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Actuarial Statistics · Estimators and their properties

Sufficiency and Cramér-Rao Lower Bound Explained

Updated 11 October 2026 · Fact-checked

A statistic is sufficient if it captures all the information about the parameter in the sample. Use the factorisation theorem to check: the likelihood must split into g(t, θ) × h(x). The Cramér-Rao lower bound says any unbiased estimator has variance at least 1 ÷ I(θ), where I is Fisher information. Estimators reaching it are minimum variance.

Understand Sufficiency and Cramér-Rao Lower Bound

A statistic is any function of the sample, such as the sample mean or the maximum. A statistic T is sufficient for a parameter θ if, once you know T, the rest of the sample tells you nothing more about θ. Formally, the conditional distribution of the sample given T does not depend on θ. You can throw away the raw data and keep T without losing information about θ.

You rarely check sufficiency from the definition. You use the factorisation criterion. Write the likelihood as a product of two parts. One part depends on the data only through t(x) and may contain θ. The other part depends on the data but not on θ. If you can do this, t(x) is sufficient. Watch the range of x. If the support depends on θ, as in Uniform(0, θ), the indicator function belongs in the first part.

Next question: how good can an unbiased estimator be? The Cramér-Rao lower bound (CRLB) gives a floor for the variance of any unbiased estimator. The floor comes from Fisher information I(θ), which measures how sharply the log-likelihood responds to θ. More information means a lower floor.

An unbiased estimator that reaches the floor has the smallest possible variance, so it is a minimum variance unbiased estimator (MVUE). Its efficiency is the bound divided by its variance, so it equals 1 when the bound is reached. Be careful: the bound is only valid under regularity conditions, mainly that the range of x does not depend on θ. Also, the bound is not always attainable. An estimator can be the MVUE and still sit above the bound.

Key rules to remember

Factorisation criterion
f(x | θ) = g(t(x), θ) × h(x)
T = t(X) is sufficient for θ if the likelihood factorises this way, with h(x) free of θ. Include any indicator for a θ-dependent support in g.
Log-likelihood score
U(θ) = ∂/∂θ log L(θ)
For a regular model, E[U(θ)] = 0.
Fisher information (two forms)
I_n(θ) = E[(∂/∂θ log L)²] = −E[∂²/∂θ² log L]
Both forms are equal under regularity conditions. The second form is usually quicker. Take the expectation after differentiating twice.
Information from an iid sample
I_n(θ) = n × I_1(θ)
I_1 comes from a single observation. Do not forget the factor n.
CRLB for θ
Var(θ̂) ≥ 1 ÷ I_n(θ)
Applies to unbiased estimators of θ under regularity conditions.
CRLB for a function g(θ)
Var(T) ≥ [g′(θ)]² ÷ I_n(θ)
Applies when T is unbiased for g(θ).
Efficiency
e(T) = CRLB ÷ Var(T)
Between 0 and 1 for an unbiased estimator in a regular model. It equals 1 when the bound is attained.

How to solve Sufficiency and Cramér-Rao Lower Bound questions

Use this order for any question on sufficiency or the CRLB. Write the likelihood first because every later step depends on it.

  1. 1Write the density or probability function for one observation, then the likelihood for the whole sample as a product.
  2. 2For sufficiency, group all terms with θ and split off the terms with no θ. Identify t(x) and name g and h. Check the range of x for θ.
  3. 3State the sufficient statistic clearly, for example Σxᵢ or max xᵢ. Any one-to-one function of it is also sufficient.
  4. 4For the CRLB, check the regularity condition: the range of x must not depend on θ.
  5. 5Take log L and differentiate twice with respect to θ.
  6. 6Take the negative expectation of the second derivative to get I_n(θ). Use E[X] or E[ΣX] with the true parameter. Check that I_n = n × I_1.
  7. 7Compute the bound as 1 ÷ I_n(θ), or [g′(θ)]² ÷ I_n(θ) if the target is g(θ).
  8. 8Compare with Var of the estimator. Show it is unbiased first. If the variances are equal, state that it attains the bound and is the MVUE. Efficiency = bound ÷ variance.

Quickest way: Fast CRLB for one-parameter families

When to use it: Use when the question asks for the CRLB or efficiency of a standard distribution such as Poisson, exponential, binomial or normal with one unknown parameter.

  1. Write log f(x) for one observation only. Drop constants that do not involve θ.
  2. Differentiate twice and take −E to get I_1(θ). Multiply by n to get I_n.
  3. Invert: the bound is 1 ÷ (n × I_1). For a function g(θ), multiply by [g′(θ)]².
  4. Compare with the known variance of the sample mean (or the stated estimator). Standard results: Var(X̄) = σ²/n.
  5. Write one line stating whether the bound is reached. For sufficiency, scan the likelihood for a term of the form θ^(Σx) or e^(−θΣx). The sum is then the sufficient statistic.

Common mistakes in Sufficiency and Cramér-Rao Lower Bound

  • Using I_1(θ) instead of I_n(θ) in the bound, so the answer has no n.

    You compute information from a single observation and then stop.

    Fix: After finding I_1, write I_n = n × I_1 on a new line. Check that the final bound shrinks as n grows.

  • Applying the CRLB to Uniform(0, θ) or another model whose range depends on θ.

    The calculation looks routine, so the regularity condition is not checked.

    Fix: Always check the range of x first. If it depends on θ, say the CRLB does not apply and move on.

  • Using the plain bound 1 ÷ I_n for an estimator of g(θ) such as 1/θ or σ².

