UPSC Statistics 2026 — Paper I

(a(i)) 'Show that' = give a complete proof of the general Bonferroni inequality P(A1 ∩ A2 ∩ ... ∩ An) ≥ Σ P(Ai) − (n − 1) for arbitrary (not necessarily independent) events. Only this single result has to be proved; no numerical work is asked. (a(ii)) Show, for mutually exclusive A and B with P(A), P(B) > 0, that P(A | A ∪ B) = P(A)/(P(A) + P(B)). A three-line conditional-probability argument is all that is required. (b) For the bivariate p.m.f. p(x, y) = (x² + 2y)/80 on x = 0, 1, 2; y = 1, 2, 3, 4, obtain (i) both marginal p.m.f.s of X and of Y, and (ii) the numerical value of P(X ≤ 1, Y > 2). Both sub-parts must be answered. (c) Using Chebyshev's inequality only, establish the bound P[0 < X < 2(λ + 1)] > λ/(λ + 1) when X has density f(x) = e^{−x} x^λ / λ!, x > 0 (λ a positive integer). The mean and variance of the distribution must be computed first. (d(i)) 'Describe how' — explain the mechanism by which the Rao-Blackwell theorem is used to construct a UMVUE: conditioning an unbiased estimator on a sufficient statistic, and the additional role of completeness (Lehmann-Scheffé) in making the result uniformly best and unique. (d(ii)) Show that ∏Xi is sufficient for θ when the sample is i.i.d. from f(x, θ) = θ x^{θ−1}, 0 < x < 1, θ > 0. The expected route is the Fisher-Neyman factorisation theorem applied to the likelihood. (e) Derive the likelihood ratio test for the one-sided hypotheses H0: θ ≤ θ0 versus H1: θ > θ0 from a random sample of size n from Poisson(θ): construct λ(x), simplify it in both regions, show it is monotone in the sample total, and state the resulting critical region and how its size is fixed.

• De Morgan's law (∩Ai)^c = ∪Ai^c
• Boole's inequality P(∪Ai) ≤ ΣP(Ai)
• P(A^c) = 1 − P(A)
• Final inequality P(∩Ai) ≥ ΣP(Ai) − (n − 1)
• Name: Bonferroni inequality
• Definition of conditional probability P(E|F) = P(E ∩ F)/P(F)
• A ∩ (A ∪ B) = A
• P(A ∪ B) = P(A) + P(B) for mutually exclusive events
• Final result P(A)/(P(A) + P(B))
• p_X(x) = (x² + 5)/20, x = 0, 1, 2
• p_Y(y) = (5 + 6y)/80, y = 1, 2, 3, 4
• P(X ≤ 1, Y > 2) = 3/8 = 0.375
• Each marginal sums to 1
• X and Y are not independent
• Gamma / Erlang distribution with shape λ+1, scale 1
• E(X) = λ + 1
• V(X) = λ + 1, σ = √(λ+1)
• Chebyshev's inequality P(|X − μ| < kσ) ≥ 1 − 1/k²
• k = √(λ+1) and final bound λ/(λ+1)
• Rao-Blackwell theorem
• U* = E(U | T), T sufficient
• V(U*) ≤ V(U), equality iff U is a function of T
• Completeness / Lehmann-Scheffé theorem
• UMVUE is unique (a.s.) when a complete sufficient statistic exists
• Likelihood θ^n (∏xi)^{θ−1}
• Fisher-Neyman factorisation theorem
• T = ∏_{i=1}^{n} Xi, g(T,θ) = θ^n T^{θ−1}, h(x) = 1
• Support 0 < x < 1 is free of θ
• Equivalent sufficient statistic −Σ ln Xi (one-to-one function)
• T = ΣXi ~ Poisson(nθ) is the sufficient statistic
• λ(x) = sup_{θ≤θ0} L(θ) / sup_{θ} L(θ)
• λ = 1 if x̄ ≤ θ0; λ = e^{n(x̄−θ0)}(θ0/x̄)^{T} if x̄ > θ0
• Reject H0 when ΣXi ≥ k
• P(T ≥ k | θ = θ0) ≤ α; monotone likelihood ratio / Karlin-Rubin, UMP

