formulas but calculation errors in final numerical values; correct integral setup but evaluation errorsMajor computational errors: incorrect antiderivatives, failure to apply chain rule in (a), wrong factorial calculations in (b), or inability to evaluate elementary exponential integrals in (c)
Interpretation20%10Provides probabilistic interpretation of results: explains why PDF in (a) has discontinuity at y=4 due to folding of distribution; interprets (b) as Bayesian hierarchical model with clear meaning of each parameter; explains geometric meaning of regions in (c) and why conditional probability simplifies nicelyStates final answers without contextual interpretation, or provides generic statements about distributions without connecting to specific problem structureNo interpretation provided, or nonsensical probabilistic statements; fails to recognize that negative correlation in (b) or that conditional probability in (c) must be between 0 and 1
Final answer & units20%10All six final answers clearly boxed/statemented: (a) complete piecewise CDF and PDF; (b) E(X)=8, Var(X)=4, E(Y)=17/12, Var(Y)=185/144, Cov(X,Y)=17/12; (c) P(Y<1)=1-2/e+1/e², P(λX<Y)=λ/(2λ-1), P(Y>3X|Y>2X)=1/3; probabilities in [0,1], variances non-negativeMost answers present but some missing or in wrong format; correct expressions but unsimplified; or correct numerical values without showing symbolic forms where requestedMissing multiple final answers, or answers with wrong units (e.g., probability >1), or illegible presentation; no indication of which answer corresponds to which sub-part

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