CSIR-UGC NET - Mathematical Sciences (statistics option) · India · National Testing Agency for CSIR
Original practice questions written to the published syllabus · not past papersHow to estimate well and know it: sufficiency and completeness, unbiasedness and minimum variance, the Cramer-Rao bound, the Rao-Blackwell and Lehmann-Scheffe theorems, the methods of moments, maximum likelihood and minimum chi-square, and confidence intervals.
In the exams: AICE-JRF Ag Stat Unit 1 (heavy: each theorem named); GATE ST 'Estimation' (heavy); IIT JAM MS (heavy); UPSC ISS Statistics I (heavy); ISI and CSIR-NET (present)
How our questions for CSIR-UGC NET spread across the units of Statistical Inference I: Estimation. Difficulty: 3 easy, 15 medium, 18 hard.
| Completeness, ancillarity and Basu's theorem | 7 | |
| Iterative maximum likelihood: Newton-Raphson and Fisher scoring | 6 | |
| Cramer-Rao inequality and Fisher information | 4 | |
| Maximum likelihood: properties, invariance and asymptotic normality | 4 | |
| Sufficiency, the factorisation theorem and minimal sufficiency | 3 | |
| Minimum variance unbiased estimation; Rao-Blackwell and Lehmann-Scheffe theorems | 3 | |
| Minimum chi-square and other methods | 3 | |
| Point estimation: unbiasedness, consistency and efficiency | 2 | |
| Interval estimation: pivotal quantities and coverage | 2 | |
| Large-sample confidence intervals | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
X₁, ..., X₅ are iid uniform on (0, θ). Using Basu's theorem (X₍ₙ₎ is complete sufficient and X₍₁₎/X₍ₙ₎ is ancillary), what is E(X₍₁₎/X₍₅₎)?
D. 0.20
Independence gives E(X₍₁₎) = E(X₍₁₎/X₍₅₎) E(X₍₅₎), so E(X₍₁₎/X₍₅₎) = [θ/(n + 1)]/[nθ/(n + 1)] = 1/n = 0.20. 1/6 = 0.1667 is E(X₍₁₎)/θ, and 5/6 is E(X₍₅₎)/θ.
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
What does the Lehmann-Scheffé theorem state?
B. An unbiased estimator that is a function of a complete sufficient statistic is the unique UMVUE
Completeness ensures uniqueness of the unbiased function of T, and Rao-Blackwell ensures it beats every other unbiased estimator. Sufficiency without completeness is not enough, and the MLE is often biased (uniform maximum, normal σ̂²).
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
Counts are iid Poisson(λ). What is the Cramér-Rao lower bound for an unbiased estimator of ψ(λ) = e^(−λ), at λ = 2 with n = 10?
D. 0.0037
Bound = [ψ′(λ)]²/[nI(λ)] = e^(−2λ)/(n/λ) = λe^(−2λ)/n = 2e^(−4)/10 = 0.0037. 0.2000 is the bound for λ itself; omitting the square of ψ′ gives 0.0271; dividing by λ instead of multiplying gives 0.0009.
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
For a random sample from the Laplace (double exponential) density (1/2)e^(−|x − θ|), what is the MLE of θ?
D. The sample median
Maximising L means minimising Σ|xᵢ − θ|, which is achieved by the median. The mean minimises squared deviations (normal case); the midrange is relevant for uniform location.
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
X₁, ..., Xₙ are iid uniform on (θ − 1/2, θ + 1/2). Which statement about the MLE of θ is correct?
B. It is not unique: any value in [X₍ₙ₎ − 1/2, X₍₁₎ + 1/2] maximises L
L(θ) = 1 when X₍ₙ₎ − 1/2 ≤ θ ≤ X₍₁₎ + 1/2 and 0 otherwise: a flat maximum over an interval. The midrange is one natural choice within it.
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
For multinomial data, minimising the likelihood-ratio statistic G² = 2ΣOⱼ log(Oⱼ/Eⱼ(θ)) over θ is equivalent to which estimation method?
C. Maximum likelihood
G² = 2[ΣOⱼ log Oⱼ − ΣOⱼ log Eⱼ(θ)], and the first term does not involve θ, so minimising G² maximises ΣOⱼ log pⱼ(θ), the multinomial log-likelihood.
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
What is the Newton-Raphson update for maximising a log-likelihood ℓ(θ) of one parameter?
C. θ₍ₖ₊₁₎ = θ₍ₖ₎ − ℓ′(θ₍ₖ₎) / ℓ″(θ₍ₖ₎)
Expanding ℓ′(θ) ≈ ℓ′(θₖ) + (θ − θₖ)ℓ″(θₖ) = 0 gives the update with a minus sign; near a maximum ℓ″ < 0, so the step moves in the direction of the score. θ − ℓ/ℓ′ solves ℓ = 0, the wrong equation.
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
For a generalised linear model with canonical link, such as logistic regression for germination data, which statement is correct?
C. Fisher scoring and Newton-Raphson coincide, each step being a weighted least-squares fit
With the canonical link the observed and expected information are equal, so the two algorithms are the same and take the iteratively reweighted least squares form. Starting values are still needed (usually from the data themselves).
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
θ̃ is a √n-consistent estimator of θ. What can be said of the one-step estimator θ̃ + ℓ′(θ̃)/[nI(θ̃)]?
C. It is asymptotically efficient, with the same limiting distribution as the MLE
A single scoring (or Newton) step from a √n-consistent start yields an estimator with √n(θ₁ − θ) → N(0, 1/I(θ)), the MLE's limit. It is not exactly the MLE, and not unbiased in finite samples.
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
Newton-Raphson for the Cauchy location MLE, started far from the data, converges to a value that is not the global maximum. What is the reason and the standard remedy?
B. The score equation has multiple roots; restart from the median and compare log-likelihoods
The Cauchy log-likelihood is not concave and can have several local maxima. A consistent start (the median) plus a comparison of log-likelihoods at the roots finds the global maximum. The sample mean is useless for the Cauchy, and no moment estimator exists.
Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics
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