    You remember the formula for θ and forget the numerator.

    Fix: Identify the parameter being estimated. Use [g′(θ)]² ÷ I_n(θ) whenever the target is not θ itself.

  • Leaving θ in h(x) during factorisation, or dropping the indicator function.

    You treat the support as unimportant.

    Fix: Put every θ term, including indicators such as I(max x ≤ θ), into g. h must be free of θ.

  • Forgetting the minus sign or expectation in −E[∂²/∂θ² log L], giving a negative or random information.

    You stop after the second derivative.

    Fix: Information is always positive. Differentiate twice, change sign, then take the expectation.

  • Treating an unbiased estimator that does not reach the CRLB as 'not the MVUE'.

    You assume the bound is always achievable.

    Fix: Say only that it is not efficient in the CRLB sense. Being above the bound does not by itself rule out being the MVUE.

Worked examples

Example 1

X₁, …, Xₙ is a random sample from a Poisson(λ) distribution. (a) Show that Σxᵢ is sufficient for λ. (b) Find the CRLB for unbiased estimators of λ. (c) Show that X̄ attains the bound.

Show the solution
  1. Likelihood: L(λ) = Π [e^(−λ) λ^(xᵢ) ÷ xᵢ!] = e^(−nλ) λ^(Σxᵢ) × [1 ÷ Π xᵢ!].
  2. (a) Take g(t, λ) = e^(−nλ) λ^t with t = Σxᵢ, and h(x) = 1 ÷ Π xᵢ!, which has no λ. By the factorisation criterion, Σxᵢ is sufficient for λ.
  3. (b) Log-likelihood: log L = −nλ + (Σxᵢ) log λ − Σ log(xᵢ!).
  4. First derivative: −n + Σxᵢ ÷ λ. Second derivative: −Σxᵢ ÷ λ².
  5. I_n(λ) = −E[−ΣXᵢ ÷ λ²] = nλ ÷ λ² = n ÷ λ.
  6. The range of x does not depend on λ, so the bound applies: CRLB = 1 ÷ I_n = λ ÷ n.
  7. (c) E[X̄] = λ, so X̄ is unbiased. Var(X̄) = λ ÷ n, which equals the CRLB.

Answer: Σxᵢ is sufficient. The CRLB is λ ÷ n. X̄ is unbiased with variance λ ÷ n, so it attains the bound and has efficiency 1. It is the MVUE of λ.

Example 2

A random sample of size n is taken from an exponential distribution with density f(x) = θe^(−θx), x > 0. (a) Find the CRLB for unbiased estimators of μ = 1/θ. (b) Show that X̄ attains it. (c) Evaluate the bound when n = 25 and θ = 0.5.

Show the solution
  1. Log of one density: log f = log θ − θx.
  2. First derivative in θ: 1/θ − x. Second derivative: −1/θ².
  3. I_1(θ) = −E[−1/θ²] = 1/θ². So I_n(θ) = n ÷ θ².
  4. The target is g(θ) = 1/θ, so g′(θ) = −1/θ² and [g′(θ)]² = 1/θ⁴.
  5. (a) CRLB = [g′(θ)]² ÷ I_n = (1/θ⁴) ÷ (n/θ²) = 1 ÷ (nθ²).
  6. (b) E[X] = 1/θ, so X̄ is unbiased for 1/θ. Var(X) = 1/θ², so Var(X̄) = 1 ÷ (nθ²). This equals the bound.
  7. (c) With n = 25 and θ = 0.5: nθ² = 25 × 0.25 = 6.25. The bound is 1 ÷ 6.25 = 0.16.

Answer: The CRLB for estimators of 1/θ is 1 ÷ (nθ²). X̄ is unbiased and has exactly this variance, so it is the MVUE. For n = 25 and θ = 0.5 the bound is 0.16.

Exam tips

  • Show the factorisation clearly: write g(t, θ) and h(x) on separate lines and say h has no θ. Examiners award marks for that statement.
  • In CRLB questions, state the regularity condition and unbiasedness before you compute. These are easy marks that many students skip.
  • For MCQs, test the answer for sense: a bound must fall as n grows and must be positive. This removes options with a missing n or a wrong sign quickly.
  • Check what the question asks you to estimate. If it is a function of θ, such as 1/θ or σ², use the g′(θ) form.
  • On the written paper, finish with a one-line conclusion: 'attains the CRLB, so it is the MVUE'. Do not stop at the algebra.

Practice questions from Estimators and their properties

Sufficiency and Cramér-Rao Lower Bound: frequently asked questions

How do I use the factorisation theorem to find a sufficient statistic?

Write the likelihood and split it into two factors. One factor depends on the data only through t(x) and may contain θ. The other depends on the data but not on θ. Then t(x) is sufficient. Include any θ-dependent range, such as an indicator, in the first factor.

How do I find the Cramér-Rao lower bound?

Write the log-likelihood, differentiate twice with respect to θ, take the negative expectation to get Fisher information I_n(θ), and invert it. For an estimator of g(θ), multiply by [g′(θ)]². The bound applies to unbiased estimators in regular models.

What is a minimum variance unbiased estimator?

It is an unbiased estimator with the smallest variance among all unbiased estimators of the same quantity. If an unbiased estimator has variance equal to the CRLB, it is certainly an MVUE. An MVUE can exist without reaching the bound.

Why does the CRLB fail for Uniform(0, θ)?

The CRLB needs regularity conditions, mainly that the range of the variable does not depend on θ. For Uniform(0, θ) the range is 0 to θ, so the conditions fail and the bound cannot be used.