Technopedia

The technologist’s encyclopedia

Estimation

Estimation is the task of inferring the value of an unknown parameter from observed data. It is the most basic form of statistical inference: given a sampling model X1,…,Xn∼PθX_1, \ldots, X_n \sim P_\theta and observed data x1,…,xnx_1, \ldots, x_n, produce a number (point estimate) or an interval (confidence interval) for θ\theta.

Authoritative sources

Maximum likelihood estimation

Given a likelihood L(θ)=∏i=1nf(xi;θ)L(\theta) = \prod_{i=1}^n f(x_i; \theta), the MLE is

θ^MLE=arg⁡max⁡θ L(θ)=arg⁡max⁡θ ℓ(θ)\hat{\theta}_{\text{MLE}} = \arg\max_\theta \, L(\theta) = \arg\max_\theta \, \ell(\theta)

where ℓ(θ)=∑ilog⁡f(xi;θ)\ell(\theta) = \sum_i \log f(x_i; \theta) is the log-likelihood.

Details

The MLE has several key properties:

The score function is S(θ)=∂ℓ/∂θS(\theta) = \partial \ell / \partial \theta. Under regularity conditions its expectation is zero, E[S(θ)]=0E[S(\theta)] = 0, and its variance is the Fisher information: I(θ)=Var(S(θ))=−E[∂2ℓ/∂θ2]I(\theta) = \text{Var}(S(\theta)) = -E[\partial^2 \ell / \partial \theta^2]. The MLE is the value of θ\theta at which the score vanishes, S(θ^)=0S(\hat{\theta}) = 0. The Fisher information thus measures the curvature of the log-likelihood and controls the asymptotic variance of the MLE.

Example. For Xi∼N(μ,σ2)X_i \sim N(\mu, \sigma^2) with known σ2\sigma^2, the MLE is μ^=Xˉ\hat{\mu} = \bar{X} and Var(μ^)=σ2/n\text{Var}(\hat{\mu}) = \sigma^2/n.

The MLE is a special case of the more general framework of Distributions: the likelihood is the density of the data viewed as a function of the parameter.

Connection to Bayesian inference. The MLE coincides with the Bayesian MAP (maximum a posteriori) estimate when the prior is flat (uniform). See Bayesian inference for the full posterior-based approach.

θ ℓ(θ) θ̂_MLE S(θ̂) = 0

Method of moments

Equate sample moments to population moments:

Xˉ=Eθ[X],1n∑i(Xi−Xˉ)2=Varθ[X]\bar{X} = E_\theta[X], \quad \frac{1}{n}\sum_i (X_i - \bar{X})^2 = \text{Var}_\theta[X]

Solve for θ\theta.

Details

The method of moments is simpler than MLE but generally less efficient. It works by matching the first kk sample moments to the first kk theoretical moments, where kk is the number of parameters.

Example. For Xi∼Gamma(α,β)X_i \sim \text{Gamma}(\alpha, \beta) (shape–rate), set Xˉ=α/β\bar{X} = \alpha/\beta and s2=α/β2s^2 = \alpha/\beta^2 to get α^=Xˉ2/s2\hat{\alpha} = \bar{X}^2/s^2 and β^=Xˉ/s2\hat{\beta} = \bar{X}/s^2.

Cramér–Rao lower bound

For any unbiased estimator TT of g(θ)g(\theta):

Var(T)≥[g′(θ)]2n I(θ)\text{Var}(T) \geq \frac{[g'(\theta)]^2}{n \, I(\theta)}
Details

The Cramér–Rao bound gives a lower bound on the variance of any unbiased estimator. An estimator that achieves the bound is called efficient. The MLE is asymptotically efficient: its variance approaches the bound as n→∞n \to \infty.

The bound can be violated for finite samples (e.g., the sample mean for a uniform distribution), and it does not apply to biased estimators without modification.

Example. For Xi∼N(μ,σ2)X_i \sim N(\mu, \sigma^2) with known σ2\sigma^2, I(μ)=1/σ2I(\mu) = 1/\sigma^2 and the bound is σ2/n\sigma^2/n, which is achieved by Xˉ\bar{X}.

n (sample size) Var(θ̂) CRLB = σ²/n Var(θ̂_MLE) gap → 0 as n → ∞

Bias–variance tradeoff

The mean squared error of an estimator θ^\hat{\theta} decomposes as

MSE(θ^)=Bias(θ^)2+Var(θ^)\text{MSE}(\hat{\theta}) = \text{Bias}(\hat{\theta})^2 + \text{Var}(\hat{\theta})

where Bias(θ^)=E[θ^]−θ\text{Bias}(\hat{\theta}) = E[\hat{\theta}] - \theta.

