Sufficiency, completeness, UMVUE, Cramér–Rao
Why this is asked: Factorisation gives sufficiency; Lehmann–Scheffé turns a complete sufficient statistic plus unbiasedness into the UMVUE. Cramér–Rao gives a bound that is often not attained.
Sufficiency
Factorisation theorem: T is sufficient ⇔ f(x|.
| Family | Sufficient statistic |
|---|---|
| , both unknown | |
| Bernoulli/Binomial/Poisson | |
| Uniform | |
| Uniform | |
| Shifted exponential , | |
| Exponential family | the natural statistic |
Minimal sufficient: use the ratio criterion — f(x|| is free ⇔ T(x) = T(y).
Completeness
T is complete if for all . Full-rank exponential families and the order statistics of Uniform are complete. **Uniform is not is sufficient but has zero expectation without being zero.
The UMVUE machine
- Rao–Blackwell: conditioning any unbiased estimator on a sufficient T improves it.
- Lehmann–Scheffé: if T is complete sufficient and g(T) is unbiased, then g(T) is the unique UMVUE.
Standard results:
| Model | Parameter | UMVUE |
|---|---|---|
| Uniform | ||
| Shifted exp | ||
| X̄ | ||
| on (0,1) | ||
| Bernoulli(p) | p | X̄ |
Cramér–Rao
nI for unbiased T of , under regularity. Equality ⇔ the model is exponential family with T the natural statistic.
Regularity fails when the support depends on uniform, shifted exponential) — there the UMVUE can beat the "bound", which is why those examples appear so often.
Key takeaways
- Factorisation ⇒ sufficient; ratio criterion ⇒ minimal.
- Complete + sufficient + unbiased = UMVUE (Lehmann–Scheffé).
- Cramér–Rao does not apply when the support moves with .
See it move
The trap here
“A sufficient statistic is complete” — false
or for Uniform
Sufficient but not complete: symmetry gives non-zero functions with zero expectation.
Next: MLE and method of moments
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Open this in the full syllabus view · Unit 4