Editing Consistent estimator (section)

=== Sample mean of a normal random variable ===

Suppose one has a sequence of [[Independence (probability theory)|statistically independent]] observations {''X''<sub>1</sub>, ''X''<sub>2</sub>, ...} from a [[Normal distribution|normal ''N''(''μ'', ''σ''<sup>2</sup>)]] distribution. To estimate ''μ'' based on the first ''n'' observations, one can use the [[sample mean]]: ''T<sub>n</sub>''&nbsp;=&nbsp;(''X''<sub>1</sub> + ... + ''X<sub>n</sub>'')/''n''. This defines a sequence of estimators, indexed by the sample size ''n''.

From the properties of the normal distribution, we know the [[sampling distribution]] of this statistic: ''T''<sub>''n''</sub> is itself normally distributed, with mean ''μ'' and variance ''σ''<sup>2</sup>/''n''. Equivalently, <math style="vertical-align:-.3em">\scriptstyle (T_n-\mu)/(\sigma/\sqrt{n})</math> has a standard normal distribution:
: <math>
    \Pr\!\left[\,|T_n-\mu|\geq\varepsilon\,\right] = 
    \Pr\!\left[ \frac{\sqrt{n}\,\big|T_n-\mu\big|}{\sigma} \geq \sqrt{n}\varepsilon/\sigma \right] = 
    2\left(1-\Phi\left(\frac{\sqrt{n}\,\varepsilon}{\sigma}\right)\right) \to 0
  </math>
as ''n'' tends to infinity, for any fixed {{nowrap|''ε'' > 0}}. Therefore, the sequence ''T<sub>n</sub>'' of sample means is consistent for the population mean&nbsp;''μ'' (recalling that <math>\Phi</math> is the [[Cumulative distribution function|cumulative distribution]] of the standard normal distribution).