Chapter 9.4, Problem 38E

Let Y₁, Y₂, … , Y_n denote a random sample from a normal distribution with mean μ and variance σ².

a If μ is unknown and σ² is known, show that $\bar{Y}$ is sufficient for μ.

b If μ is known and σ² is unknown, show that ${\displaystyle {\sum }_{i=1}^n{\left(Y_i-\mu \right)}^2}$ is sufficient for σ².

c If μ and σ² are both unknown, show that ${\displaystyle {\sum }_{i=1}^n{Y}_i}$ and ${\displaystyle {\sum }_{i=1}^n{Y}_i^2}$ are jointly sufficient for μ and σ². [Thus, it follows that $\bar{Y}$ and ${{\displaystyle {\sum }_{i=1}^n\left(Y_i-\bar{Y}\right)}}^2$ or $\bar{Y}$ and S² are also jointly sufficient for μ and σ².]

To determine

Prove that if µ is unknown and ${\sigma }^2$ is known, $\overline{Y}$ is sufficient for µ.

Answer to Problem 38E

It is proved that if µ is unknown and ${\sigma }^2$ is known, $\overline{Y}$ is sufficient for µ.

Explanation of Solution

Let $Y_1,Y_2,\mathrm{...},Y_n$ denote a random sample from a normal distribution with the mean µ and variance ${\sigma }^2$.

The likelihood function is given as follows:

$\begin{array}{c}L=\frac{1}{{\left(2\pi \right)}^{\frac{n}{2}}{\sigma }^n}\exp \left[\frac{-1}{2{\sigma }^2}{\displaystyle \sum _{i-1}^n{\left(y_i-\mu \right)}^2}\right]\\ ={\left(2\pi \right)}^{-\frac{n}{2}}{\sigma }^{-n}\exp \left[\frac{-1}{2{\sigma }^2}\left({\displaystyle \sum _{i=1}^n{y}_i^2}-2\mu n\overline{y}+n{\mu }^2\right)\right]\\ ={\left(2\pi \right)}^{-\frac{n}{2}}{\sigma }^{-n}\exp \left[\frac{-1}{2{\sigma }^2}{\displaystyle \sum _{i-1}^n\left(y_i^2\right)}\right]\exp \left[\frac{1}{2{\sigma }^2}\left(2\mu n\overline{y}-n{\mu }^2\right)\right]\\ =h\left(y\right)g\left(\overline{y},\mu \right)\end{array}$

Where $h\left(y\right)={\left(2\pi \right)}^{-\frac{n}{2}}{\sigma }^{-n}\exp \left[\frac{-1}{2{\sigma }^2}{\displaystyle \sum _{i-1}^n\left(y_i^2\right)}\right]$ and $g\left(\overline{y},\mu \right)=\exp \left[\frac{1}{2{\sigma }^2}\left(2\mu n\overline{y}-n{\mu }^2\right)\right]$.

By Theorem 9.4 (Factorization theorem), it can be said that if µ is unknown and ${\sigma }^2$ is known, $\overline{Y}$ is sufficient for µ.

To determine

Prove that if µ is known and ${\sigma }^2$ is unknown, ${\displaystyle \sum _{i=1}^n{\left(y_i-\mu \right)}^2}$ is sufficient for ${\sigma }^2$.

Answer to Problem 38E

It is proved that if µ is known and ${\sigma }^2$ is unknown, ${\displaystyle \sum _{i=1}^n{\left(y_i-\mu \right)}^2}$ is sufficient for ${\sigma }^2$.

Explanation of Solution

The likelihood function is given as follows:

$\begin{array}{c}L=\frac{1}{{\left(\sqrt{2\pi }\sigma \right)}^n}\exp \left[\frac{-1}{2{\sigma }^2}{\displaystyle \sum _{i-1}^n{\left(y_i-\mu \right)}^2}\right]\\ =h\left(y\right)g\left({\displaystyle \sum _{i=1}^n{\left(y_i-\mu \right)}^2},{\sigma }^2\right)\end{array}$

Where $h\left(y\right)=1$ and $g\left({\displaystyle \sum _{i=1}^n{\left(y_i-\mu \right)}^2},{\sigma }^2\right)=\frac{1}{{\left(\sqrt{2\pi }\sigma \right)}^n}\exp \left[\frac{-1}{2{\sigma }^2}{\displaystyle \sum _{i-1}^n{\left(y_i-\mu \right)}^2}\right]$.

By Theorem 9.4 (Factorization theorem), it can be said that if µ is known and ${\sigma }^2$ is unknown, ${\displaystyle \sum _{i=1}^n{\left(y_i-\mu \right)}^2}$ is sufficient for ${\sigma }^2$.

To determine

Prove that if both µ and ${\sigma }^2$ are unknown, ${\displaystyle \sum _{i=1}^n\left(Y_i\right)}$ and ${\displaystyle \sum _{i=1}^n\left(Y_i^2\right)}$ are jointly sufficient for µ and ${\sigma }^2$.

Answer to Problem 38E

It is proved that if both µ and ${\sigma }^2$ are unknown, ${\displaystyle \sum _{i=1}^n\left(Y_i\right)}$ and ${\displaystyle \sum _{i=1}^n\left(Y_i^2\right)}$ are jointly sufficient for µ and ${\sigma }^2$.

Explanation of Solution

The likelihood function is given as follows:

$\begin{array}{c}L=\frac{1}{{\left(\sqrt{2\pi }\sigma \right)}^n}\exp \left[\frac{-1}{2{\sigma }^2}{\displaystyle \sum _{i-1}^n{\left(y_i-\mu \right)}^2}\right]\\ =\frac{1}{{\left(\sqrt{2\pi }\sigma \right)}^n}\exp \left[\frac{-1}{2{\sigma }^2}{\displaystyle \sum _{i-1}^n\left(y_i^2-2\mu y_i+{\mu }^2\right)}\right]\\ =h\left(y\right)g\left({\displaystyle \sum _{i=1}^n{y}_i,{\displaystyle \sum _{i=1}^n{y}_i^2},\mu },{\sigma }^2\right)\end{array}$

Where $h\left(y\right)=1$ and $g\left({\displaystyle \sum _{i=1}^n{y}_i,{\displaystyle \sum _{i=1}^n{y}_i^2},\mu },{\sigma }^2\right)=\frac{1}{{\left(\sqrt{2\pi }\sigma \right)}^n}\exp \left[\frac{-1}{2{\sigma }^2}{\displaystyle \sum _{i-1}^n\left(y_i^2-2\mu y_i+{\mu }^2\right)}\right]$.

By Theorem 9.4 (Factorization theorem), it can be said that if both µ and ${\sigma }^2$ are unknown, ${\displaystyle \sum _{i=1}^n\left(Y_i\right)}$ and ${\displaystyle \sum _{i=1}^n\left(Y_i^2\right)}$ are jointly sufficient for µ and ${\sigma }^2$.