Actuarial Statistics · Estimators and their properties

Consistency of Estimators: How to Show an Estimator Is Consistent

Updated 11 October 2026 · Fact-checked

An estimator θ̂ₙ is consistent if it converges in probability to the true θ as the sample size n grows. To prove it, show that bias → 0 and variance → 0 as n → ∞. Equivalently, show MSE → 0. This is a sufficient condition, not a necessary one.

Understand Consistency of Estimators

An estimator is a rule that turns sample data into a guess for a parameter. Its value changes from sample to sample. Consistency asks a simple question: if you collect more and more data, does the estimator settle down on the true value?

Formally, θ̂ₙ is consistent for θ if, for every ε > 0, P(|θ̂ₙ − θ| > ε) → 0 as n → ∞. This is called convergence in probability. The chance of being more than any fixed distance away from θ shrinks to zero.

Proving this directly from the definition is often hard. So we use a shortcut. By Markov's inequality applied to (θ̂ₙ − θ)² (a form of Chebyshev's inequality), P(|θ̂ₙ − θ| > ε) ≤ E[(θ̂ₙ − θ)²] ÷ ε² = MSE(θ̂ₙ) ÷ ε². If MSE → 0, the probability is squeezed to zero and the estimator is consistent. Since MSE = Var(θ̂ₙ) + bias², MSE → 0 exactly when both bias → 0 and variance → 0.

This gives the standard exam test: show bias → 0 (the estimator is at least asymptotically unbiased) and variance → 0. Then conclude consistency.

Keep the difference between unbiased and consistent clear. Unbiased is about the average of θ̂ₙ at a fixed n. Consistent is about behaviour as n grows. An unbiased estimator need not be consistent, for example using only the first observation X₁ to estimate the mean (when σ² > 0). A consistent estimator need not be unbiased, for example the divisor-n sample variance.

The MSE condition is sufficient, not necessary. For example, let Yₙ be independent of the sample, with Yₙ = n with probability 1/n and Yₙ = 0 otherwise. Then X̄ₙ + Yₙ is consistent for μ, because P(Yₙ ≠ 0) = 1/n → 0. But E(Yₙ²) = n, so its MSE is σ² ÷ n + n, which does not tend to 0.

Key rules to remember

Definition of consistency: θ̂ₙ is consistent for θ if P(|θ̂ₙ − θ| > ε) → 0 as n → ∞, for every ε > 0; This is convergence in probability, written θ̂ₙ →p θ.
Mean squared error: MSE(θ̂) = E[(θ̂ − θ)²] = Var(θ̂) + [bias(θ̂)]²; Where bias(θ̂) = E(θ̂) − θ.
Sufficient condition for consistency: bias(θ̂ₙ) → 0 and Var(θ̂ₙ) → 0 as n → ∞ ⇒ θ̂ₙ is consistent; Equivalent to MSE(θ̂ₙ) → 0. Sufficient, not necessary.
Markov/Chebyshev link: P(|θ̂ₙ − θ| > ε) ≤ MSE(θ̂ₙ) ÷ ε²; Markov's inequality applied to (θ̂ₙ − θ)², a form of Chebyshev's inequality. This is why MSE → 0 forces convergence in probability.
Variance of the sample mean: Var(X̄) = σ² ÷ n; For independent observations with common variance σ². Used to show X̄ is consistent for μ.

How to solve Consistency of Estimators questions

Use this method for any question asking you to show or test whether an estimator is consistent.

1Write down the estimator θ̂ₙ clearly and state the true parameter θ. State assumptions, such as independent identically distributed observations.
2Find E(θ̂ₙ) and hence the bias, E(θ̂ₙ) − θ.
3Find Var(θ̂ₙ) as a function of n.
4Let n → ∞. Check whether bias → 0 and whether Var → 0. Show the limits explicitly.
5If both tend to zero, state that MSE → 0 and conclude the estimator is consistent, quoting the condition.
6If the variance does not tend to zero, the test fails. Say that this condition does not prove consistency, then check the definition directly if the question asks.
7Write a clear final sentence with your conclusion.

Quickest way: Bias and variance limit check

When to use it: Use when the estimator is a function of the sample mean or a simple statistic whose mean and variance you can write down quickly, especially in multiple-choice questions.

Ask whether the estimator averages over all n observations. If it uses a fixed number of observations, variance will not shrink, so it is not consistent.
Check whether the bias has a factor like 1/n or (n−1)/n. If so, bias → 0.
Check whether the variance has a factor like 1/n. If so, variance → 0.
Both limits zero means consistent. Pick the option that matches.
Beware of estimators that are unbiased for every n but have variance that does not shrink.

