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Part IA Easter Term

Method of Moments

Idea

Match sample moments with population moments to estimate parameters.

Moments

  • First moment: E[X]E[X] (mean)
  • Second moment: E[X2]E[X^2]
  • kkth moment: E[Xk]E[X^k]

Sample Moments

  • Xˉ=1ni=1nXi\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i (first sample moment)
  • 1ni=1nXi2\frac{1}{n}\sum_{i=1}^n X_i^2 (second sample moment)

Method

  1. Express population moments in terms of parameters
  2. Solve by matching with sample moments

Example: Normal Distribution

XN(μ,σ2)X \sim N(\mu, \sigma^2). Parameters: μ,σ\mu, \sigma.

First moment: E[X]=μ=XˉE[X] = \mu = \bar{X}

Second moment: E[X2]=μ2+σ2=1nXi2E[X^2] = \mu^2 + \sigma^2 = \frac{1}{n}\sum X_i^2

Solution:

  • μ^=Xˉ\hat{\mu} = \bar{X}
  • σ^2=1nXi2Xˉ2=Sn2\hat{\sigma}^2 = \frac{1}{n}\sum X_i^2 - \bar{X}^2 = S_n^2

Example: Exponential

XExp(λ)X \sim \text{Exp}(\lambda). Parameter: λ\lambda.

E[X]=1λ=XˉE[X] = \frac{1}{\lambda} = \bar{X}

λ^=1Xˉ=nXi\hat{\lambda} = \frac{1}{\bar{X}} = \frac{n}{\sum X_i}

Example: Uniform

XUnif[0,θ]X \sim \text{Unif}[0, \theta]. Parameter: θ\theta.

E[X]=θ2=XˉE[X] = \frac{\theta}{2} = \bar{X}

θ^=2Xˉ\hat{\theta} = 2\bar{X}

Properties

  • Simple to compute
  • Consistent (moment estimators converge)
  • Not always efficient
  • May not work if moments don’t exist

Summary

StepAction
1Write population moments in parameters
2Equate to sample moments
3Solve for parameters