(a(i)) Two linked tasks: (a) obtain the m.g.f. of the symmetric triangular density f(x) = x on [0,1), 2 − x on [1,2), and (b) use that m.g.f. (not direct integration of f) to get E(X) and V(X). (a(ii)) Given the characteristic function φ_X(t) = e^{−t²/2}, recover the density of X by explicit use of the Fourier inversion formula (not by simply quoting the uniqueness theorem for the normal law). (b(i)) Three computations on f(x) = c(4x − x²), 0 ≤ x ≤ 4: find the normalising constant c, then P(1 ≤ X < 3) and P(X > 2). All three answers must be numerical. (b(ii)) This is Q.2(b)(ii). For f(x) = (3x/4)(2 − x), 0 < x < 2, write the complete cumulative distribution function F(x) (including the values outside the support) and evaluate P(1/2 < X < 3/2). (c(i)) Find the distribution of Y = −log_e(1 − X) when X ~ U(0, 1). Name the resulting distribution and give its density — this is the standard inverse-transform result. (c(ii)) 'Check whether' — test the given sequence of independent variables against the weak law of large numbers: compute the means and variances, apply Chebyshev's sufficient condition n^{−2} Σ V(X_i) → 0, and give a definite verdict. (c(iii)) Apply the central limit theorem to the sample mean of 64 items (population mean 100 g, variance 16 g²) and compute P(99 < X̄ < 101) using the standard normal table printed in the paper.

• M(t) = (e^{t} − 1)²/t², t ≠ 0; M(0) = 1
• Symmetric triangular distribution on (0, 2) = sum of two independent U(0,1)
• E(X) = 1
• E(X²) = 7/6
• V(X) = 1/6 ≈ 0.167
• Inversion formula f(x) = (1/2π)∫e^{−itx}φ(t)dt
• Completing the square −t²/2 − itx = −(1/2)(t+ix)² − x²/2
• Gaussian integral value √(2π)
• f(x) = (1/√(2π))e^{−x²/2}
• X ~ N(0, 1); uniqueness theorem for characteristic functions
• c = 3/32
• Antiderivative 2x² − x³/3
• P(1 ≤ X < 3) = 11/16 = 0.6875
• P(X > 2) = 1/2 = 0.5
• Symmetry of f(x) about x = 2 (mean repair time 2 hours)
• F(x) = (x²/4)(3 − x) for 0 ≤ x ≤ 2
• F(x) = 0 for x ≤ 0, F(x) = 1 for x ≥ 2
• F(3/2) = 27/32, F(1/2) = 5/32
• P(1/2 < X < 3/2) = 11/16 = 0.6875
• f is symmetric about x = 1 (a Beta(2,2) scaled to (0,2))
• F_Y(y) = 1 − e^{−y}, y > 0
• f_Y(y) = e^{−y}, y > 0
• Exponential distribution with mean 1 (rate 1)
• F_X(u) = u for U(0,1)
• Inverse probability integral transform
• E(X_k) = 1
• V(X_k) = n − 1
• Chebyshev's condition n^{−2} Σ V(X_i) → 0
• (n − 1)/n → 1 ≠ 0
• Conclusion: WLLN does not hold
• Central limit theorem, X̄ ~ N(μ, σ²/n) approximately
• SE = σ/√n = 0.5 g
• z = ±2
• P(0 < Z < 2) = 0.4772 (table value)
• P(99 < X̄ < 101) ≈ 0.9544

(a) A five-part computation on the joint p.d.f. f(x,y) = (3/4)(xy + x²/2) on 0 < x < 1, 0 < y < 2: (i) both marginal densities, (ii) E(X) and V(X), (iii) E(Y) and V(Y), (iv) Cov(X,Y) and (v) the correlation coefficient. Every sub-part carries marks and none may be skipped. (b) Apply two non-parametric two-sample tests to the six battery lifetimes of each brand: (i) the Kolmogorov-Smirnov two-sample test (critical value D(6,6,0.05) = 4/6 supplied) and (ii) the median test. Both tests must be carried out with the arithmetic shown and a decision stated for each. (c) Three linked tasks: prove that a sufficient statistic always exists for the one-parameter exponential family; deduce that ΣXi is sufficient for the Poisson parameter; and state (with reason) whether that statistic is complete.

