ICAR AICE-JRF/SRF Probability Theory: practice questions

ICAR AICE-JRF/SRF - Agricultural Statistics, Computer Application, Bioinformatics (PhD admission and fellowship) · India · National Testing Agency for ICAR

Original practice questions written to the published syllabus · not past papers

10worked here, free

45written to this exam in this subject

11syllabus units

What this subject covers

Probability 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)

Sample spaces, sigma-fields and the axioms of probability
Conditional probability, independence and Bayes' theorem
Random variables, distribution functions, mass and density functions
Expectation, moments and moment-generating, probability-generating and characteristic functions
Joint, marginal and conditional distributions; conditional expectation
Functions of random variables and transformations (Jacobian method)
Inequalities: Markov, Chebyshev, Jensen and Cauchy-Schwarz
Modes of convergence and their relations; Slutsky's lemma; Borel-Cantelli lemma
Weak and strong laws of large numbers
Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method
Elements of measure theory: measurable functions and Lebesgue integral

What ICAR AICE-JRF/SRF asks from it

ICAR AICE-JRF/SRF Agricultural Statistics (Unit 1)

Axioms and Bayes' theorem, generating functions, Chebyshev's inequality, modes of convergence, WLLN and SLLN conditions, Lindeberg-Lévy and Liapounov CLTs: theorem statements, conditions and one-step applications.

Where the 45 questions fall

How our questions for ICAR AICE-JRF/SRF spread across the units of Probability Theory. Difficulty: 6 easy, 22 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
Weak and strong laws of large numbers	5
Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method	5
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
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

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Expectation, moments and moment-generating, probability-generating and characteristic functions · medium

For the standard Cauchy distribution, with density 1/(π(1 + x²)), which statement is true?

Its MGF M(t) exists near zero, but its variance is infinite
Its characteristic function does not exist, because no moments exist
Its mean is 0 by symmetry, and its variance is infinite
Its characteristic function e^(−|t|) exists, but its mean does not

Answer and working

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

2 · Expectation, moments and moment-generating, probability-generating and characteristic functions · medium

Which statement about a characteristic function φ(t) = E(e^(itX)) is NOT true for every random variable X?

φ is uniformly continuous on the real line
φ(−t) is the complex conjugate of φ(t)
|φ(t)| ≤ 1 for every real value of t
φ(t) is a real number for every value of t

Answer and working

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

3 · Functions of random variables and transformations (Jacobian method) · medium

Z is standard normal and Y = Z². Which is the density of Y for y > 0?

(1/2)(2πy)^(−1/2) e^(−y/2)
(1/2) e^(−y/2)
(2π)^(−1/2) e^(−y²/2)
(2πy)^(−1/2) e^(−y/2)

Answer and working

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

4 · Modes of convergence and their relations; Slutsky's lemma; Borel-Cantelli lemma · medium

Xₙ converges in distribution to a constant c. What can be concluded?

Xₙ converges to c in mean square as well
Xₙ converges to c in probability as well
Xₙ converges to c almost surely as well
Nothing more, since the limit is only in law

Answer and working

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

5 · Modes of convergence and their relations; Slutsky's lemma; Borel-Cantelli lemma · medium

Events Aₙ have P(Aₙ) = 1/n², with no assumption about dependence. What is P(Aₙ occurs infinitely often)?

It cannot be determined without independence
1, by the second Borel-Cantelli lemma
π²/6, the sum of the probabilities
0, by the first Borel-Cantelli lemma

Answer and working

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

6 · Weak and strong laws of large numbers · medium

For iid X₁, X₂, ..., Kolmogorov's strong law states that X̄ₙ converges almost surely to a finite constant if and only if:

Var(X₁) > 0, and then the constant is the median
the Xᵢ are bounded, and then the constant is 0
E(X₁²) < ∞, and then the constant is E(X₁)
E|X₁| < ∞, and then the constant is E(X₁)

Answer and working

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

7 · Weak and strong laws of large numbers · easy

What is the essential difference between the weak and the strong law of large numbers?

The weak law holds for small samples; the strong law holds only for samples over thirty
The weak law needs a finite variance; the strong law needs only identically distributed variables
The weak law gives convergence in probability; the strong law gives almost-sure convergence
The weak law concerns sums; the strong law concerns the normal limit of sums

Answer and working

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

8 · Central limit theorems (Lindeberg-Levy, Liapounov) and the delta method · easy

Which conditions does the Lindeberg-Lévy central limit theorem require of X₁, X₂, ...?

Independent, identically distributed, with finite positive variance
Independent, not necessarily identical, with a Liapounov moment condition
Independent and normal, with means and variances allowed to vary
Identically distributed with finite mean, any dependence allowed

Answer and working

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

9 · Elements of measure theory: measurable functions and Lebesgue integral · hard

For a non-negative measurable function f, the Lebesgue integral ∫ f dμ is defined as:

The supremum of ∫ s dμ over non-negative simple functions s ≤ f
The infimum of ∫ s dμ over non-negative simple functions s ≥ f
The limit of Riemann sums over ever finer partitions of the domain
The value of the antiderivative of f evaluated at the end-points

Answer and working

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

10 · Elements of measure theory: measurable functions and Lebesgue integral · hard

Which statement about the Cantor set C in [0, 1] is correct?

C is countable and has Lebesgue measure 2/3
C is uncountable and has Lebesgue measure 0
C is countable and has Lebesgue measure 0
C is uncountable and has Lebesgue measure 1/3

Answer and working

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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