Q1 50M Compulsory prove Probability theory and distributions
(a) Let E, F and G be three pairwise independent events such that P(E∩F) = 0·1 and P(F∩G) = 0·3. Prove that P(Eᶜ∪G) ≥ 11/12. 10 marks
(b) If X and Y are non-negative independent random variables and their joint moment generating function is given by M_{X,Y}(t₁,t₂) = e^{t₁+2e^{t₂}-3}; t₁ > 0, t₂ > 0, then show that 2P(X+Y=2) = 9P(X+Y=0). 10 marks
(c) If X₁, X₂, ... be a sequence of i.i.d. U(0, 1) random variables, then find the value of Lim_{n→∞} P(∑_{i=1}^{n} X_i ≤ n/2 + √(n/144)) [use √3 = 1·74, Φ(0·29) = 0·6141, Φ(1) = 0·8413]. 10 marks
(d) Let X₁, X₂, ..., Xₙ be a random sample from normal N(μ, σ²) distribution. Obtain sufficient statistic for parameters (μ, σ²) when both the parameters are unknown. If σ² is known, what will be sufficient statistic for parameter μ ? 10 marks
(e) A random variable X has the following distribution under H₀ and H₁ :
x : 1 2 3 4 5 6
f₀(x) : 0·01 0·01 0·01 0·01 0·01 0·95
f₁(x) : 0·05 0·04 0·03 0·02 0·01 0·85
Find the best test of size 0·03 and its probability of type-II error for testing H₀ : f = f₀ versus H₁ : f = f₁. Is it unbiased test ? Why ? 10 marks
हिंदी में पढ़ें
(a) मान लीजिए E, F और G तीन युग्मतः स्वतंत्र घटनाएँ इस प्रकार हैं कि P(E∩F) = 0·1 और P(F∩G) = 0·3 है। सिद्ध कीजिए कि P(Eᶜ∪G) ≥ 11/12 है। 10
(b) यदि X और Y ऋणेतर स्वतंत्र यादृच्छिक चर हैं तथा जिनका संयुक्त आघूर्ण जनक फलन M_{X,Y}(t₁,t₂) = e^{t₁+2e^{t₂}-3}; t₁ > 0, t₂ > 0 है, तब दर्शाइए कि, 2P(X+Y=2) = 9P(X+Y=0) है। 10
(c) यदि X₁, X₂, ... स्वतंत्र और सर्वथा बंटित U(0, 1) यादृच्छिक चरों का एक अनुक्रम है, तब Lim_{n→∞} P(∑_{i=1}^{n} X_i ≤ n/2 + √(n/144)) का मान ज्ञात कीजिए। [प्रयोग कीजिए, √3 = 1·74, Φ(0·29) = 0·6141, Φ(1) = 0·8413] 10
(d) मान लीजिए कि X₁, X₂, ..., Xₙ प्रसामान्य N(μ, σ²) बंटन से लिया गया एक यादृच्छिक प्रतिदर्श है। जब दोनों प्राचल अज्ञात हैं, तो प्राचलों (μ, σ²) के लिए पर्याप्त प्रतिदर्शज निकालिए। यदि σ² ज्ञात है, तो प्राचल μ के लिए पर्याप्त प्रतिदर्शज क्या होगा ? 10
(e) H₀ तथा H₁ के अंतर्गत एक यादृच्छिक चर X का बंटन निम्नलिखित है :
x : 1 2 3 4 5 6
f₀(x) : 0·01 0·01 0·01 0·01 0·01 0·95
f₁(x) : 0·05 0·04 0·03 0·02 0·01 0·85
H₀ : f = f₀ विरुद्ध H₁ : f = f₁ का परीक्षण करने के लिए 0·03 आमाप वाला श्रेष्ठतम परीक्षण तथा उसकी द्वितीय प्रकार की त्रुटि की प्रायिकता प्राप्त कीजिए। क्या यह परीक्षण अनभिनत है ? क्यों ? 10
Answer approach & key points
This question demands rigorous mathematical proofs and derivations across five sub-parts. Begin with (a) by establishing bounds using pairwise independence and probability inequalities; for (b) extract marginal MGFs to identify distributions and compute probabilities; for (c) apply CLT with given variance 1/12; for (d) use factorization theorem for sufficient statistics; for (e) construct Neyman-Pearson test via likelihood ratios. Allocate time proportionally: ~18 min each for (a), (b), (c) and ~12 min each for (d), (e).
- For (a): Apply P(Eᶜ∪G) = 1 - P(E∩Gᶜ) and use pairwise independence to bound P(E∩G) ≤ 1/12, yielding the inequality
- For (b): Factor joint MGF to identify X~Poisson(1) and Y~Poisson(2), then compute P(X+Y=2) and P(X+Y=0) using sum of Poissons
- For (c): Apply CLT with E[X_i]=1/2, Var(X_i)=1/12, standardize to get Φ(0.29)=0.6141 as limit
- For (d): State factorization theorem, show (X̄, S²) is sufficient for (μ,σ²), and X̄ alone when σ² known
- For (e): Compute likelihood ratios f₁/f₀, order critical region by ratios, find size 0.03 test using x=4,5,6, compute β=0.88, verify unbiasedness
Q2 50M solve Random variables and distributions
(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 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
Answer approach & key points
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.