• f_X(x) = (3/2)x + (3/4)x², 0 < x < 1
• f_Y(y) = (3y + 1)/8, 0 < y < 2
• E(X) = 11/16, V(X) = 67/1280 ≈ 0.0523
• E(Y) = 5/4, V(Y) = 13/48 ≈ 0.2708
• Cov(X,Y) = −1/192 ≈ −0.00521 and ρ ≈ −0.0437
• H0: identical distributions / equal average life
• D = 3/6 = 0.5 versus D(6,6,0.05) = 4/6
• Pooled median = 45
• 2 × 2 median-test table: A (1, 5), B (4, 2)
• Both tests: do not reject H0 at the 5% level
• Chi-square critical value χ²(1, 0.05) = 3.841 (or Fisher's exact test for the small counts)
• One-parameter exponential family f(x;θ) = a(θ)b(x)exp{c(θ)d(x)}
• Fisher-Neyman factorisation theorem
• T = Σ d(Xi) is sufficient
• Poisson: c(θ) = ln θ, d(x) = x ⇒ T = ΣXi sufficient, T ~ Poisson(nθ)
• T is complete (full-rank exponential family); Lehmann-Scheffé ⇒ X̄ is the UMVUE of θ

(a) Four linked demands for an i.i.d. N(μ, σ²) sample: derive the MLE of σ², test it for unbiasedness, propose an unbiased alternative and check its consistency, and finally compare the mean squared errors of the two estimators. (b) With X1,...,Xn i.i.d. N(θ, σ²) and conjugate prior θ ~ N(μ, τ²) (σ², μ, τ² all known), derive the posterior and hence the Bayes estimator of θ under (i) squared error loss and (ii) absolute error loss. Both loss functions must be treated. (c) Three demands: describe Wald's sequential probability ratio test in general; specialise it to H0: θ = θ0 versus H1: θ = θ1 for Poisson(θ) observations; and obtain the operating characteristic (OC) function of that test.

• σ̂²_MLE = (1/n)Σ(Xi − X̄)²
• E(σ̂²_MLE) = (n − 1)σ²/n, bias = −σ²/n
• s² = Σ(Xi − X̄)²/(n − 1), unbiased
• Σ(Xi − X̄)²/σ² ~ χ²_{n−1}; V(s²) = 2σ⁴/(n − 1)
• MSE(σ̂²) = (2n − 1)σ⁴/n² < MSE(s²) = 2σ⁴/(n − 1)
• Conjugate normal-normal prior
• Posterior θ | x ~ N(μ_n, σ_n²)
• μ_n = (n τ² x̄ + σ² μ)/(n τ² + σ²) = (n x̄/σ² + μ/τ²)/(n/σ² + 1/τ²)
• σ_n² = σ²τ²/(n τ² + σ²)
• Squared error loss → posterior mean; absolute error loss → posterior median; both equal μ_n by symmetry
• Wald's SPRT; A = (1 − β)/α, B = β/(1 − α)
• λ_m = e^{−m(θ1 − θ0)}(θ1/θ0)^{ΣXi}
• Straight-line boundaries for S_m = ΣXi against m
• OC function L(θ) = (A^h − 1)/(A^h − B^h)
• θ = h(θ1 − θ0)/[(θ1/θ0)^h − 1]; L(θ0) = 1 − α, L(θ1) = β

