CUET-PG - Statistics · India · National Testing Agency
Original practice questions written to the published syllabus · not past papersProbability from its axioms to the limit theorems: random variables, expectation, generating functions, inequalities and modes of convergence, ending with the laws of large numbers and the central limit theorem that justify most of applied statistics.
In the exams: AICE-JRF Ag Stat Unit 1 (heavy); GATE ST 'Probability' and limit theorems (heavy); IIT JAM MS (heavy); UPSC ISS Statistics I (heavy); ISI admission (heavy); CSIR-NET Mathematical Sciences Unit 4 (present)
ICAR AIEEA-PG Group 07, CUET-PG Statistics, SSC JSO and state statistical services
Classical and conditional probability, addition and multiplication rules, Bayes' theorem, expectation and variance, Chebyshev's inequality and the normal approximation.
How our questions for CUET-PG spread across the units of Probability Theory. Difficulty: 35 easy, 42 medium, 8 hard.
| Sample spaces, sigma-fields and the axioms of probability | 10 | |
| Expectation, moments and moment-generating, probability-generating and characteristic functions | 10 | |
| Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz | 9 | |
| Conditional probability, independence and Bayes' theorem | 8 | |
| Random variables, distribution functions, mass and density functions | 8 | |
| Joint, marginal and conditional distributions; conditional expectation | 8 | |
| Functions of random variables and transformations (Jacobian method) | 8 | |
| Modes of convergence and their relations; Slutsky's lemma; Borel-Cantelli lemma | 7 | |
| Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method | 7 | |
| Weak and strong laws of large numbers | 6 | |
| Elements of measure theory: measurable functions and Lebesgue integral | 4 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A test for a seed-borne virus has sensitivity 0.95 and specificity 0.9. The virus is present in 2% of lots. A lot tests positive. What is the probability that it is infected?
A. 0.162
Bayes: P(infected | +) = (0.02 × 0.95)/(0.02 × 0.95 + 0.98 × 0.10) = 0.019/0.117 = 0.162. 0.950 is the sensitivity P(+ | infected), the commonest confusion; 0.019 is the joint probability P(infected and +), not the conditional.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
X is uniform on (0, 0.5), so its density equals 2 on that interval. Which statement is correct?
B. This is valid: a density must integrate to 1, and its values are allowed to exceed 1
A pdf must be non-negative and integrate to 1; 2 × 0.5 = 1, so it is a valid density. Density values are not probabilities: P(X = x) = 0 for a continuous variable, and probabilities come from areas.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
Which statement about a characteristic function φ(t) = E(e^(itX)) is NOT true for every random variable X?
D. φ(t) is a real number for every value of t
φ(0) = 1, |φ| ≤ 1, uniform continuity and φ(−t) = conj(φ(t)) hold for all distributions. φ is real-valued only when the distribution is symmetric about 0; an exponential variable, for example, has φ(t) = 1/(1 − it), which is complex.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
(X, Y) has joint density f(x, y) = x + y on 0 < x < 1, 0 < y < 1. What is the marginal density of X?
B. x + 1/2 for 0 < x < 1
f_X(x) = ∫₀¹ (x + y) dy = x + 1/2. It integrates to 1/2 + 1/2 = 1 as it must. A marginal cannot contain y; dropping the y-integral gives x, which integrates to only 1/2.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
Z is standard normal and Y = Z². Which is the density of Y for y > 0?
D. (2πy)^(−1/2) e^(−y/2)
F_Y(y) = P(−√y ≤ Z ≤ √y) = 2Φ(√y) − 1; differentiating gives 2φ(√y) × 1/(2√y) = (2πy)^(−1/2)e^(−y/2), the chi-square(1) density. The version with the extra 1/2 counts only one branch. (1/2)e^(−y/2) is chi-square with 2 df.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
Plant height has mean 50 cm and standard deviation 4 cm; the distribution is unknown. What does Chebyshev's inequality give as an upper bound for P(|X − 50| ≥ 10)?
D. 0.16
P(|X − μ| ≥ c) ≤ σ²/c² = 16/100 = 0.16. 0.40 uses σ/c without squaring; 0.84 is the lower bound for the complement; 0.0124 is the exact value only if X were normal.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
Xₙ converges in distribution to a constant c. What can be concluded?
B. Xₙ converges to c in probability as well
For a degenerate limit, P(|Xₙ − c| > ε) ≤ 1 − Fₙ(c + ε) + Fₙ(c − ε) → 0, because c ± ε are continuity points. Almost-sure and mean-square convergence need more (moments may not even exist).
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
For iid X₁, X₂, ..., Kolmogorov's strong law states that X̄ₙ converges almost surely to a finite constant if and only if:
D. E|X₁| < ∞, and then the constant is E(X₁)
Kolmogorov's SLLN: X̄ₙ → μ a.s. iff E|X₁| < ∞, with μ = E(X₁). A finite second moment is sufficient but not necessary; boundedness is far stronger than needed.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
What is the essential difference between the weak and the strong law of large numbers?
C. The weak law gives convergence in probability; the strong law gives almost-sure convergence
The laws differ in the mode of convergence of X̄ₙ to μ. Khintchine's WLLN needs no variance, the SLLN needs E|X| < ∞, and neither is about a particular sample size or about normality, which is the CLT.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
√n(Tₙ − θ) converges in distribution to N(0, σ²), and g′(θ) = 0 while g″(θ) ≠ 0. Which statement is correct?
A. √n(g(Tₙ) − g(θ)) → 0 in probability, and n(g(Tₙ) − g(θ)) has a scaled chi-square(1) limit
First-order delta method gives variance g′(θ)²σ² = 0, a degenerate limit. The second-order expansion g(Tₙ) − g(θ) ≈ g″(θ)(Tₙ − θ)²/2 gives n(g(Tₙ) − g(θ)) → (g″(θ)σ²/2)χ²₁.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
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