- 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
Q3 50M calculate Probability distributions and estimation theory
(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 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)
(b)(i) 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)
(b)(ii) 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)
(c) 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 पढ़ें
(a) 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)
(b)(i) 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)
(b)(ii) 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)
(c) 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)
Answer approach & key points
Calculate the required quantities systematically across all three parts. For part (a), identify distributions and apply variance decomposition for compound random variables; for (b)(i), derive the MLE using the Laplace distribution's median property; for (b)(ii), establish the Cramer-Rao bound and verify attainment; for (c), use method of moments or MLE for consistency. Allocate approximately 40% time to part (a) given its 20 marks, 30% to part (c) for 15 marks, and 30% to part (b) combining 5+10 marks. Present derivations stepwise with clear notation before substituting numerical values.
- Part (a): Correctly identify X ~ Binomial(25, 4/5) and Y|X=x ~ Binomial(x, 4/5), then apply law of total variance to find Var(X+25Y) using E[Var(Y|X)] + Var(E[Y|X]) components
- Part (b)(i): Recognize f(x,θ) as Laplace distribution with location parameter θ, hence MLE of θ equals the sample median (4.5 or between 4 and 5)
- Part (b)(ii): Derive Cramer-Rao lower bound as θ(1-θ)/n, identify sample mean as UMVUE, and prove it attains the bound by showing equality in Cauchy-Schwarz
- Part (c): For Beta(1,θ), derive method of moments estimator θ̂ = (1-X̄)/X̄ or MLE, prove consistency via weak law of large numbers, and compute asymptotic variance
- Correct application of variance formulas: Var(X) = np(1-p) for binomial, and careful handling of the 25Y scaling factor in part (a)
- Proper justification of why sample mean is UMVUE in (b)(ii) using completeness and sufficiency of T = ΣXi, or direct variance calculation
- Verification that estimator in (c) is consistent by showing plim(θ̂) = θ as n → ∞, with explicit variance expression involving θ and n
Q4 50M derive Sequential probability ratio test and non-parametric tests
(a) Let X₁, X₂, ... be a sequence of random variables from Bernoulli distribution with mean θ, 0<θ<1. Derive SPRT for testing H₀ : θ = θ₀ versus H₁ : θ = θ₁ = 1 – θ₀, 0<θ₀<1. Also obtain expressions for OC function and ASN function. (20 marks)
(b) A random sample of size n is taken from the exponential distribution with mean θ>0. Given that n₁ observations out of n observations are less than 'a'. Show that minimum Chi-square estimate and maximum likelihood estimate of θ are same. (15 marks)
(c) The life of 6 items of brand-A and 6 items of brand-B are given below:
A : 40 62 55 35 48 88
B : 50 70 65 30 45 92
Using Kolmogorov-Smirnov test, test whether the distribution of life of both the brands are same or not at 5% level of significance.
[Given that D₍₆, ₆, ₀.₀₅₎ = 2/3] (15 marks)
हिंदी में पढ़ें
(a) मान लीजिए X₁, X₂, ... माध्य θ, 0<θ<1 वाले बर्नौली-बंटन से लिए गए यादृच्छिक चरों का एक अनुक्रम है । H₀ : θ = θ₀ विरुद्ध H₁ : θ = θ₁ = 1 – θ₀, 0<θ₀<1, के परीक्षण के लिए SPRT व्युत्पन्न कीजिए । इस परीक्षण के OC फलन तथा ASN फलन के व्यंजकों को भी प्राप्त कीजिए । (20 अंक)
(b) माध्य θ>0 वाले चरघातांकी बंटन से आमाप n का एक यादृच्छिक प्रतिदर्श लिया गया है । दिया गया है कि, n प्रेक्षणों में से n₁ प्रेक्षण 'a' से छोटे हैं । दर्शाइए कि θ का न्यूनतम काई-वर्ग आकलक तथा अधिकतम संभाविता आकलक समान है । (15 अंक)
(c) ब्रांड-A की 6 वस्तुओं और ब्रांड-B की 6 वस्तुओं के जीवनकाल नीचे दिये गये हैं :
A : 40 62 55 35 48 88
B : 50 70 65 30 45 92
कोलमोगोरोव-स्मिरनोव परीक्षण का उपयोग करते हुए, 5% सार्थकता स्तर पर परीक्षण कीजिए कि दोनों ब्रांड का जीवन का बंटन समान है या नहीं ।
[दिया गया है कि D₍₆, ₆, ₀.₀₅₎ = ²/₃] (15 अंक)
Answer approach & key points
Derive the SPRT for Bernoulli in part (a) with proper likelihood ratio development, spending ~40% time on this 20-mark component; for (b) prove equivalence of MCSE and MLE through differentiation of chi-square and likelihood functions (~30%); for (c) apply K-S test with correct empirical CDF construction and comparison against critical value D₍₆,₆,₀.₀₅₎ = 2/3 (~30%). Structure: state hypotheses → show derivations/computations → conclude with statistical decisions.
- Part (a): Derive Wald's SPRT using likelihood ratio Λₙ = (θ₁/θ₀)^ΣXᵢ · ((1-θ₁)/(1-θ₀))^(n-ΣXᵢ) with continuation region A < Λₙ < B
- Part (a): Obtain OC function L(θ) ≈ (A^(h(θ))-1)/(A^(h(θ))-B^(h(θ))) and ASN Eθ(N) ≈ L(θ)lnA + (1-L(θ))lnB / Eθ(Z) where Z = ln[f₁(X)/f₀(X)]
- Part (b): Set up Pearson's chi-square with cells [0,a) and [a,∞), minimize Σ(Oᵢ-Eᵢ)²/Eᵢ to get MCSE, show same equation as MLE score function
- Part (c): Construct ordered empirical distribution functions F₆(x) and G₆(x) for both brands, compute D₆,₆ = sup|F₆(x)-G₆(x)|
- Part (c): Compare calculated D statistic with critical value 2/3, conclude whether to reject H₀ of identical distributions at 5% level