CSIR-UGC NET - Mathematical Sciences (statistics option) · India · National Testing Agency for CSIR
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)
How our questions for CSIR-UGC NET spread across the units of Probability Theory. Difficulty: 3 easy, 21 medium, 17 hard.
| Elements of measure theory: measurable functions and Lebesgue integral | 9 | |
| Modes of convergence and their relations; Slutsky's lemma; Borel-Cantelli lemma | 8 | |
| Sample spaces, sigma-fields and the axioms of probability | 4 | |
| Expectation, moments and moment-generating, probability-generating and characteristic functions | 4 | |
| Random variables, distribution functions, mass and density functions | 3 | |
| Weak and strong laws of large numbers | 3 | |
| Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method | 3 | |
| Conditional probability, independence and Bayes' theorem | 2 | |
| Functions of random variables and transformations (Jacobian method) | 2 | |
| Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz | 2 | |
| Joint, marginal and conditional distributions; conditional expectation | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Two fair coins are tossed. A = first coin shows a head, B = second coin shows a head, C = the two coins show the same face. Which statement is correct?
C. A, B and C are pairwise independent but not mutually independent
Each event has probability 1/2 and each pairwise intersection has probability 1/4 = (1/2)(1/2), so the events are pairwise independent. But A ∩ B ∩ C is 'both heads', with probability 1/4, not 1/8, so they are not mutually independent (Bernstein's example). Pairwise independence does not imply mutual independence.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
For the standard Cauchy distribution, with density 1/(π(1 + x²)), which statement is true?
D. Its characteristic function e^(−|t|) exists, but its mean does not
∫|x|/(π(1 + x²)) dx diverges, so E(X) does not exist; symmetry does not rescue it because E|X| = ∞. E(e^(tX)) is infinite for every t ≠ 0, so there is no MGF. The CF always exists; here it is e^(−|t|), not differentiable at 0.
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
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
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
Events Aₙ have P(Aₙ) = 1/n², with no assumption about dependence. What is P(Aₙ occurs infinitely often)?
D. 0, by the first Borel-Cantelli lemma
Σ1/n² = π²/6 < ∞, so by the first Borel-Cantelli lemma P(lim sup Aₙ) = 0. That lemma needs no independence. The second lemma needs both a divergent sum and independence. A sum of probabilities is not itself a probability.
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
In measure-theoretic terms, the probability density function of a continuous random variable X is:
D. The Radon-Nikodym derivative of the law P_X with respect to Lebesgue measure
P_X is absolutely continuous with respect to Lebesgue measure λ, so the Radon-Nikodym theorem gives f = dP_X/dλ with P(X ∈ B) = ∫_B f dλ. The reverse derivative dλ/dP_X need not even exist. A pmf is the corresponding derivative with respect to counting measure.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
For a non-negative measurable function f, the Lebesgue integral ∫ f dμ is defined as:
A. The supremum of ∫ s dμ over non-negative simple functions s ≤ f
The construction runs: integral of a simple function Σaᵢμ(Aᵢ); then the supremum over simple functions below f; then f⁺ − f⁻ for general f. Riemann sums partition the domain; the Lebesgue approach in effect partitions the range.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
Which statement about the Cantor set C in [0, 1] is correct?
B. C is uncountable and has Lebesgue measure 0
The removed intervals have total length 1/3 + 2/9 + 4/27 + ... = 1, so C has measure 0; yet its points correspond to ternary expansions using only 0 and 2, an uncountable set. So 'measure zero' does not mean 'countable'.
Further reading: Rohatgi and Saleh, An Introduction to Probability and Statistics; Casella and Berger, Statistical Inference
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