India · National Testing Agency for CSIR · phd · unverified source
Computer-based, 3 hours, Parts A (general aptitude), B and C (subject); statisticians answer from the statistics unit.
Built from Laisha's knowledge because the official page could not be opened or was not found during research (web search quota exhausted); confirm against the body's current notification before relying on it. Qualifies for JRF and lectureship in mathematical sciences, including statistics departments of universities. The NTA CSIR-NET page fetched did not show the syllabus.
Official source: https://csirnet.nta.ac.in/
Original questions written to this exam's published syllabus and pattern. Not past papers.
1. The least upper bound property distinguishes R from Q because:
A. Some bounded sets of rationals have no rational supremum
{q ∈ Q : q² < 2} is bounded above in Q but its supremum √2 is not rational.
Real Analysis for Statistics · Sets, countability and the real number system
2. Which statement is true?
A. Any 4 vectors in R³ are linearly dependent
More than n vectors in Rⁿ are dependent; three vectors form a basis only if independent.
Linear Algebra & Matrix Theory for Statistics · Vector spaces, subspaces, linear independence, basis and dimension
3. The problem dy/dx = 3y^(2/3), y(0) = 0 has more than one solution because:
B. ∂f/∂y is unbounded at y = 0
Picard–Lindelöf needs a bounded ∂f/∂y (Lipschitz condition); both y = 0 and y = x³ solve the problem.
Differential Equations, Finite Differences & Numerical Analysis · Ordinary differential equations of first order: exact, linear and Bernoulli equations
4. A collection 𝓕 of subsets of Ω contains Ω. Which further pair of closure properties makes 𝓕 a σ-field?
B. Closure under complements and under countable unions
A σ-field needs Ω, complements and countable unions; countable intersections then follow by De Morgan. Closure under complements and finite unions only defines a field (algebra), which is weaker. Arbitrary uncountable unions are not required, and without complements the collection need not be a σ-field.
Probability Theory · Sample spaces, sigma-fields and the axioms of probability
5. X and Y are independent Poisson variables with means 2 and 3. What is P(X = 2 | X + Y = 4)?
C. 0.3456
Given X + Y = 4, X ~ binomial(4, 2/5): P(X = 2) = C(4, 2)(0.4)²(0.6)² = 0.3456. 0.2707 is the unconditional P(X = 2); 0.3750 uses p = 1/2, ignoring the unequal means.
Probability Distributions & Sampling Distributions · Discrete distributions: Bernoulli, binomial, Poisson, geometric, negative binomial, hypergeometric and multinomial
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