Common mistakes in Consistency of Estimators

Saying an unbiased estimator is automatically consistent.
Both ideas sound like 'getting the right answer', so they get mixed up.
Fix: Unbiased concerns the mean at fixed n. Always check that the variance also tends to zero.
Saying a biased estimator cannot be consistent.
Students think any bias means the estimator misses the target.
Fix: Only bias that does not vanish matters. If bias → 0 as n → ∞ and variance → 0, the estimator is consistent.
Checking only that the variance tends to zero.
Variance of the mean σ²/n is familiar, so the bias check is forgotten.
Fix: State both limits every time. For example, an estimator with variance → 0 but a constant bias is not consistent.
Treating the MSE condition as necessary as well as sufficient.
The condition is taught as the standard test, so it feels like an 'if and only if'.
Fix: Say MSE → 0 implies consistency. The reverse need not hold. If the test fails, use the definition.
Leaving bias and variance as expressions without taking limits.
Students stop after computing E and Var and assume the conclusion is obvious.
Fix: Write 'as n → ∞, bias → 0 and Var → 0' and then state the conclusion in words.

Worked examples

Example 1

X₁, …, Xₙ are independent with mean μ and variance σ², where σ² > 0. Show that the sample mean X̄ is a consistent estimator of μ. Then show that T = X₁ is unbiased for μ but not consistent.

Show the solution

For X̄: E(X̄) = μ, so bias = 0 for every n.
Var(X̄) = σ² ÷ n, which tends to 0 as n → ∞.
Since bias → 0 and variance → 0, MSE(X̄) = σ² ÷ n → 0, so X̄ is consistent for μ.
For T = X₁: E(T) = μ, so T is unbiased.
Var(T) = σ², which does not depend on n, so MSE(T) = σ² does not tend to 0. The MSE condition fails, but that alone does not prove T is not consistent, because the condition is only sufficient. So use the definition.
T = X₁ has the same distribution for every n, so P(|T − μ| > ε) = P(|X₁ − μ| > ε) does not depend on n.
Since σ² > 0, X₁ is not degenerate at μ, so P(X₁ ≠ μ) > 0. Hence there is some ε > 0 with P(|X₁ − μ| > ε) = p > 0.
For that ε, the probability is the constant p for every n. It does not tend to 0, so T is not consistent.

Answer: X̄ is consistent for μ because bias = 0 and Var = σ²/n → 0. T = X₁ is unbiased but, when σ² > 0, it is not consistent: P(|X₁ − μ| > ε) is constant in n and positive for some ε, so it does not tend to 0.

Example 2

A random sample X₁, …, Xₙ is taken from a distribution with mean μ and variance σ². The estimator S′² = (1/n) Σ(Xᵢ − X̄)² is used for σ². Given that E(S′²) = (n − 1)σ² ÷ n and that its variance tends to 0 as n → ∞, show that S′² is consistent for σ².

Show the solution

Bias = E(S′²) − σ² = (n − 1)σ²/n − σ² = −σ²/n.
As n → ∞, −σ²/n → 0, so the bias tends to 0.
The variance is given to tend to 0.
So MSE = Var + bias² → 0 + 0 = 0.
By the MSE condition, S′² is consistent for σ².

Answer: The bias is −σ²/n → 0 and the variance → 0, so MSE → 0 and S′² is consistent for σ². It is biased for every finite n, so it is consistent but not unbiased.

Exam tips

Always write both limits: bias → 0 and variance → 0. Marks are usually split between them.
In 'compare unbiased and consistent' questions, give one example of each failing: X₁ for the mean (unbiased, not consistent) and the divisor-n variance (consistent, biased).
Use MSE = Var + bias² to save time when a question gives you MSE directly.
In multiple-choice questions, look for whether the estimator uses all n observations. A fixed-size subset will not be consistent.
Quote the condition in words, such as 'MSE tends to zero, so consistent', rather than just writing symbols.

Practice questions from Estimators and their properties

Consistency of Estimators: frequently asked questions

How do I show an estimator is consistent in the exam?

Find the bias and variance of θ̂ₙ as functions of n. Show that both tend to zero as n → ∞. Then state that MSE → 0, so the estimator is consistent.

What is the difference between an unbiased and a consistent estimator?

Unbiased means E(θ̂) = θ at a given sample size. Consistent means θ̂ₙ converges in probability to θ as n grows. Neither property implies the other.

Is the MSE condition necessary for consistency?

No. It is sufficient only. If MSE → 0, the estimator is consistent. A consistent estimator can still have MSE that does not tend to zero. For example, X̄ₙ + Yₙ, where Yₙ is independent of the sample and equals n with probability 1/n and 0 otherwise, converges in probability to μ but has MSE σ²/n + n.

Can a biased estimator be consistent?

Yes. If the bias vanishes as n → ∞ and the variance also tends to zero, the estimator is consistent. The divisor-n sample variance is a standard example.