IIT JAM - Mathematical Statistics (MS) · India · IITs (JAM 2026 organised by IIT Bombay)
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)
GATE Statistics (ST) and IIT JAM Mathematical Statistics
Multi-step numericals and NAT items: conditional probability and Bayes, CDF manipulation, MGFs and PGFs, conditional expectation and total variance, Jacobian transformations, Slutsky and delta method limits, CLT approximations; MSQs on which modes of convergence imply which.
How our questions for IIT JAM spread across the units of Probability Theory. Difficulty: 20 easy, 38 medium, 19 hard.
| Functions of random variables and transformations (Jacobian method) | 10 | |
| Expectation, moments and moment-generating, probability-generating and characteristic functions | 9 | |
| Joint, marginal and conditional distributions; conditional expectation | 9 | |
| Modes of convergence and their relations; Slutsky's lemma; Borel-Cantelli lemma | 9 | |
| Random variables, distribution functions, mass and density functions | 7 | |
| Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz | 7 | |
| Weak and strong laws of large numbers | 7 | |
| Sample spaces, sigma-fields and the axioms of probability | 6 | |
| Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method | 6 | |
| Elements of measure theory: measurable functions and Lebesgue integral | 4 | |
| Conditional probability, independence and Bayes' theorem | 3 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A seed company buys 50%, 30% and 20% of its seed from growers G1, G2 and G3, whose seed fails to germinate with probabilities 0.02, 0.03 and 0.05. A seed fails. What is the probability that it came from G3?
A. 0.345
P(failure) = 0.010 + 0.009 + 0.010 = 0.029. Bayes: P(G3 | fail) = 0.2 × 0.05 / 0.029 = 0.345. 0.200 is the prior share, 0.050 is P(fail | G3) and 0.010 is the joint probability.
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
Which conditions does the Lindeberg-Lévy central limit theorem require of X₁, X₂, ...?
A. Independent, identically distributed, with finite positive variance
Lindeberg-Lévy: iid with 0 < σ² < ∞. The non-identical case with a (2 + δ)-th moment condition is Liapounov's theorem. Normal summands need no limit theorem at all, since their sum is exactly normal.
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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