(a) Two demands for n independent observations with common mean μ and common variance σ²: obtain the best linear unbiased estimator (BLUE) of μ, and obtain an unbiased quadratic estimator of σ². Both derivations are needed — quoting Ȳ and s² without proof is not enough. (b) Three demands: explain what the classification problem in discriminant analysis is; define Fisher's linear discriminant function for two known multivariate normal populations with a common dispersion matrix; and compute it numerically for μ⁽¹⁾ = (24, 12)', μ⁽²⁾ = (18, 10)' and Σ = [[9, −5], [−5, 8]]. (c) Within the Gauss-Markov set-up Y = Xβ + u with E(Y) = Xβ and D(Y) = σ²I, prove that E[Y'{I − X(X'X)⁻¹X'}Y] = σ²(n − p) — the result that makes the residual mean square an unbiased estimator of σ². (d) Derive — not merely quote — the variance of the sample mean ȳ under simple random sampling without replacement, ending with V(ȳ) = ((N − n)/(Nn))S² = (1 − f)S²/n. (e) For a three-stratum population (N_h = 200, 300, 500; S_h² = 25, 36, 49; C_h = 4, 9, 16) and a total sample of n = 100, compute the stratum sample sizes under (i) proportional allocation and (ii) optimum allocation. Since costs per unit are given, 'optimum' means the cost-constrained optimum n_h ∝ N_h S_h/√C_h.

• Σa_i = 1 (unbiasedness) and V(Σa_iY_i) = σ²Σa_i²
• a_i = 1/n; BLUE = Ȳ
• V(Ȳ) = σ²/n
• E[Σ(Yi − Ȳ)²] = (n − 1)σ²
• s² = Σ(Yi − Ȳ)²/(n − 1) unbiased for σ²
• Gauss-Markov setting: no normality required
• Misclassification probabilities P(2|1), P(1|2); regions R1, R2
• U = (μ⁽¹⁾ − μ⁽²⁾)'Σ⁻¹X
• |Σ| = 47, Σ⁻¹ = (1/47)[[8, 5], [5, 9]]
• U(x) = (58x1 + 48x2)/47 ≈ 1.234x1 + 1.021x2
• Cut-off = (1/2)(μ⁽¹⁾+μ⁽²⁾)'Σ⁻¹(μ⁽¹⁾−μ⁽²⁾) = 1746/47 = 37.1
• Mahalanobis distance D² = 444/47 = 9.45
• Hat matrix H = X(X'X)⁻¹X', symmetric and idempotent
• E(Y'AY) = tr(A D(Y)) + (EY)'A(EY)
• (I − H)X = 0
• tr(AB) = tr(BA); tr(H) = p; tr(I − H) = n − p
• σ̂² = RSS/(n − p) is unbiased for σ²
• S² = Σ(Yi − Ȳ)²/(N − 1); σ² = (N−1)S²/N
• Cov(y_i, y_j) = −S²/N, i ≠ j
• V(ȳ) = ((N − n)/(Nn))S² = (1 − f)S²/n
• f = n/N, finite population correction (1 − f)
• Unbiased estimator v(ȳ) = (1 − f)s²/n
• N = 1000; S_h = 5, 6, 7; √C_h = 2, 3, 4
• Proportional: n_h = nN_h/N → 20, 30, 50
• Optimum with cost: n_h ∝ N_hS_h/√C_h
• N_hS_h/√C_h = 500, 600, 875; Σ = 1975
• Optimum: unrounded 25.3, 30.4, 44.3; rounded to total 100 as 25, 31, 44 (25, 30, 45 also accepted)

(a) derive: given > assumptions > stepwise derivation > result > check | (b) calculate: given > formula > substitution > result with units > interpretation | (c) discuss: intro > 3-4 dimensions > example > balanced close Full marks: Rigorous proofs, correct computations, clear interpretation of all parts

