exercise:974197a009: Difference between revisions
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(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Show that, for the sample mean <math>\bar x</math> and sample variance <math>s^2</math> as defined in Exercise Exercise, <ul><li> <math>E(\bar x) = \mu</math>. </li> <li> <math>E\bigl((\bar x - \mu)^2\bigr) = \sigma^2/n</m...") |
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Show that, for the sample mean <math>\bar x</math> and sample variance <math>s^2</math> as defined in [[exercise:2efc106914 |Exercise]], | |||
<ul style="list-style-type:lower-alpha"><li> <math>E(\bar x) = \mu</math>. | |||
variance <math>s^2</math> as defined in | |||
<ul><li> <math>E(\bar x) = \mu</math>. | |||
</li> | </li> | ||
<li> <math>E\bigl((\bar x - \mu)^2\bigr) = \sigma^2/n</math>. | <li> <math>E\bigl((\bar x - \mu)^2\bigr) = \sigma^2/n</math>. | ||
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and take expectations of both sides, using part (b) when necessary. | and take expectations of both sides, using part (b) when necessary. | ||
</li> | </li> | ||
<li> Show that if, in the definition of <math>s^2</math> in | <li> Show that if, in the definition of <math>s^2</math> in [[exercise:2efc106914 |Exercise]], we | ||
replace the coefficient <math>1/n</math> by the coefficient <math>1/(n-1)</math>, then <math>E(s^2) = \sigma^2</math>. | replace the coefficient <math>1/n</math> by the coefficient <math>1/(n-1)</math>, then <math>E(s^2) = \sigma^2</math>. | ||
(This shows why many statisticians use the coefficient <math>1/(n-1)</math>. The number <math>s^2</math> | (This shows why many statisticians use the coefficient <math>1/(n-1)</math>. The number <math>s^2</math> | ||
Latest revision as of 22:14, 14 June 2024
Show that, for the sample mean [math]\bar x[/math] and sample variance [math]s^2[/math] as defined in Exercise,
- [math]E(\bar x) = \mu[/math].
- [math]E\bigl((\bar x - \mu)^2\bigr) = \sigma^2/n[/math].
- [math]E(s^2) = \frac {n-1}n\sigma^2[/math]. Hint: For (c) write
[[math]] \begin{eqnarray*} \sum_{i = 1}^n (x_i - \bar x)^2 & = & \sum_{i = 1}^n \bigl((x_i - \mu) - (\bar x - \mu)\bigr)^2 \\ & = & \sum_{i = 1}^n (x_i - \mu)^2 - 2(\bar x - \mu) \sum_{i = 1}^n (x_i - \mu) + n(\bar x - \mu)^2 \\ & = & \sum_{i = 1}^n (x_i - \mu)^2 - n(\bar x - \mu)^2, \end{eqnarray*} [[/math]]and take expectations of both sides, using part (b) when necessary.
- Show that if, in the definition of [math]s^2[/math] in Exercise, we replace the coefficient [math]1/n[/math] by the coefficient [math]1/(n-1)[/math], then [math]E(s^2) = \sigma^2[/math]. (This shows why many statisticians use the coefficient [math]1/(n-1)[/math]. The number [math]s^2[/math] is used to estimate the unknown quantity [math]\sigma^2[/math]. If an estimator has an average value which equals the quantity being estimated, then the estimator is said to be unbiased. Thus, the statement [math]E(s^2) = \sigma^2[/math] says that [math]s^2[/math] is an unbiased estimator of [math]\sigma^2[/math].)