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IIT JAM Statistical Inference I: Estimation: practice questions

IIT JAM - Mathematical Statistics (MS) · India · IITs (JAM 2026 organised by IIT Bombay)

Original practice questions written to the published syllabus · not past papers
10worked here, free
84written to this exam in this subject
11syllabus units

What this subject covers

How 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)

What IIT JAM asks from it

GATE Statistics (ST) and IIT JAM Mathematical Statistics

Numerical and multiple-select items: UMVUEs of functions of parameters, Fisher information and CRLB values, MLEs including uniform and non-regular cases, invariance, consistency checks, and pivotal confidence intervals.

Where the 84 questions fall

How our questions for IIT JAM spread across the units of Statistical Inference I: Estimation. Difficulty: 32 easy, 38 medium, 14 hard.

Cramer-Rao inequality and Fisher information9
Maximum likelihood: properties, invariance and asymptotic normality9
Sufficiency, the factorisation theorem and minimal sufficiency8
Minimum variance unbiased estimation; Rao-Blackwell and Lehmann-Scheffe theorems8
Method of moments8
Minimum chi-square and other methods8
Iterative maximum likelihood: Newton-Raphson and Fisher scoring8
Point estimation: unbiasedness, consistency and efficiency7
Interval estimation: pivotal quantities and coverage7
Completeness, ancillarity and Basu's theorem6
Large-sample confidence intervals6

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 · Minimum variance unbiased estimation; Rao-Blackwell and Lehmann-Scheffe theorems · easy

Four observations from uniform (0, θ) have maximum 9.0. What is the UMVUE estimate of θ?

  1. 9.00
  2. 18.00
  3. 7.20
  4. 11.25
Answer and working

D. 11.25

(n + 1)X₍ₙ₎/n = 5 × 9/4 = 11.25, unbiased and a function of the complete sufficient X₍ₙ₎. 9.00 is the MLE, biased downward; 7.20 corrects in the wrong direction.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

2 · Minimum variance unbiased estimation; Rao-Blackwell and Lehmann-Scheffe theorems · easy

What does the Rao-Blackwell theorem guarantee for an unbiased estimator U and a sufficient statistic T?

  1. E(U | T) is the unique UMVUE, whatever the family of distributions
  2. E(U | T) is unbiased with variance ≤ Var(U) for every θ
  3. E(U | T) attains the Cramér-Rao lower bound for every θ
  4. E(U | T) is consistent, even if U itself is not consistent
Answer and working

B. E(U | T) is unbiased with variance ≤ Var(U) for every θ

Conditioning on a sufficient statistic keeps unbiasedness and cannot raise variance (law of total variance). Uniqueness and UMVUE status need completeness as well (Lehmann-Scheffé); attaining the CRLB is not guaranteed.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

3 · Cramer-Rao inequality and Fisher information · hard

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?

  1. 0.0009
  2. 0.2000
  3. 0.0271
  4. 0.0037
Answer and working

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

4 · Method of moments · medium

Five observations from uniform (0, θ) have mean 3.1 and maximum 6.5. What is the method-of-moments estimate, and what is wrong with it?

  1. 3.1, which is too small because it ignores the factor 2 in E(X) = θ/2
  2. 6.2, which is impossible because it is below the observed maximum of 6.5
  3. 7.8, which is fine but biased upward by the factor (n + 1)/n
  4. 6.5, which is fine because the moment estimate equals the maximum
Answer and working

B. 6.2, which is impossible because it is below the observed maximum of 6.5

E(X) = θ/2 gives θ̂ = 2X̄ = 6.2, yet θ must be at least 6.5. Moment estimates can fall outside the parameter space; the MLE X₍ₙ₎ = 6.5 or the UMVUE 7.8 respect it.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

5 · Maximum likelihood: properties, invariance and asymptotic normality · hard

X₁, ..., Xₙ are iid uniform on (θ − 1/2, θ + 1/2). Which statement about the MLE of θ is correct?

  1. It is unique and equals the sample maximum X₍ₙ₎
  2. It is not unique: any value in [X₍ₙ₎ − 1/2, X₍₁₎ + 1/2] maximises L
  3. It does not exist, because the likelihood is unbounded near the sample extremes
  4. It is unique and equals the sample mean X̄
Answer and working

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

6 · Minimum chi-square and other methods · medium

For multinomial data, minimising the likelihood-ratio statistic G² = 2ΣOⱼ log(Oⱼ/Eⱼ(θ)) over θ is equivalent to which estimation method?

  1. Least squares on the frequencies
  2. The method of moments
  3. Maximum likelihood
  4. Minimum Pearson chi-square
Answer and working

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

7 · Interval estimation: pivotal quantities and coverage · easy

Ten plots of a new wheat variety give mean 4.2 t/ha and s = 0.5 t/ha. With t₀.₀₂₅,₉ = 2.262, what is the 95% confidence interval for the mean yield?

  1. 4.09 to 4.31 t/ha
  2. 3.07 to 5.33 t/ha
  3. 3.89 to 4.51 t/ha
  4. 3.84 to 4.56 t/ha
Answer and working

D. 3.84 to 4.56 t/ha

x̄ ± t s/√n = 4.2 ± 2.262 × 0.5/√10 = 4.2 ± 0.358. Using z = 1.96 ignores that σ is estimated; omitting √n gives the range for single plots, not for the mean.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

8 · Large-sample confidence intervals · medium

Why is the Wilson score interval preferred to the Wald interval for a binomial proportion?

  1. It is exact, guaranteeing at least nominal coverage for every p and n
  2. It is always narrower than the Wald interval at the same confidence level and sample size
  3. It needs no normal approximation at all, unlike the Wald interval
  4. It inverts the score test, keeps coverage near nominal for small n and stays inside [0, 1]
Answer and working

D. It inverts the score test, keeps coverage near nominal for small n and stays inside [0, 1]

The Wilson interval uses the null standard error √(p(1 − p)/n) and solves for p, which corrects the Wald interval's undercoverage near 0 and 1. It is still an approximation and is not always narrower; exact (Clopper-Pearson) intervals guarantee coverage.

Further reading: Casella and Berger, Statistical Inference; Lehmann and Casella, Theory of Point Estimation; Rohatgi and Saleh, An Introduction to Probability and Statistics

9 · Iterative maximum likelihood: Newton-Raphson and Fisher scoring · easy

What is the Newton-Raphson update for maximising a log-likelihood ℓ(θ) of one parameter?

  1. θ₍ₖ₊₁₎ = θ₍ₖ₎ − ℓ(θ₍ₖ₎)/ℓ′(θ₍ₖ₎)
  2. θ₍ₖ₊₁₎ = θ₍ₖ₎ + ℓ″(θ₍ₖ₎)/ℓ′(θ₍ₖ₎)
  3. θ₍ₖ₊₁₎ = θ₍ₖ₎ − ℓ′(θ₍ₖ₎) / ℓ″(θ₍ₖ₎)
  4. θ₍ₖ₊₁₎ = θ₍ₖ₎ + ℓ′(θ₍ₖ₎)/ℓ″(θ₍ₖ₎)
Answer and working

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

10 · Iterative maximum likelihood: Newton-Raphson and Fisher scoring · hard

For a generalised linear model with canonical link, such as logistic regression for germination data, which statement is correct?

  1. Fisher scoring converges but Newton-Raphson diverges, because the observed Hessian is random
  2. Neither method applies; the EM algorithm must be used for every GLM
  3. Fisher scoring and Newton-Raphson coincide, each step being a weighted least-squares fit
  4. Newton-Raphson needs no starting values, because the likelihood is quadratic
Answer and working

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

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