• Estimability: some d′Y has E(d′Y) = l′β, i.e. A′d = l
• Existence: c = projection of d on V(A′) satisfies A′c = l, so E(c′Y) = l′β
• Uniqueness: c₁, c₂ ∈ V(A′) with A′c₁ = A′c₂ = l ⇒ c₁ − c₂ lies in V(A′) and is orthogonal to it ⇒ c₁ = c₂
• For any unbiased d′Y: A′(d − c) = 0 ⇒ c′(d − c) = 0 ⇒ V(d′Y) = σ²c′c + σ²(d − c)′(d − c) ≥ V(c′Y)
• Equality iff d = c, so c′Y is the unique BLUE
• T² = N(X̄ − μ₀)′S⁻¹(X̄ − μ₀) with S = Σ(Xᵢ − X̄)(Xᵢ − X̄)′/(N − 1)
• p = 1: T² = N(x̄ − μ₀)²/s² = t² with N − 1 d.f.
• X̄ = (8, 6)′, S = [4 −3; −3 9], |S| = 27
• S⁻¹ = (1/27)[9 3; 3 4], X̄ − μ₀ = (−1, 1)′
• T² = 3 × 7/27 = 7/9 ≈ 0.778
• Model y = Σ βⱼξⱼ(x) + e with Σᵢ ξⱼ(xᵢ)ξ_r(xᵢ) = 0 for j ≠ r
• Construction by Gram-Schmidt on 1, x, x², … over the design points; tabulated values for equally spaced x
• β̂ⱼ = Σyᵢξⱼ(xᵢ)/Σξⱼ²(xᵢ), V(β̂ⱼ) = σ²/Σξⱼ²(xᵢ), estimates uncorrelated
• Advantages: earlier estimates unchanged when a degree is added; SS for each degree separates in ANOVA; no multicollinearity

(a(i)) Give the usual estimator of the population mean under linear systematic sampling and derive its variance expressed in terms of the intraclass correlation coefficient ρ_w between units of the same systematic sample. (a(ii)) 'Discuss the effect' — explain how the sign and magnitude of the intraclass correlation ρ_w govern the efficiency of the systematic-sampling estimator of the mean, with the precise condition under which systematic sampling beats SRSWOR. (b(i)) Write the linear model, with ALL its assumptions, for the given two-factor RANDOM-effects data (Factor A at 2 levels, Factor B at 3 levels, 2 observations per cell) when the experimenter suspects an A × B interaction. Figure/table: the data are A1: (6,8), (10,12), (14,16) and A2: (12,14), (16,18), (20,22) across B1, B2, B3. (b(ii)) State the hypotheses and carry out the complete two-way ANOVA with interaction for the given 2 × 3 × 2 random-effects data, presenting the full ANOVA table (sources, d.f., SS, MS, F) and the conclusions. Figure/table: A1: B1 (6,8), B2 (10,12), B3 (14,16); A2: B1 (12,14), B2 (16,18), B3 (20,22). (c) Three demands for the BIBD with v = 4, b = 6, r = 3, k = 2, λ = 1, adjusted treatment totals Q1 = 12, Q2 = 8, Q3 = 4, Q4 = 0 and MSE = 2.5: (i) verify the parameter relations, (ii) estimate the treatment effects and their variances, and (iii) compute the efficiency factor relative to an RBD with the same number of replications.

• N = nk, random start, k possible samples each with probability 1/k
• ȳ_sy = (1/n)Σ_j y_{i+jk}, unbiased for Ȳ when N = nk
• V(ȳ_sy) = (1/k)Σ_i(ȳ_i − Ȳ)²
• Intraclass correlation ρ_w within a systematic sample
• V(ȳ_sy) = ((N − 1)S²/(Nn))[1 + (n − 1)ρ_w]
• −1/(n − 1) ≤ ρ_w ≤ 1
• V(ȳ_sy) = (σ²/n)[1 + (n − 1)ρ_w]
• ρ_w > 0 ⇒ less efficient; ρ_w < 0 ⇒ more efficient; ρ_w ≈ 0 ⇒ comparable to SRSWOR
• Break-even condition ρ_w < −1/(N − 1)
• Periodicity matching the sampling interval k is the worst case
• Single systematic sample gives no unbiased variance estimator (use multiple random starts)
• y_ijk = μ + a_i + b_j + (ab)_ij + e_ijk
• i = 1,2 (a = 2); j = 1,2,3 (b = 3); k = 1,2 (n = 2); N = 12
• a_i ~ N(0, σ²_A), b_j ~ N(0, σ²_B), (ab)_ij ~ N(0, σ²_AB), e_ijk ~ N(0, σ²_e)
• Mutual independence of all random components
• V(y_ijk) = σ²_A + σ²_B + σ²_AB + σ²_e
• CF = 2352; SS(Total) = 248; SS(A) = 108; SS(B) = 128; SS(AB) = 0; SS(Error) = 12
• d.f. 1, 2, 2, 6, 11; MS 108, 64, 0, 2
• Random model EMS: main effects tested against MS_AB
• F(2, 6; 0.05) = 5.14; F(1, 8; 0.05) = 5.32; F(2, 8; 0.05) = 4.46
• Interaction not significant; both main effects significant
• Variance components σ̂²_A = 18, σ̂²_B = 16, σ̂²_AB = 0, σ̂²_e = 2
• vr = bk = 12; λ(v − 1) = r(k − 1) = 3; b ≥ v (Fisher's inequality)
• τ̂_i = kQ_i/(λv) = Q_i/2
• τ̂ = (6, 4, 2, 0)
• V(τ̂_i) = k(v − 1)σ²/(λv²) = 0.9375; V(τ̂_i − τ̂_j) = 2kσ²/(λv) = 2.5
• Efficiency factor E = λv/(rk) = 2/3 = 0.667
• ΣQ_i = 0 is a required property of adjusted treatment totals