Details

The decomposition shows that an estimator’s total error has two independent sources: systematic error (bias) and random fluctuation (variance). The Cramér–Rao bound constrains only the variance of unbiased estimators; a biased estimator can have lower MSE by trading some bias for reduced variance.

Example. For Xi∼N(μ,σ2)X_i \sim N(\mu, \sigma^2) with known σ2\sigma^2, the sample mean Xˉ\bar{X} is unbiased and achieves the CRLB. The estimator c Xˉc\,\bar{X} with 0<c<10 < c < 1 is biased (bias =(1−c)μ= (1-c)\mu), but its MSE is (1−c)2μ2+c2σ2/n(1-c)^2\mu^2 + c^2\sigma^2/n, which can be smaller than σ2/n\sigma^2/n for large ∣μ∣/(σ/n)|\mu|/(\sigma/\sqrt{n}).

This tradeoff is the key lens for comparing estimators beyond the Cramér–Rao bound and connects directly to the bias–variance decomposition in Bayesian inference (posterior mean vs. MAP).

Sufficient statistics

A statistic T(X)T(X) is sufficient for θ\theta if the conditional distribution of XX given T(X)T(X) does not depend on θ\theta.

Details

Fisher’s factorization theorem. T(X)T(X) is sufficient for θ\theta if and only if the likelihood factors as

f(x;θ)=g(T(x),θ) h(x)f(x; \theta) = g(T(x), \theta) \, h(x)

for some functions gg and hh (where hh does not depend on θ\theta).

Rao–Blackwell theorem. If TT is any unbiased estimator of g(θ)g(\theta) and SS is a sufficient statistic, then T∗=E[T∣S]T^* = E[T \mid S] is also unbiased and has variance no greater than that of TT. Conditioning on a sufficient statistic can only reduce variance.

Example. For Xi∼N(μ,σ2)X_i \sim N(\mu, \sigma^2) with both parameters unknown, (Xˉ,∑i(Xi−Xˉ)2)(\bar{X}, \sum_i (X_i - \bar{X})^2) is sufficient. For Xi∼Bernoulli(p)X_i \sim \text{Bernoulli}(p) the sufficient statistic is ∑iXi\sum_i X_i.

Sufficiency is the bridge between the raw data and the minimal information needed for estimation; it underlies the construction of UMVU (uniformly minimum-variance unbiased) estimators.

Confidence intervals

A 1−α1-\alpha confidence interval for θ\theta is a random interval [L(X),U(X)][L(X), U(X)] such that

Pθ(θ∈[L,U])≥1−α∀ θP_\theta\bigl(\theta \in [L, U]\bigr) \geq 1 - \alpha \quad \forall\, \theta
Details

Construction methods:

The coverage probability is a property of the procedure, not of the realized interval. A 95% CI means that 95% of intervals generated by repeated sampling contain the true parameter.

Example. For Xˉ∼N(μ,σ2/n)\bar{X} \sim N(\mu, \sigma^2/n) with known σ\sigma, the pivot is Z=(Xˉ−μ)/(σ/n)∼N(0,1)Z = (\bar{X} - \mu)/(\sigma/\sqrt{n}) \sim N(0,1) and the 95% CI is Xˉ±1.96 σ/n\bar{X} \pm 1.96\,\sigma/\sqrt{n}.

Interactive simulation

Example problem

Problem. A factory produces bolts whose diameters are N(μ,0.012)N(\mu, 0.01^2) cm. A sample of n=25n = 25 bolts has mean diameter xˉ=1.004\bar{x} = 1.004 cm. (a) Find the MLE of μ\mu. (b) Construct a 95% confidence interval for μ\mu. (c) What is the Cramér–Rao lower bound on Var(μ^)\text{Var}(\hat{\mu})?

Solution

(a) The MLE is μ^=xˉ=1.004\hat{\mu} = \bar{x} = 1.004 cm.

(b) Since σ=0.01\sigma = 0.01 is known, the 95% CI is

1.004±1.96⋅0.0125=1.004±0.003921.004 \pm 1.96 \cdot \frac{0.01}{\sqrt{25}} = 1.004 \pm 0.00392

so the interval is [1.0001,1.0079][1.0001, 1.0079] cm.

(c) The Fisher information for one observation is I(μ)=1/σ2=10000I(\mu) = 1/\sigma^2 = 10000 and for n=25n = 25 observations is nI=250000nI = 250000. The Cramér–Rao bound is

Var(μ^)≥1nI(μ)=σ2n=0.000125=4×10−6\text{Var}(\hat{\mu}) \geq \frac{1}{nI(\mu)} = \frac{\sigma^2}{n} = \frac{0.0001}{25} = 4 \times 10^{-6}

The sample mean achieves this bound, confirming it is efficient.