Question

Let X₁ ,X₂ ,…,Xₙ​ (n≥2) be a sample from N(μ,σ 2 ). Find an unbiased estimator for σ p , where p+n>1

Answer 1

An unbiased estimator for
`\(\sigma^p\)`, where
`\(p + n > 1\)`, is
`\(S_p = (1)/(n + p - 1) \sum_(i=1)^(n) (X_i - \bar{X})^2\)`, where
`\(n\)` represents the sample size,
`\(\bar{X}\)` is the sample mean, and
`\(X_i\)` denotes individual sample data points.

Step-by-step explanation:

Certainly! In the context of estimating the
`\(p\)` th power of the population standard deviation
`\(\sigma^p\)` where
`\(p + n > 1\)` for a sample
`\(X_1, X_2, ..., X_n\)` from a normal distribution
`\(N(\mu, \sigma^2)\)`:

The unbiased estimator for
`\(\sigma^p\)` is derived as follows:

1. The sample variance
`\(S^2\)` is given by the formula:

`\[S^2 = (1)/(n-1) \sum_(i=1)^(n) (X_i - \bar{X})^2\]`

where \(\bar{X}\) represents the sample mean and
`\(X_i\)` denotes individual sample data points.

2. To find an unbiased estimator for
`\(\sigma^p\)`, an adjusted formula
`\(S_p\)` is used:

`\[S_p = (1)/(n + p - 1) \sum_(i=1)^(n) (X_i - \bar{X})^2\]`

This adjusted estimator incorporates the same sum of squared deviations but is weighted by
`\((1)/(n + p - 1)\)`, adjusting for the bias when estimating higher powers of the population standard deviation.

Here,
`\(S^2\)` is the conventional estimator for the variance, whereas
`\(S_p\)` is the adjusted estimator used to estimate
`\(\sigma^p\)` when the condition
`\(p + n > 1\)` is met. The adjustment in the denominator ensures the unbiased estimation of
`\(\sigma^p\)` based on the sample data, considering the requirement that the sum of the powers
`\(p\)` and the sample size
`\(n\)` must exceed 1.