(a) Four demands: define the multivariate normal distribution; list its important properties; find the marginal distributions of W1 = (X1 − X2 + X3 − X4)/4 and W2 = (X1 + X2 − X3 − X4)/4 for mutually independent X1,...,X4 ~ N_p(μ, Σ); and obtain their joint distribution. (b) For the N = 4 population y = (10, 20, 30, 40) with size measures x = (1, 2, 3, 4) and πPS inclusion probabilities π_i = n x_i/Σx_i (n = 2): (i) estimate the population total from the sample {2, 4} by the Horvitz-Thompson estimator, and (ii) with π24 = 0.3 estimate the variance of that estimator and comment on the possibility of a negative variance estimate. (c(i)) 'Distinguish between' — bring out the difference between complete (total) confounding and partial confounding in factorial experiments: what is confounded, in which replicates, and what information is recoverable in each case. (c(ii)) 'Illustrate how' — construct a partially confounded 2⁴ factorial laid out in 2² = 4 blocks per replicate over 2 replicates, with no main effect confounded, and state explicitly which effects are confounded in each replicate.

• φ_X(t) = exp{it'μ − (1/2)t'Σt} and/or the N_p density
• AX + b ~ N_q(Aμ + b, AΣA')
• Zero covariance implies independence for jointly normal vectors
• W1 ~ N_p(0, Σ/4) and W2 ~ N_p(0, Σ/4)
• Cov(W1, W2) = 0; W1 and W2 independent
• (W1', W2')' ~ N_{2p}(0, blockdiag(Σ/4, Σ/4))
• π_i = nx_i/Σx_i = 0.2, 0.4, 0.6, 0.8; Σπ_i = n = 2
• Ŷ_HT = Σ_{i∈s} y_i/π_i = 100
• Sen-Yates-Grundy variance estimator v_YG = Σ_{i<j}[(π_iπ_j − π_ij)/π_ij](y_i/π_i − y_j/π_j)²
• π2π4 = 0.32 versus π24 = 0.30
• v_YG = 0 (alternatively the HT form gives 1666.67)
• Non-negativity condition π_iπ_j ≥ π_ij (Sen-Yates-Grundy condition)
• Confounding with blocks; smaller, more homogeneous blocks
• Complete confounding: same effect in all replicates, relative information = 0
• Partial confounding: different effects in different replicates, relative information (r − c)/r
• Partially confounded effects are estimated only from the replicates in which they are unconfounded
• Complete confounding used for high-order interactions of no interest
• 2⁴ = 16 treatments, blocks of size 4, 4 blocks per replicate, 2 replicates
• Three effects confounded per replicate: two independent plus their generalised interaction
• Replicate I: ABC, BCD and their GI = AD
• Replicate II: ABD, ACD and their GI = BC
• No main effect confounded; partially confounded effects retain 1/2 information
• Principal block of Replicate I: (1), bc, abd, acd