You have an unbiased estimator T of the mean μ of a normal distribution, and S = X̄ is the sufficient statistic. After Rao-Blackwellization, φ = E[T | X̄]. Which claim about φ is correct?
Aφ is biased because conditioning on S changes the expected value of T
Bφ has higher variance than T because averaging over the conditioning introduces extra randomness
Cφ is unbiased with variance ≤ Var(T); if T = X₁, then φ = X̄ with variance σ²/n instead of σ²
Dφ equals T with probability one, so the theorem has no content in this case
The law of total expectation preserves unbiasedness: E[φ] = E[E[T|S]] = E[T] = μ. The variance decomposition Var(T) = Var(E[T|S]) + E[Var(T|S)] shows variance can only decrease or stay the same — the extra term E[Var(T|S)] is the noise in T irrelevant to θ. In the normal example, E[X₁ | X̄] = X̄, reducing variance from σ² to σ²/n. Option A mistakes the law of total expectation; option B reverses the variance inequality.
Question 2 Multiple Choice
What role does the sufficient statistic play in the Rao-Blackwell argument?
AIt provides a lower bound on the variance of any unbiased estimator
BIt captures all parameter information in the data, so conditioning on it removes noise irrelevant to θ without losing signal
CIt automatically guarantees the conditioned estimator will be the UMVUE without any additional assumptions
DIt replaces the unknown parameter θ with a known quantity, enabling exact variance calculations
Sufficiency means the conditional distribution of the data given S does not depend on θ — S has already extracted all the θ-relevant information. So conditioning T on S averages away the part of T's variability that is unrelated to θ (the 'bad variance'), while keeping the part that tracks θ (the 'good variance'). Option C is wrong: completeness of S is additionally required for the result to be the UMVUE (Lehmann-Scheffé theorem).
Question 3 True / False
If T is an unbiased estimator and S is a sufficient statistic, then E[T|S] has variance no greater than Var(T).
TTrue
FFalse
Answer: True
This follows directly from the variance decomposition: Var(T) = Var(E[T|S]) + E[Var(T|S)]. Since E[Var(T|S)] ≥ 0, we have Var(E[T|S]) ≤ Var(T). Equality holds when T is already a function of S — when no irrelevant noise exists to remove. This is the core guarantee of the Rao-Blackwell theorem.
Question 4 True / False
The Rao-Blackwell theorem guarantees that conditioning any unbiased estimator on a sufficient statistic usually produces the UMVUE.
TTrue
FFalse
Answer: False
Rao-Blackwellization guarantees an improvement (or no change) in variance, but the result is the UMVUE only when the sufficient statistic is also complete. The Lehmann-Scheffé theorem provides the extra step: if the sufficient statistic is complete, then any unbiased function of it is the unique UMVUE. Without completeness, the conditioned estimator may be improvable further, and multiple different unbiased functions of S with different variances could exist.
Question 5 Short Answer
Explain intuitively why conditioning an unbiased estimator T on a sufficient statistic S reduces variance.
Think about your answer, then reveal below.
Model answer: Any estimator T has two kinds of variability: fluctuations that carry information about θ (captured by S) and random noise unrelated to θ. Since S extracts all the θ-relevant information, E[T|S] averages away the irrelevant noise while preserving the signal. The variance decomposition makes this precise: Var(T) = Var(E[T|S]) + E[Var(T|S)], where E[Var(T|S)] is the 'wasted variance' not connected to θ. Conditioning eliminates this term.
The intuition is that T is doing two jobs: tracking θ and varying randomly for reasons unrelated to θ. A sufficient statistic has already done the first job optimally. Conditioning T on S extracts only what T knows about θ, discarding the rest. The result is an estimator with the same expected value but smaller variance — the same signal with less noise.