- #1

sanctifier

- 58

- 0

## Homework Statement

[itex] X_{1},\; X_{2},\;\;...,\;\; X_{n} [/itex] are sampled from a normal random variable x of mean [itex] \mu =0 [/itex] and variance [itex] 4 \sigma ^2 [/itex]

Question: What is the efficient estimator of [itex] \sigma ^2 [/itex]

## Homework Equations

Nothing Special.

## The Attempt at a Solution

When [itex] \sigma [/itex] is given, the joint probability density function (p.d.f.) of [itex] X_{1},\; X_{2},\;\;...,\;\; X_{n} [/itex] is

[itex] f(X| \sigma ^2) = \frac{1}{( \sqrt[2]{2 \pi }2 \sigma )^n}exp\{- \frac{1}{2}\sum_{i=1}^{n} \frac{{X_i}^2}{4\sigma^2} \} [/itex]

Where [itex] X [/itex] denotes a vector of components [itex] X_{1},\; X_{2},\;\;...,\;\; X_{n} [/itex].

Let [itex] \lambda (X|\sigma^2) = -n*ln( \sqrt{2 \pi } 2\sigma)- \frac{1}{2}\sum_{i=1}^{n} \frac{{X_i}^2}{4\sigma^2} [/itex]

Namely, [itex] \lambda (X|v) = -n*ln( \sqrt{8 \pi v} )- \frac{1}{2}\sum_{i=1}^{n} \frac{{X_i}^2}{4v} [/itex] when [itex] v=\sigma^2 [/itex]

[itex] \lambda' (X|v) = \frac{d\lambda (X|v)}{dv} = - \frac{n}{2v} + \frac{n\overline{X} _n}{2v^2} [/itex]

Where [itex] \overline{X} _n = \sum_{i=1}^{n} X_i / n [/itex]

Hence, the efficient estimator of [itex] \sigma^2 [/itex] is

[itex] \overline{X} _n = \frac{2v^2}{n} \lambda' (X|v) + v = \frac{2\sigma^4}{n}\lambda' (X|\sigma^2) + \sigma^2 [/itex]

Is this correct?

Last edited: